551, Fall '02                 B1a.    Direct Products                 p. 1

Math 551 – Algebra – Fall 2002

Richard Lyons Rutgers University New Brunswick, New Jersey, USA

B. Abelian Groups and Modules over Principal Ideal Domains

1. Abelian Groups.

1a. Direct Products

The nicest way a group G can decompose is as the direct product of a number of other groups (which are then isomorphic to subgroups of G). We give a criterion here for G to be isomorphic to G_{1} × ⋯ × G_{r} for certain subgroups G_{1}, …, G_{r} of G.

Definition. Let G_{1}, …, G_{r} be groups. The (external) direct product G_{1} . × · · · . × G_{r} is the group whose underlying set is the Cartesian product G_{1} × ⋯ × G_{r}, with the operation

(g_{1}, …, g_{r})(h_{1}, …, h_{r}) = (g_{1}h_{1}, …, g_{r}h_{r}),

the product in the i-th place being formed in the group G_{i}.

A similar definition can be made for an infinite family of groups {G_{i}}_{i∈I}: the external direct product

∏ ·_{i∈I} G_{i}

is the group whose underlying set is the Cartesian product of the G_{i}. The elements of this set are functions f on I such that f(i) ∈ G_{i} for each i; the multiplication is pointwise, i.e. (ff′)(i) = f(i)f′(i).

(External) direct products come with a family of projection homomorphisms

π_{j} : G_{1} . × · · · . × G_{r} → G_{j}, j = 1, …, r

or more generally

π_{j} : ∏ ·_{i∈I} G_{i} → G_{j}, j ∈ I,

and satisfy the universal property that for any group G and family of homomorphisms γ_{j} : G → G_{j}, j ∈ I, there is a unique homomorphism

γ : G → ∏ ·_{i∈I} G_{i} such that π_{j}γ = γ_{j} ∀j ∈ I.

Namely, for any g ∈ G, γ(g) has γ_{j}(g) for its j-th coordinate.

There is another internal notion of direct product, at least for the case of direct products of finitely many groups:

p. 2                 B1.    Abelian Groups.                   551, Fall '02

Definition. Let G be a group and G_{1}, …, G_{r} be subgroups of G. We write

G = G_{1} × ⋯ × G_{r}

and say that G is the direct product of its subgroups G_{1}, …, G_{r} if and only if the mapping

ϕ : G_{1} . × · · · . × G_{r} → G, ϕ(g_{1}, …, g_{r}) = g_{1} ⋯ g_{r}

is an isomorphism.

It is not always the case that ϕ is even a homomorphism; the requirements that it be a homomorphism, and that it be injective and surjective, each carry weight. When G = G_{1} × ⋯ × G_{r}, we are free to think of G interchangeably with the external direct product.

Theorem. Let G_{1}, …, G_{r} be subgroups of a group G. Then (a), (b) and (c) are equivalent.

(a) G = G_{1} × ⋯ × G_{r}.

(b) (1) G_{i} ◁ G for all i;

(2) G = ⟨G_{1}, …, G_{r}⟩;

(3) For each i = 1, …, r, if we put G^{i} = ∏_{j≠i} G_{j} (in order), then G^{i} ∩ G_{i} = 1.

(c) (1) For any i ≠ j, and any g_{i} ∈ G_{i} and g_{j} ∈ G_{j}, g_{i}g_{j} = g_{j}g_{i};

(2) G = G_{1} ⋯ G_{r};

(3) G^{i} ∩ G_{i} = 1 for all i (with G^{i} defined as in (b)).

Proof. (a) implies (b): Let Γ = G_{1} . × · · · . × G_{r}, and let Γ_{i} be the subgroup consisting of all elements of Γ all of whose coordinates are 1, except possibly the i-th coordinate. Clearly Γ_{i} ≤ Γ and ϕ maps Γ_{i} isomorphically onto G_{i}. Now if (a) holds, then Γ ≅_{ϕ} G, and so to prove the various statements of (b) it suffices to check the corresponding statements for Γ and the Γ_{i}. First, Γ_{i} is the intersection of the kernels of the projections π_{j}, j ≠ i, so Γ_{i} ◁ Γ. A typical element (g_{1}, …, g_{r}) ∈ Γ is the product of the elements (g_{1}, 1, …, 1)(1, g_{2}, 1, …, 1) ⋯ so Γ = Γ_{1} ⋯ Γ_{r}. Under ϕ, G^{i} corresponds to the kernel of the projection π_{i}, and Γ_{i} ∩ ker π_{i} = 1.

(b) implies (c): For g_{i} ∈ G_{i}, g_{j} ∈ G_{j} and i ≠ j, [g_{i}, g_{j}] =^{g_{i}} g_{j}(g_{j}^{−1}) = g_{i}(^{g_{j}}g_{i}^{−1}) ∈ G_{i}∩G_{j} ≤ G_{i} ∩ G^{i} = 1. This implies (1). Then any word in elements of G_{1}, …, G_{r} can be shuffled to a word g_{1} ⋯ g_{r}, so (b2) implies (c2). Trivially (b3) implies (c3).

(c) implies (a): The three properties imply in turn that ϕ is a homomorphism, ϕ is surjective, and ϕ is injective. Namely,

\begin{aligned}ϕ[(g_{1}, …, g_{r})(h_{1}, …, h_{r})] & = ϕ(g_{1}h_{1}, …, g_{r}h_{r}) = g_{1}h_{1} ⋯ g_{r}h_{r} = g_{1} ⋯ g_{r} · h_{1} ⋯ h_{r} \\ & = ϕ(g_{1}, …, g_{r})ϕ(h_{1}, …, h_{r}) \end{aligned} ;

551, Fall '02       B1b.    Direct Sums and Free Abelian Groups       p. 3

ϕ(Γ) is a subgroup containing G_{1}, …, G_{r} so equals G; and if g = (g_{1}, …, g_{r}) ∈ ker ϕ, then g_{1} ⋯ g_{r} = 1, and for each i if we solve for g_{i} we get g_{i} ∈ G_{i} ∩ G^{i} = 1, so ker ϕ = 1. QED

The case of two factors is important:

G = H × K ⟺ H ◁ G, K ◁ G, H ∩ K = 1 and G = ⟨H, K⟩.

Of course H ×K has the normal subgroup H ×1 ≅ H, and the quotient H ×K/H ×1 ≅ K (by the first isomorphism theorem applied to the projection onto K). We routinely identify these subgroups and quotients with H and K. What distinguishes the direct product among all “extensions of H by K”) are the facts that (1) H has a complement (a subgroup K such that HK = G and H ∩ K = 1), and (2) that complement commutes elementwise with H.

Not every subgroup of a direct product is a direct product of subgroups. However, the following is fundamental.

Proposition. Suppose that H_{1} ◁ G_{1} and H_{2} ◁ G_{2}. Then H_{1} × H_{2} ◁ G_{1} × G_{2}, and

(G_{1} × G_{2})/(H_{1} × H_{2}) ≅ G_{1}/H_{1} × G_{2}/H_{2}.

Proof. Define ϕ : G_{1} × G_{2} → G_{1}/H_{1} × G_{2}/H_{2} by ϕ(g_{1}, g_{2}) = (g_{1}H_{1}, g_{2}H_{2}). It is easily checked that this is a surjective homomorphism whose kernel is H_{1} × H_{2}. Now apply the first isomorphism theorem.

Finite direct products have another universal property, for which they are also called “sums”, particularly in the context of abelian groups. Namely, given groups G_{1}, …, G_{n}, there are canonical injections

ι_{j} : G_{j} → G_{1} × ⋯ × G_{n},

with ι_{j}(g) being the n-tuple whose j-th coordinate is g and all of whose other coordinates are 1. The images of the various ι_{j} are “supported” in different coordinates and so commute elementwise with one another. The universal property is that for any group H, and any n-tuple of homomorphisms ϕ_{i} : G_{i} → H such that

[ϕ_{i}(g_{i}), ϕ_{j}(g_{j})] = 1 for all i ≠ j, all g_{i} ∈ G_{i} and all g_{j} ∈ G_{j},

there is a unique homomorphism

Φ : G_{1} × ⋯ × G_{n} → H, (g_{1}, …, g_{n}) ↦ ϕ_{1}(g_{1})ϕ_{2}(g_{2}) ⋯ ϕ_{n}(g_{n})

such that ϕ_{i} = Φ ∘ ι_{i} for each i = 1, …, n. The unique mapping Φ is sometimes notated ϕ_{1} × ⋯ × ϕ_{n}, so that by definition

ϕ_{1} × ⋯ × ϕ_{n}(g_{1}, …, g_{n}) = ϕ_{1}(g_{1})ϕ_{2}(g_{2}) ⋯ ϕ_{n}(g_{n}).

For example the mapping constructed in the proof of the preceding proposition is π_{H_{1}}×π_{H_{2}}. The commutator condition [ϕ_{i}(g_{i}), ϕ_{j}(g_{j})] = 1 is needed to prove (actually equivalent to the statement) that ϕ_{1} × ⋯ × ϕ_{n} is a homomorphism.

1b. Direct Sums and Free Abelian Groups

For small abelian groups there are decisive structure theorems, giving essentially unique decompositions as direct products of cyclic groups. By “decisive” is meant that the theorems are strong enough in many cases to enable one to check whether a given statement about (small) abelian groups is true, to check whether two abelian groups arising in different contexts are isomorphic, etc. The word “small” here could be interpreted in two senses: either finitely generated, or of finite exponent. We shall prove the decomposition theorem for the first of these.

It is customary to use additive notation for abelian groups, and we shall do so. Nevertheless, the direct product of a family {G_{i}}_{i∈I} of abelian groups will still be denoted

∏_{i∈I} G_{i}.

However, for infinite index sets I, it is the direct sum (or coproduct) which best suits the theory of abelian groups.

Definition. Let {G_{i}}_{i∈I} be a family of abelian groups. The direct sum

∐_{i∈I} G_{i} (or ⨁_{i∈I} G_{i}

is the subgroup of the direct product ∏_{i∈I} G_{i} consisting of all elements which have only finitely many non-identity coordinates.

Then there are canonical (injective) homomorphisms ι_{j} : G_{j} → ∐_{i∈I} G_{i}, one for each j ∈ I, such that ι_{j}(g) is the element whose j-th coordinate is g and all of whose other coordinates are the identity (in the appropriate group). These mappings have the universal property that for any abelian group H and any family of homomorphisms {ϕ_{i}}_{i∈I} with ϕ_{i} : G_{i} → H, there is a unique homomorphism

Φ = ⨁_{i∈I} ϕ_{i} : ⨁ G_{i} → H such that ϕ_{i} = Φ ∘ ι_{i} for all i ∈ I;

namely ⨁_{i∈I} ϕ_{i} maps (g_{i})_{i∈I} to ∑_{i∈I} ϕ_{i}(g_{i}), this (possibly infinite) sum being well-defined since g_{i} = 0 for all but finitely many i, and hence ϕ_{i}(g_{i}) = 0 for all but finitely many i.

This direct sum therefore plays the same role in the theory of abelian groups that the free product plays in the theory of (all) groups. We have really defined an “external” direct sum, and then we say that an abelian group G is the “internal” direct sum of subgroups G_{i} if and only if the mapping ⊕_{i∈I} : G_{i} → G induced by the inclusion mappings (i.e., ⊕_{i∈I} : (g_{i})_{i∈I} ↦ ∑_{i∈I} g_{i}) is an isomorphism. Generally the distinction between internal and external is blurred, since it can always be bridged by replacing groups by isomorphic copies.

Definition. A free abelian group is the direct sum

⨁_{i∈I} Z

of copies of Z (indexed by some set I).

If we let e_{i} be the element of this direct sum which is 1 in the i-th coordinate and 0 in every other coordinate, then every element of ⨁_{i∈I} Z is uniquely expressible as a sum ∑_{i∈I} n_{i}e_{i}, where the n_{i} are all integers, all but finitely many (“almost all”) of them being 0. The set {e_{i}} is called a basis, by analogy with the theory of vector spaces.

A free abelian group has the expected (?) universal property, which as usual characterizes it: there is a (set) mapping ι : I → ⨁_{i∈I} Z, taking i to the element e_{i} which is 1 in the i-th coordinate and 0 in every other coordinate; moreover for any abelian group H and any (set) mapping ψ : I → H there exists a unique homomorphism Ψ : ⨁_{i∈I} Z → H such that Ψ ∘ ι = ψ. Namely,

Ψ(∑_{i} n_{i}e_{i} = ∑_{i} n_{i}ψ(e_{i}).

This has the following consequence, completely analogous to the corresponding result for arbitrary groups and free groups.

Theorem 0. Every abelian group is a quotient of a free abelian group. More precisely, if G is an abelian group and S ⊆ G is a subset such that ⟨S⟩ = G, then there is a surjective homomorphism

ϕ : F → G,

where F is free abelian on S; thus G ≅ F/ ker ϕ.

Pursuing the analogy with vector spaces a bit further, we may define a subset {f_{i}}_{i∈I} ⊆ H of an arbitrary abelian group H to be linearly independent if and only if whenever ∑_{i∈I} n_{i}f_{i} = 0, the n_{i} being integers almost all zero, it follows that n_{i} = 0 for all i. Then a basis is just a linearly independent generating set. It is easily seen that a basis can be equivalently defined as a subset {f_{i}} such that every element of the group is uniquely expressible as a sum ∑_{i} n_{i}f_{i}, n_{i} ∈ Z, almost all n_{i} being 0.

Proposition. An abelian group is free abelian on some set if and only if it has a basis. In that case, the cardinality of a basis is uniquely determined.

The cardinality of a basis is called the “rank” of a free abelian group.

Proof. We saw above that a free abelian group has a basis {e_{i}}. Conversely, if G has the basis {f_{i}}_{i∈I}, then G, together with the mapping i ↦ f_{i}, has the above universal property since every element of G may be uniquely expressed ∑_{i} n_{i}f_{i}, so that the analogue of Ψ above is well-defined.

Next suppose that F = ⊕_{i∈I}⟨e_{i}⟩ is free abelian with basis {e_{i}}_{i∈I}. Thus each e_{i} ≅ Z. Pick a prime p ∈ Z and set pF = {pg | g ∈ F }. It is easily checked that H ≤ F and indeed pF = ⊕_{i∈I}⟨pe_{i}⟩. Consequently

F/pF ≅ ⊕_{i∈I}⟨e_{i}⟩/⟨pe_{i}⟩ ≅ ⊕_{i∈I}Z/pZ,

the direct sum of |I| copies of Z/pZ. But this direct sum may be considered a vector space over Z/pZ, and |I| is its dimension. so is uniquely determined. Thus the rank of F is dim_{Z/pZ}(F/pF), an expression independent of the basis.

(In the finite-dimensional case it is familiar that the dimension is uniquely determined. In the infinite-dimensional case it is also true, and just as easy to prove given the Schröder-Bernstein theorem in set theory. For vector spaces over Z/pZ, one can also easily argue that if V is an infinite-dimensional vector space, then |V | = dim V. However, this last approach doesn’t work for PID’s in general, see next section.)

An important property of free abelian groups is the following splitting property (free abelian groups are “projective” abelian groups):

Theorem 1. Suppose that G is an abelian group and H is a subgroup such that G/H is free. Then G = H ⊕ K for some subgroup K ≤ G.

Proof. We prove the equivalent statement: if ϕ : G → F is a surjective homomorphism of abelian groups with F free abelian, then G = ker ϕ ⊕ K for some subgroup K ≤ G. (This implies the desired statement when applied to the projection π_{H} : G → G/H.)

Let {f_{i}}_{i∈I} be a basis of F. Choose for each i ∈ I an element g_{i} ∈ G such that ϕ(g_{i}) = f_{i}. Let K = ⟨g_{i} | i ∈ I⟩. Then G = H + K: given g ∈ G we write ϕ(g) = ∑ n_{i}f_{i}, set g′ = ∑ n_{i}g_{i} and observe that ϕ(g′) = ϕ(g). Thus ϕ(g − g′) = 0, and g = (g − g′) + g′ with g − g′ ∈ H and g′ ∈ K. Also H ∩ K = 0: if g ∈ H ∩ K, then g = ∑ n_{i}g_{i} for some integers n_{i}, and applying ϕ gives 0 = ∑ n_{i}f_{i}, so n_{i} = 0 for all i whence g = 0.

1c. Finitely Generated Free Abelian Groups

The main theorems on finitely generated abelian groups rest on (and are equivalent to) theorems on finitely generated free abelian groups.

Theorem 2. Any subgroup H of a finitely generated free abelian group G is a free abelian group. Moreover the rank of H is at most the rank of G.

Proof. Let {e_{1}, …, e_{n}} be a basis of G. The proof is by induction on n. Let G_{0} = Ze_{1} + ⋯ + Ze_{n−1} and H_{0} = H ∩ G_{0}. Then by induction H_{0} is free abelian of rank r ≤ n − 1. Moreover H/H_{0} ≅ H + G_{0}/G_{0} ≤ G/G_{0} ≅ Ze_{n} ≅ Z. Hence H/H_{0} ≅ Z or 0. In the first case there exists K ≤ H such that H = H_{0} ⊕ K, so H is free abelian of rank r + 1 ≤ n. In the second case the desired conclusion is obvious.

Corollary. Any subgroup of a finitely generated abelian group is finitely generated.

551, Fall '02         B1d.    Integer matrices and equivalence         p. 7

Proof. If H ≤ G with G finitely generated, we may write G = F/R where F is finitely generated free abelian. Then H = F_{0}/R for some R ≤ F_{0} ≤ F by the third isomorphism theorem, and the theorem implies that F_{0} is finitely generated. A fortiori, H is finitely generated.

The main theorem is formulated as a theorem about the relationship between appropriate bases of a free abelian group and a subgroup. It will be applied later in the context of Theorem 0 to the inclusion ker ϕ → F (note that Theorem 2 tells us that ker ϕ is itself finitely generated and free, given that F is). It will yield structural information about the quotient F/ ker ϕ, which as Theorem 0 shows is an arbitrary finitely generated abelian group.

The theorem has an existence and a uniqueness statement.

Fundamental theorem on finitely generated abelian groups (I: free abelian group version). Let F be a finitely generated free abelian group, of rank s, and E a subgroup of F. Then E is a finitely generated free abelian group, of rank r ≤ s. Moreover there exist bases e_{1}, …, e_{r} of E and f_{1}, …, f_{s} of F and uniquely determined positive integers m_{1}, …, m_{r} such that

a) e_{i} = m_{i}f_{i}, i = 1, …, r, and
b) m_{1}|m_{2}| ⋯ |m_{r}.

However, the bases {e_{i}} and {f_{i}} are not uniquely determined. The integers

m_{1}, m_{2}, …, m_{r}

of the theorem are sometimes called the “invariant factors” of the inclusion mapping E → F.

Before proceeding with the proof of this theorem, we make two digressions.

1d. Integer matrices and equivalence

The first digression is to interpret the Fundamental Theorem (version Ia) in matrix language. The reader should be able to supply proofs for all statements in this section. Like finite-dimensional vector spaces and linear transformations, free abelian groups of finite rank and homomorphisms between them have concrete realizations as column vectors and matrices; setting up these realizations for abelian groups is essentially identical to setting them up for vector spaces, except that the entries of the matrices and column vectors now come from Z instead of from a field.

The group Z^{n} of all integer n × 1 column vectors is clearly a free abelian group, and one of its bases is {e_{1}^{n}, ⋯, e_{n}^{n}}, where e_{i}^{n} is the n × 1 column vector whose i-th entry is 0 and all of whose other entries are 0.

Moreover, given any rank n free abelian group G and basis B = {e_{1}, …, e_{n}}, and given any g ∈ G, we write g = ∑_{i=1}^{n} m_{i}e_{i} and define [g]^{B} = [m_{1} ⋯ m_{n}]^{T}. The mapping

G → Z^{n}, g ↦ [g]^{B},

is then an isomorphism.

Now, given a homomorphism ϕ : G → H and bases B = {e_{1}, …, e_{n}} and B′ = {f_{1}, …, f_{m}} of G and H, we write ϕ(e_{j}) = ∑_{i=1}^{m} c_{ij}f_{i}, (c_{ij} ∈ Z) for each j = 1, …, n, and define

[ϕ]_{B}^{B′} = [c_{ij}]_{1≤i≤m, 1≤j≤n}.

The mapping

Hom(G, H) → Z^{m×n}, ϕ ↦ [ϕ]_{B}^{B′},  1A

is then an isomorphism of abelian groups, and

[ϕ(g)]^{B′} = [ϕ]_{B}^{B′}[g]^{B} for all g ∈ G,

with ordinary matrix multiplication on the right. Indeed, given ϕ, [ϕ]_{B}^{B′} is the unique m × n matrix for which this equation is true for all g ∈ G. This last remark (or direct computation) quickly implies in turn that for any ϕ, ψ ∈ Hom(G, H) and χ ∈ Hom(H, K), and for any bases B, B′, B′′ of G, G, H, respectively, we have

[ϕ + ψ]_{B}^{B′} = [ϕ]_{B}^{B′} + [ψ]_{B}^{B′} and [χ ∘ ϕ]_{B}^{B′′} = [χ]_{B}^{B′′}[ϕ]_{B}^{B′},

again with matrix multiplication on the right.

If B and B′ are two bases of the same free abelian group H, then

[1_{H}]_{B}^{B′}

is the “change of basis” matrix whose columns give the B-expansion of the elements of B′. In particular

[1_{H}]_{B}^{B} = I,

the identity matrix. Moreover different choices C, C′ of bases of H and K, respectively, give the related matrix

[ϕ]_{C′}^{C} = [id_{K}]_{C′}^{B′}[ϕ]_{B′}^{B}[id_{H}]_{B}^{C} = P [ϕ]_{B′}^{B}Q,  1B

where P = [id_{K}]_{C′}^{B′} and Q = [id_{H}]_{B}^{C}.

If H and K are free abelian groups of ranks m and n, respectively, and ϕ : H → K and ψ : K → H are homomorphisms, then with respect to bases B and B′ of H and K respectively we have

[ϕ]_{B′}^{B}[ψ]_{B}^{B′} = [ϕ ∘ ψ]_{B′}^{B′} and [ψ]_{B}^{B′}[ϕ]_{B′}^{B} = [ψ ∘ ϕ]_{B′}^{B′}.

Because of the isomorphisms ϕ ↦ [ϕ]_{B′}^{B} and ψ ↦ [ψ_{B}^{B′} of Hom(H, K) and Hom(K, H), respectively, with Z^{m×n} and Z^{n×m}, we conclude that

ϕ is invertible with inverse ψ if and only if [ϕ]_{B′}^{B} is invertible with inverse [ψ]_{B}^{B′}.

In particular,

[id_{H}]_{B}^{B′} = ([id_{H}]_{B′}^{B})^{−1} .

for any two bases B, B′ of H.

This permits us to establish bijections (not canonical) between the set of all bases of H and the set of all invertible matrices in Z^{m×m}. Namely, fix a basis B of H; the correspondence

B′ ↦ [id_{H}]_{B}^{B′}

is a bijection. So is the correspondence B′ ↦ [id_{H}]_{B′}^{B}.

As a consequence we can determine, given ϕ : H → K, all the matrices representing ϕ with respect to all bases of H and K. We first define an equivalence relation ∼ (called “equivalence”) on Z^{m×n} by:

A ∼ A′ ⟺ A′ = P AQ for some invertible P ∈ Z^{m×m} and Q ∈ Z^{n×n}.

Fix bases B and C of H and K respectively. Let ϕ : H → K be a homomorphism and A = [ϕ]_{C}^{B}. The matrices P and Q allowable above are precisely the matrices [id_{H}]_{B}^{B′} and [id_{H}]_{C′}^{C} as B′ and C′ range over the sets of bases of H and K, respectively. Because of (1B), we conclude:

Proposition. Let H and K be free abelian groups of finite ranks m and n, respectively. Let ϕ : H → K be a homomorphism, and let A = [ϕ]_{C}^{B} for some bases B and C of H and K, respectively. Then for any m × n matrix A′ over Z, A ∼ A′ if and only if A′ = [ϕ]_{C′}^{B′} for some bases B′ and C′ of H and K, respectively.

The following theorem therefore largely follows from the main theorem of the last section.

Fundamental theorem on finitely generated abelian groups (Ib: Z-matrix version). Let A be an integer m×n matrix. Then there exists a uniquely determined integer r ≥ 0, r ≤ min(m, n), and uniquely determined positive integers m_{1}, …, m_{r} such that

a) A ∼ diag(m_{1}, m_{2}, …, m_{r}, 0, …, 0); and
b) m_{1}|m_{2}| ⋯ |m_{r}.

Here diag(m_{1}, m_{2}, …, m_{r}, 0, …, 0) is the m × n matrix with all entries 0 except the first r entries down the main diagonal, which are m_{1}, …, m_{r}, in that order.

Proof. Let H and K be free abelian groups of ranks m and n, respectively. By the isomorphism (1A), there is a homomorphism ϕ : H → K and bases B and C of H and K, respectively, such that [ϕ]_{C}^{B} = A.

Set G = ϕ(H) ≤ K and r = rank(G). By Theorem 2, G is free, and hence by Theorem 1, H = H_{1} ⊕ ker ϕ for some subgroup H_{1}. Then ψ = ϕ|H_{1} is an isomorphism H_{1} ≅ G.

By version Ia, there exists r ≥ 0 and uniquely determined positive integers m_{1}, …, m_{r} such that there are bases C′ = {e_{1}, …, e_{n}} of K and {f_{1}, …, f_{r}} of G with f_{i} = m_{i}e_{i} for each i = 1, …, r. The elements ψ^{−1}f_{1}, …, ψ^{−1}f_{r}, together with an arbitrarily chosen basis of ker ϕ form a basis B′ of H. We then have

[ϕ]_{C′}^{B′} = diag(m_{1}, …, m_{r}, 0, …, 0),  1C

establishing existence. Conversely, if A is equivalent to such a matrix, then bases B′ and C′ exist such that (1C) holds. Writing C′ = {e_{1}, …, e_{n}} and B′ = {f_{1}, …, f_{r}, …} we have that ϕ(f_{1}), …, ϕ(f_{r}) form a basis of G, and ϕ(f_{i}) = m_{i}e_{i} for each i = 1, …, r. So the uniqueness follows from version Ia.

In using this theorem, the following observation is useful:

Proposition. Let P be a square matrix with entries in Z. Then P is invertible in Z^{m×m} if and only if det(P) is a unit in Z, i.e, det(P) = ±1.

On the one hand if P is invertible then det(P) det(P^{−1}) = det I = 1, with both det(P) and det(P^{−1}) in Z. On the other hand if det(P) is a unit in Z then the familiar formula

P^{−1} = det(P)^{−1} adj(P),

adj(P) being the transpose of the matrix of cofactors, shows that P^{−1} ∈ Z^{m×m}.

1e. Principal Ideal Domains

The second digression is to observe that the proofs we are giving generalize without change to modules over PID’s (instead of just abelian groups). We might as well formalize this.

Definition. A principal ideal domain (PID) is a commutative ring R with unit 1, such that R is an integral domain (xy = 0 implies either x = 0 or y = 0) and such that every ideal in R is principal, i.e. consists of all the R-multiples of a single element of R.

If we consider R to be an R-group (an abelian group with operators R, acting by left multiplication) then the principal ideal condition is just the condition that every R-subgroup (i.e. ideal) is “R-cyclic”–i.e., is generated as an R-group by a single element.

Examples of principal ideal domains include the following two essential examples: Z, and k[X], the polynomial ring in one variable over a field. In both cases a stronger statement is actually true: there is a division algorithm. In the case of Z, for every d ≠ 0 and every n ∈ Z, there exist q, r ∈ Z such that n = qd + r and |r| < |d|. In the case of k[X], for every d, n ∈ k[X] with d ≠ 0, there exist q, r ∈ k[X] such that n = qd + r and deg r < deg d. Thus these are “Euclidean domains”.

Definition. A Euclidean domain (ED) is an integral domain R possessing a function ϕ : R − {0} → Z^{+}, the set of nonnegative integers, such that

a) ϕ(ab) ≥ ϕ(a) for all a, b ∈ R − {0};

b) for every d, n ∈ R with d ≠ 0 there exist q, r ∈ R such that n = qd + r, and either r = 0 or ϕ(r) < ϕ(d).

551, Fall '02             B1e.    Principal Ideal Domains             p. 11

Proposition. Every ED is a PID.

Proof. Given an ideal I ⊆ R, we must find a generator for it. If I = 0, then 0 is a generator. Otherwise choose x ∈ I such that ϕ(x) ≤ ϕ(y) for all y ∈ I, y ≠ 0. Then for any u ∈ I, write u = qx + r with ϕ(r) < ϕ(x) or r = 0. But then r = u − qx ∈ I so r = 0 and u = qx ∈ Rx. Therefore I ⊆ Rx, and the reverse inclusion is obvious.

In any integral domain R we may define a | b, for a, b ∈ R, to mean: b = qa for some q ∈ R. Equivalently:

a | b ⟺ Rb ⊆ Ra.

We may also define a unit to be an element u such that Ru = R, or equivalently such that uv = 1 for some v ∈ R. Two elements a, b ∈ R are said to be associates if and only if a = ub for some unit u.

Exercise. The relation

a ∼ b ⟺ a and b are associates

is an equivalence relation.

Then the relation | is transitive, and also is “antisymmetric”, at least relative to: if a | b and b | a, then a = qb and b = q′a for some q, q′, so a = qq′a and a(1 − qq′) = 0. If a = 0, then b = 0. If a ≠ 0, then qq′ = 1 so q is a unit and a and b are associates.

We may also define a gcd of two nonzero elements a, b ∈ R to be an element d such that

a) d | a and d | b;
b) for any d′ ∈ R such that d′ | a and d′ | b, we have d′ | d.

In any integral domain, GCD’s are unique up to associates (if they exist). For if d and d′ are two gcd’s of the same a, b, then by definition d | d′ and d′ | d, so d and d′ are associates.

Proposition. Let R be a PID and let a, b ∈ R. Then there exists a gcd d of a and b in R. Moreover, Rd = Ra + Rb, so that d = ma + nb for some m, n ∈ R.

Proof. In a PID, the set of ideals is in one-to-one correspondence with the set of associateclasses of elements of R, under Rx ↔ [x], where [x] denotes the set of associates of x. Moreover x | y ⟺ Rx ⊇ Ry. Moreover, by definition a gcd is simply a “greatest lower bound” with respect to the divisibility relation. Hence we must prove that in the set of ideals, Ra + Rb is the least ideal containing Ra and Rb. But this is obvious.

The maximum condition holds in any PID:

Theorem. In a PID, the maximum condition holds on the set of all ideals.

Proof. Let R be a PID and let I_{1} ⊆ I_{2} ⊆ I_{3} ⊆ ⋯ be an infinite ascending chain of ideals of R. Set

I = ∪_{n=1}^{∞}I_{i}.

Then I is an ideal, and so I = Rx for some x ∈ R as R is a PID. But then x ∈ I_{i} for some i, and therefore I ⊆ I_{i}, so I = I_{i}. QED Remark. Virtually the same proof shows that for any commutative ring R, if every ideal of R is finitely generated then R satisfies the maximum condition on ideals. The converse to this statement is also true:

Theorem–Definition. If R is a commutative ring, then the following statements about R are equivalent:

a) Every ideal of R is finitely generated.

b) The set of ideals of R satisfies the maximum condition.

c) R is a noetherian ring.

(Part c) is just the definition of “noetherian ring”.)

To show that b) implies a), take an ideal I. If I were not finitely generated then we could find an infinite sequence of elements x_{1}, …, x_{n}, … of I such that if I_{i} is the ideal generated by x_{1}, …, x_{i}, then

I_{1} < I_{2} < ⋯ < I_{n} ⋯ < I,

contradicting the maximum condition.

The maximum condition has two equivalent formulations: a) no infinite ascending chain I_{1} < I_{2} < ⋯ of ideals exists; b) every nonempty set of ideals has a maximal element. Indeed an infinite ascending chain has no maximal element, so b) implies a). If some nonempty set of ideals had no maximal element, then an infinite ascending chain could be obtained, starting with any ideal in the set, then any ideal proving that the first wasn’t maximal in the set, then any ideal proving that that one wasn’t maximal, etc.

Now we can generalize the theorem of the previous section, with the identical proof. First of all we make the following definition.

Definition. Let R be a ring with 1. A (left) R-module is an abelian group M equipped with a “scalar multiplication”

R × M → M, written (r, m) ↦ rm,

such that the following conditions hold for all r, s ∈ R and all m, n ∈ M:

a) r(m + n) = rm + rn;
b) (r + s)m = rm + sm;
c) r(sm) = (rs)m;
d) 1m = m.

The usual trivial consequences follow, as with vector spaces: 0m = 0, r0 = 0 and r(−m) = −(rm) = (−r)m for all r ∈ R and m ∈ M.

Definition. Let M and N be R-modules. An R-homomorphism from M to N, or a homomorphism of R-modules from M to N, is a homomorphism ϕ : M → N of abelian groups such that ϕ(rm) = rϕ(m) for all r ∈ R, m ∈ M.

Thus an R-module is in particular an abelian group with operators R (satisfying the extra conditions b), c) and d) above), and an R-homomorphism is precisely a homomorphism of R-groups.

Definition. If M is an R-module, then an R-submodule of M is an additive subgroup N of M such that rN ⊆ N for all r ∈ R.

An R-submodule is thus an R-subgroup, in the terminology of groups with operators.

Thus

The Noether isomorphism theorems hold for R-modules.

and the Jordan-Hölder theorem does as well.

The terms “homomorphism”, “subgroup” have to be replaced by “R-homomorphism” and “R-submodule”. The parallelogram law reads

(N + P)/N ≅ P/(N ∩ P)

for any R-submodules N and P of an R-module M. (The analogue of the hypothesis H ≤ N_{G}(K) is automatically true since the group operation on M is commutative.)

Given an R-module M and a subset S ⊆ M, the R-submodule of M generated by S is defined to be the intersection of all submodules of M containing S, or equivalently the set of all elements of M representable as “R-linear combinations” ∑_{i} r_{i}s_{i} with all r_{i} ∈ R, all s_{i} in S, and almost all r_{i} = 0.

An R-module M is “cyclic” if and only if it is generated by a single element. We claim that M is cyclic if and only if M ≅ _{R}R/A for some ideal A of R. If M is generated by m, then

RR → M, r ↦ rm

is a surjective homomorphism and so M ≅ R/A where A = {r ∈ R | rm = 0} is the “annihilator” of m. Conversely, Conversely, _{R}R is generated as an R-module by 1, hence is cyclic, and so any quotient of it is generated by a single element as well.

The notions of direct sum and product go over as well to R-modules: the direct product and sum of R-modules are both R-modules, with scalar multiplication being defined coordinate by coordinate, e.g., in the R-module M . ⊕ N, r(m, n) = (rm, rn) by definition. The usual universal properties hold for direct sums and products. In particular, if N and P are submodules of the R-module M, then M = N ⊕ P if and only if N + P = M and N ∩ P = 0.

The ring R is itself a (left) R-module; its submodules are precisely its (left) ideals. The notation for this module is _{R}R.

Definition. Let R be a ring. A free (left) R-module (on a set I) is the direct sum of copies of the R-module _{R}R (indexed by I).

We may also define a basis of an R-module M to be a subset B ⊆ M such that each m ∈ M has a unique expression

m = ∑_{b∈B} r_{b}b

with almost all coefficients r_{b} = 0.

As with free abelian groups, M is a free (left) R-module on I if and only if it has a (left) basis indexed by I, and these conditions are equivalent to the existence of a mapping ι : I → M satisfying the usual universal mapping property.

Namely, if M = ⊕_{i∈I}_{R}R, then ι takes i ∈ I to ι(i) = the element of M which is 0 in all coordinates but the i-th coordinate, and which is 1 in that coordinate. This mapping has the usual universal property: for any R-module N and any (set) map ψ : I → N there is a unique homomorphism Ψ : M → N (of R-modules) such that Ψ ∘ ι = ψ.

As with abelian groups any two free R-modules on the same set I are isomorphic. The commutativity of R allows us to prove as well that the cardinality of I is determined by the isomorphism type of the corresponding free module.

Proposition. Let R be a PID, or indeed any commutative ring with 1. Let M be an R-module which is free on a set I and also free on a set J. Then |I| = |J|.

Proof. Choose any maximal ideal* A of R. Define AM to be the submodule of M generated by all products am, a ∈ A, m ∈ M. Set M_{I} = ⊕_{i∈I}_{R}R, so that M ≅ M_{I}. Under such an isomorphism, AM corresponds to AM_{1}, so M/AM ≅ M_{I}/AM_{I}. However, it is clear that AM_{I} = ⊕_{i∈I}A, and so M_{I}/AM_{I} ≅ ⊕_{i∈I}_{R}R/A.

Now M/AM is an R-module and ax = 0 for all a ∈ A and x ∈ M/AM. We may therefore try (and succeed) making M/AM into an R/A-module by defining

(r + A)x = rx

for all r ∈ R. The fact that Ax = 0 makes this a good definition, and the module axioms may be checked without incident.

In the same way we may make M_{1}/AM_{1} an R/A-module, and our R-isomorphism between M/AM and M_{1}/AM_{1} is also an R/A-isomorphism, because of the way the R/A-module structure has been defined on these modules. Therefore, as R/A-modules, we have

M/AM ≅ M_{1}/AM_{1} ≅ ⊕_{i∈I}_{R/A}(R/A)

But since A is a maximal ideal, R/A is a field, and the three objects above are vector spaces over R/A. Thus |I| = dim_{R/A}(M/AM), which equals |J| by a similar argument.

QED

Moreover, the universal property as usual implies:

* That is, A < R and there exist no ideals B of R such that A < B < R. The existence of maximal ideals in a (commutative) ring follows quickly from Zorn’s Lemma, q.v. In the case of PID’s, we know that the set of all ideals satisfies the maximum condition and so maximal ideals of course exist.

Theorem 0^{R}. Let R be any ring. Then every (left) R-module is a quotient of a free (left) R-module. More precisely, if M is an R-module and S ⊆ G is a subset such that ⟨S⟩ = M, then there is a surjective homomorphism

ϕ : F → M,

where F is a free R-module on S; thus M ≅ F/ ker ϕ.

Now we can repeat the development of the previous section.

Theorem 1^{R}. Suppose that G is an R-module and H is a submodule such that G/H is free. Then G = H ⊕ K for some R-submodule K ≤ G.

Proof. Choose a basis B for G/H and choose an arbitrary preimage of each element of B, thereby forming a subset C ⊂ G. Let K be the R-submodule generated by C. The freeness of G/H implies that π_{H}|K is an isomorphism between K and G/H. This in turn means that H ∩ K = 0 and H + K = G, as required.

Theorem 2^{R}. Let R be a PID. Any submodule H of a finitely generated free R-module G is a free R-module. Moreover the rank of H is at most the rank of G.

Proof. Let {e_{1}, …, e_{n}} be a basis of G. The proof is by induction on n. Let G_{0} = Re_{1} + ⋯ + Re_{n−1} and H_{0} = H ∩ G_{0}. Then by induction H_{0} is a free R-module of rank r ≤ n − 1. Moreover H/H_{0} ≅ H + G_{0}/G_{0} ≤ G/G_{0} ≅ Re_{n} ≅ R. Hence H/H_{0} ≅ R or 0. In the first case there exists K ≤ H such that H = H_{0} ⊕ K, so H is a free R-module of rank r + 1 ≤ n. In the second case the desired conclusion is obvious.

Corollary. Let R be a PID. Then any submodule of a finitely generated R-module is finitely generated.

Proof. If H ≤ G with G finitely generated, we may write G = F/R where F is finitely generated free. Then H = F_{0}/R for some R ≤ F_{0} ≤ F by the third isomorphism theorem, and the theorem implies that F_{0} is finitely generated. A fortiori, H is finitely generated.

The main theorem is formulated as a theorem about the relationship between appropriate bases of a free module over a PID R, and a submodule. It will be applied in the next section in the context of Theorem 0 to the inclusion ker ϕ → F (note that Theorem 2 tells us that ker ϕ is itself finitely generated and free, given that F is). It will yield structural information about the quotient F/ ker ϕ, which as Theorem 0 shows is an arbitrary finitely generated R-module.

The theorem has an existence and a uniqueness statement. We shall prove the existence now; the uniqueness assertion will be argued after we formulate and prove the main theorem (see Section 1g) for arbitrary finitely generated modules over the PID R.

Fundamental theorem on finitely generated modules over a PID (Ia: free module version). Let R be a PID. Let M be a finitely generated free R-module, of rank s, and N a submodule of M. Then N is a finitely generated free R-module, of rank r ≤ s. Moreover there exist bases h_{1}, …, h_{r} of N and e_{1}, …, e_{s} of M, and nonzero elements m_{1}, …, m_{r} ∈ R, the m_{i}’s being uniquely determined up to associates, such that

a) f_{i} = m_{i}e_{i}, i = 1, …, r, and
b) m_{1}|m_{2}| ⋯ |m_{r}.

Again, the bases {e_{i}} and {h_{i}} are not uniquely determined.

Proof of existence. First some observations about (R-module) homomorphisms M → RR. These form an R-module Hom_{R}(M,_{R}R) under (ϕ + ψ)(m) = ϕ(m) + ψ(m) and (rϕ)(m) = r(ϕ(m)) for all m ∈ M and r ∈ R. Checking that ϕ + ψ is a homomorphism requires the commutativity of addition in the image R; checking that rϕ is a homomorphism (in particular, that it preserves scalar multiplication) requires the commutativity of multiplication in R; for any r, s ∈ R and m ∈ M, and ϕ ∈ Hom_{R}(M,_{R}R),

(rϕ)(sm) = r(ϕ(sm)) = r(sϕ(m)) = (rs)ϕ(m) = (sr)ϕ(m) = s(rϕ(m)) = s[(rϕ)(m)].

The identity element of Hom_{R}(M,_{R}R) is the 0 mapping 0(m) = 0 for all m ∈ M; the inverse of ϕ is −ϕ, defined by (−ϕ)(m) = −(ϕ(m)) for all m ∈ M. Furthermore, every choice of an (ordered) basis B = {e_{1}, …, e_{n}} of M gives rise to coordinate mappings ϕ_{i} : M → R (depending on B), namely with ϕ_{i}(∑_{j} n_{j}g_{j}) = n_{i}; each such coordinate mapping lies in Hom_{R}(M,_{R}R).

Now to the proof of the theorem. By Theorem 2, N is free abelian of rank r ≤ n.

We consider the set of all images

Φ = {ϕ(N) | ϕ ∈ Hom_{R}(M,_{R}R).}

Each ϕ(N) is an ideal of R, and so the maximum condition implies that we may select and fix ϕ ∈ Hom(M,_{R}R) such that ϕ(N) is maximal in Φ, i.e., whenever ϕ′ ∈ Hom(M,_{R}R) and ϕ′(N) ≥ ϕ(N), then ϕ′(N) = ϕ(N). As an ideal of R,

ϕ(N) = Rm_{1}

for some m_{1} ∈ R.

If m_{1} = 0, then ϕ(N) = 0 for all ϕ ∈ Hom(M,_{R}R). But for any basis of M, each basiscoordinate function is a homomorphism, so annihilates N. Therefore N = 0, and the theorem holds with r = 0. So we may assume that

m_{1} ≠ 0.

Next let us show that ϕ is surjective, i.e.,

ϕ(M) = R.

Of course ϕ(M) = Ra for some a, since ϕ(M) is an ideal of R. Since R is an integral domain, every element of Ra has the form ra for a unique r ∈ R. Therefore we may write

ϕ(m) = ψ(m)a

where ψ is a well-defined function M → R. The additivity of ϕ, and the fact that R is an integral domain so obeys the cancellation law, implies that ψ is additive, and similarly, ψ is an R-homomorphism. From the above equation, ϕ(N) = aψ(N) ⊆ ψ(N). The maximality of ϕ(N) implies therefore that ψ(N) = aψ(N). Therefore a is a unit, and so ϕ(M) = Ra = R, as claimed.

Now we can choose and fix h_{1} ∈ N, as any element of N such that ϕ(h_{1}) = m_{1}. The homomorphism ϕ|N : N → Rm_{1} maps the submodule Rh_{1} of N onto Rm_{1} = ϕ(N), and so

N = Rh_{1} ⊕ ker(ϕ|N) = Rh_{1} ⊕ (K ∩ N),

where we have put

K = ker(ϕ).

Likewise ϕ : M → R is surjective, and if we choose any f_{1} ∈ M such that ϕ(f_{1}) = 1, we similarly get

M = Rf_{1} ⊕ K

The next objective is to show that f_{1} may be replaced by e_{1} ∈ M satisfying h_{1} = m_{1}e_{1}. The submodule K of M is free, and we choose a basis f_{2}, …, f_{n} of it. Then f_{1}, …, f_{n} form a basis of M, and so h_{1} = r_{1}f_{1} + ⋯ + r_{n}f_{n} for some r_{i} ∈ R. Applying ϕ we get m_{1} = r_{1} + 0 = r_{1}. Thus h_{1} = m_{1}f_{1} + r_{2}f_{2} + ⋯ + r_{n}f_{n}.

We show that m_{1}|r_{i} for all i ≥ 2. Let ϕ_{i} be the i-th coordinate function on M with respect to f_{1}, …, f_{n}. Then ϕ_{2}(h_{1}) = r_{2}. Let d be a gcd of m_{1} and r_{2} and write d = am_{1} + br_{2}, a, b ∈ R. Then (aϕ + bϕ_{2})(h_{1}) = d. The image of aϕ + bϕ_{2} thus contains Rd and hence Rm_{1}. The maximality of Rm_{1} then implies that Rd = Rm_{1}, so m_{1} divides r_{2}. Similarly it divides r_{i} for all i ≥ 2, and we write r_{i} = m_{1}s_{i}.

Set e_{1} = f_{1} + s_{2}f_{2} + ⋯ + s_{n}f_{n}. Then h_{1} = m_{1}f_{1} + m_{1}s_{2}f_{2} + ⋯ = m_{1}e_{1}. Moreover, ϕ(e_{1}) = 1 since ϕ(f_{i}) = 0 for all i ≥ 2. Therefore

M = Re_{1} ⊕ K and h_{1} = m_{1}e_{1}.

Clearly K has rank n − 1. If K = 0 (i.e., n = 1), then there is nothing more to prove. If K ≠ 0, then by induction on n, applied to K ∩ N ⊆ K, there are bases e_{2}, …, e_{n} of K and h_{2}, …, h_{r} of K ∩ N and positive integers m_{2}, …, m_{r} ∈ R such that h_{j} = m_{j}e_{j}, 2 ≤ j ≤ r, and m_{2} | ⋯ | m_{r}. Then e_{1}, e_{2}, …, e_{n} form a basis of M, and the h_{i} form a basis of N, so to complete the proof of existence it remains only to check that

m_{1} | m_{2}.

Let ϕ_{i} now be the coordinate functions on M with respect to e_{1}, …, e_{n}. Then as h_{1} +h_{2} = m_{1}e_{1} + m_{2}e_{2}, we have ϕ_{1}(h_{1} + h_{2}) = m_{1} and ϕ_{2}(h_{1} + h_{2}) = m_{2}. Thus for suitable a, b ∈ R, (aϕ_{1} + bϕ_{2})(h_{1} + h_{2}) = d, a gcd of m_{1} and m_{2}. As before, Rm_{1} ⊆ Rd ⊆ im(aϕ_{1} + bϕ_{2}), so the maximality of Rm_{1} implies that Rm_{1} = Rd, so m_{1}|m_{2}. This completes the proof of existence in Theorem Ia.

We have only proved existence here; the uniqueness assertion will be proved later.

The association classes [m_{1}], [m_{2}], …, [m_{r}], or more sloppily the elements m_{1}, …, m_{r} ∈ R, are sometimes called the “invariant factors” of the inclusion mapping E → F.

Exercise. State and prove the analogue of version Ib for matrices with entries in PID.

1f. Finitely Generated Abelian Groups and Modules over PID’s

In the next two sections we apply the previous theorem to determine the structure of finitely generated modules over a PID. We also complete the (uniqueness) proof of the previous theorem!

In any abelian group G, the set

T (G) = {g ∈ G | g has finite order}

is a subgroup of G, since mg = nh = 0 implies mn(g − h) = 0. It is called the torsion subgroup of G. The group G is called torsion-free if and only if T (G) = 0.

Likewise if M is a module over a PID R, we define

T (M) = {m ∈ M | for some 0 ≠ r ∈ R, rm = 0}.

In a similar way we see that T (M) is a submodule of M. (This actually only requires R to be an integral domain.) The module M is called torsion-free if and only if T (M) = 0.

Exercise. T (M) is “fully invariant”, that is, for every homomorphism ϕ : M → M of R-modules, we have ϕ(T (M)) ≤ T (M).

Proposition. Let R be a PID (or any integral domain). Let M be an R-module. Then T (M) is a submodule of M, and M/T (M) is torsion-free.

Proof. If rm = sn = 0 with r ≠ 0 ≠ s, then rs(m ± n) = 0 with rs ≠ 0. Hence T (M) is a submodule of M. If x + T (M) ∈ T (M/T (M)), then rx ∈ T (M) for some r ≠ 0. Therefore srx = 0 for some s ≠ 0, and as sr ≠ 0, this gives x ∈ T (M). This proves that T (M/T (M)) = 0.

Fundamental theorem on finitely generated modules over a PID (II: invariant factor version). Let R be a PID. Let M be finitely generated module over a PID R. Then M = T (M) ⊕ N for some submodule N ≤ M. Moreover N is a free R-module, and there exist m_{1}, …, m_{r} ∈ R such that no m_{i} is a unit, m_{1} | ⋯ | m_{r} and

T (M) ≅ Z_{m_{1}} ⊕ ⋯ ⊕ Z_{m_{r}}.

The rank of N, and the association classes [m_{1}], …, [m_{r}] are uniquely determined (by M).

This theorem implies, and indeed is equivalent to the following three results:

551, Fall '02     B1g.    Primary Decomposition; PID's are UFD's     p. 19

Theorem 3. Finitely generated torsion-free modules over a PID are free modules.

Theorem 4. If M is a finitely generated module over a PID, then M ≅ T (M)⊕M/T (M).

Theorem 5. If M is a finitely generated torsion module over a PID (“torsion” means M = T (M)), then M ≅ R/m_{1}R⊕⋯ R/m_{r}R for some r ≥ 0 and some nonunits m_{1}, …, m_{r} ∈ R such that m_{1} | ⋯ | m_{r}. The association classes [m_{1}], …, [m_{r}] are uniquely determined by these conditions.

Exercise. Demonstrate the equivalence just asserted.

Proof The “existence” statements are an application of version I of the theorem. Namely, by Theorem 0 there is a finitely generated free module F and a submodule K such that M ≅ F/K. By Theorem 1, K is free, and by the main theorem there are bases {e_{i}}1 ≤ i ≤ s and {f_{i}}_{1≤i≤r} of F and K respectively such that r ≤ s and f_{i} = m_{i}e_{i} for each i ≤ r, where m_{i} ∈ R and m_{1} | ⋯ | m_{r}. Letting F_{i} = ⟨e_{i}⟩, and setting m_{i} = 0 for r < i ≤ s we have

M ≅ F/K = (⊕_{i=1}^{s}F_{i})/(⊕_{i=1}^{s}m_{i}F_{i}) ≅ ⊕_{i=1}^{s}F_{i}/m_{i}F_{i} ≅ (⊕_{i=1}^{r}R/m_{i}R) ⨁(⊕_{i=r+1}^{s}R).

Obviously the first summand corresponds to T (M), and the second summand is free of rank s−r. Furthermore, for those m_{i} which are units, R/m_{i}R = 0, so we may delete them from the final expression. This proves everything but uniqueness.

Now s − r is the rank of M/T (M), so is uniquely determined by M. It remains to show that the m_{i} in the statement of the theorem (i.e., the nonunits) are uniquely determined up to associates.

Notice that our argument shows more, namely that the m_{i} coming from a submodule N ≤ M in version I are the same as the m_{i} coming from the module M/N in version II, except that enough units are added to bring the rank of N up to r = s − (s − r), the rank of M minus the torsion-free rank of M/N. Consequently knowing the rank of M and the invariant factors of M/N determines the m_{i} in version I. Therefore

uniqueness in version II implies uniqueness in version I.

What remains for us to do, therefore, is prove uniqueness in version II.

1g. Primary Decomposition; PID’s are UFD’s

A further decomposition can be made, using the Chinese Remainder Theorem. For the case R = Z this theorem states that Z_{n} is isomorphic to the direct product (or sum!) of the groups Z_{p_{i}^{e_{i}}}, where n = ∏_{i} p_{i}^{e_{i}} is the factorization of n into powers of distinct primes. (In a homework exercise you showed that the direct product of two finite cyclic groups of relatively prime orders is a cyclic group, and this implies the C.R.T. for the integers.) By applying the C.R.T. to each of the summands Z_{m_{i}} in the fundamental theorem on finitely generated abelian groups, we get a decomposition of T (G) as the direct sum of cyclic groups of prime power order. Moreover, a uniqueness statement can be obtained here as well. But before doing this for PID’s, we need the C.R.T. for them. Indeed we need to prove the fundamental theorem of arithmetic for them!

Definition. Let R be an integral domain, and let p ∈ R. Then p is prime if and only if p ≠ 0, p is not a unit, and whenever a, b ∈ R are such that p | ab, then either p | a or p | b.

Definition. Let R be an integral domain, and let p ∈ R. Then p is irreducible if and only if p ≠ 0, p is not a unit, and in any factorization p = ab, a, b ∈ R, either a or b is a unit.

These two notions, identical if R = Z, are conceptually different and are not identical in general. For example, in R = Z[\sqrt{−5}], the ring of all integer combinations a + b\sqrt{−5}, if we put z = 1 + 2\sqrt{−5}, then

3 · 7 = z · z̄

the bar denoting complex conjugation. Thus 3 | z · z̄, although it is easily checked that 3 does not divide z or z̄. Hence 3 is not prime. However, 3 is irreducible. Indeed the norm mapping N : R → Z, N(x) = xx̄, is multiplicative, and N(3) = 9. A factorization 3 = xy would give N(x)N(y) = 9. However, using N(a + b\sqrt{−5}) = a^{2} + 5b^{2}, we see that no element of R has norm 3, so either N(x) = 1 or N(y) = 1, which in turn implies that x or y is ±1, a unit. Hence “irreducible” does not imply “prime” in general. (The problem is that the F.T. of Arithmetic does not extend from Z to R. )

Lemma. In any integral domain, prime implies irreducible.

Proof. Suppose that p = ab and p is prime. Then p | ab, so p divides one of the factors, which we may as well assume is a. Thus a = pc for some c. Now p = pcb, so 1 = cb as we are in an integral domain. Therefore b is a unit. QED

Lemma. In a PID, irreducible implies prime.

Proof. The key is that GCD’s of two elements exist in a PID (Section 1d), and are linear combinations of the two elements. Suppose that R is a PID and p ∈ R is irreducible, and p | ab. Suppose that p does not divide a. Let d = gcd(p, a), well-defined up to associates. If [d] = [p], then p | a, contradiction. But by irreducibility the only other divisors of p are units, so [d] = [1]. Hence there are m, n ∈ R such that mp + na = 1. Then b = bmp + nab, and as p | ab we get p | b. QED

Definition. A unique factorization domain (UFD) is an integral domain R such that for every x ∈ R such that x ≠ 0, there exists a unit u, an integer n ≥ 0, primes p_{1}, …, p_{n}, no two of which are associates, and positive integers e_{1}, …, e_{n} such that

x = up_{1}^{e_{1}} ⋯ p_{n}^{e_{n}};

moreover, this decomposition is unique, except for “trivial changes” of the following sorts: rearrange the order of terms, and replace some p_{i} by an associated element (changing u in the process as well).

Theorem. Every PID is a UFD.

Proof. Let R be a PID. The existence of a factorization uses only the property that the ideals of R satisfy the maximum condition: Suppose by way of contradiction that some element x ≠ 0 of R has no factorization as a product of a unit and irreducibles, and among all counterexamples choose one such that the ideal Rx is maximal. Now x is itself not irreducible, otherwise it would have the factorization x = x. So x = yz for some nonunits y and z. Clearly x ∈ Ry, so Rx ⊆ Ry. Likewise Rx ⊆ Rz. If both these inclusions are proper, then y and z would have factorizations, which when put together would give a factorization of x, contradiction. Therefore Rx = Ry, say. But then x and y are associates and so z is a unit, contradiction.

The uniqueness uses only the fact that every irreducible is prime. Suppose that

x = u ∏_{i} p_{i}^{e_{i}} = v ∏_{j} q_{j}^{f_{j}}

with the p_{i} distinct irreducibles, u a unit, the e_{i} positive integers, and similar conditions on the right side. Then p_{1} | x. Since p_{1} is prime and divides the right side, p_{1} | q_{j} for some j. But q_{j} is irreducible and so [p_{1}] = [q_{j}]. We may change notation so that [p_{1}] = [q_{1}], so that p_{1} = wq_{1}, w a unit. We may now cancel one p_{1} from the left, cancel one q_{1} from the right and replace v by vw^{−1}, and then complete the proof by induction on ∑_{i} e_{i}. QED

Corollary. In a PID, the gcd of two elements a and b is the product of all the “common” factors in the prime factorizations of a and b.

Here “common” means “up to associates”; recall that gcd’s are only well-defined up to associates, anyway.

Chinese Remainder Theorem. Let R be a commutative ring with 1. Let I_{1}, …, I_{n} be ideals of R such that for each i ≠ j, I_{i} + I_{j} = R. Define I_{1} ⋯ I_{n} to be the ideal of R generated by all products r_{1} ⋯ r_{n} with r_{i} ∈ I_{i} for each i. Then there is an isomorphism of R-modules

\frac{R}{I_{1} ⋯ I_{n}} ≅ \frac{R}{I_{1}} ⊕ ⋯ ⊕ \frac{R}{I_{n}} .

The hypothesis I_{i} + I_{j} = R is absolutely essential here!

Proof. The R-module homomorphism

ϕ : R → \frac{R}{I_{1}} ⊕ ⋯ ⊕ \frac{R}{I_{n}} , r ↦ (r + I_{1}, …, r + I_{n})

has kernel I_{1} ∩ ⋯ ∩ I_{n}. We finish the proof by showing

a) I_{1} ∩ ⋯ ∩ I_{n} = I_{1} ⋯ I_{n};
b) ϕ is onto.

Indeed,

R = R · R ⋯ R = (I_{1} + I_{2})(I_{1} + I_{3}) ⋯ (I_{1}I_{n}) ⊆ I_{1} + (I_{2} ⋯ I_{n}).

In particular every coset of I_{1} in R contains an element of I_{2} ⋯ I_{n}, which shows that the image of ϕ contains the first direct factor R/I_{1}. Similarly it contains the others, so ϕ is onto. To prove a), it is obvious that the product of ideals lies in their intersection, by definition of ideal. The displayed equation has an analogue R = I_{i} + I_{1} ⋯ Î_{i} ⋯ I_{n} for each i (Î_{i} means to omit this term). Multiplying these n equations we find that

R = Î_{1}I_{2} ⋯ I_{n} + I_{1}Î_{2} ⋯ I_{n} + ⋯ + I_{1}I_{2} ⋯ Î_{n}.

Putting J = ∩_{i=1}^{n}I_{i} we see that the product of J with any one of these summands lies in I_{1}I_{2} ⋯ I_{n}. Hence

J = RJ ⊆ I_{1}I_{2} ⋯ I_{n}.

QED

Corollary. Let R be a PID and let x = up_{1}^{e_{1}} ⋯ p_{n}^{e_{n}} be the primary decomposition of x ∈ R. Then

R/Rx ≅ R/Rp_{1}^{e_{1}} ⊕ ⋯ ⊕ R/Rp_{n}^{e_{n}}.

It is vital here that the p_{i} be non-associated with one another.

Proof. Since the p_{i} are non-associated with one another, gcd(p_{i}^{e_{i}}, p_{j}^{e_{j}}) = 1 for i ≠ j by the last corollary. So Rp_{i}^{e_{i}} + Rp_{j}^{e_{j}} = R. Now apply the previous result with I_{i} = Rp_{i}^{e_{i}}. QED

Now we can formulate the third and last version of our main theorem on finitely generated modules over a PID.

Fundamental theorem on finitely generated modules over a PID (III: elementary divisor version). Let R be a PID. Let M be finitely generated module over a PID R. Then M = T (M) ⊕ N for some submodule N ≤ M. Moreover N is a free R-module, and there exist prime powers p_{1}^{e_{1}}, …, p_{s}^{e_{s}} in R such that

T (M) ≅ Z_{p_{1}^{e_{1}}} ⊕ ⋯ ⊕ Z_{p_{s}^{es}} .

The rank of N is uniquely determined by M, as are the association classes [p_{1}^{e_{1}}], …, [p_{s}^{e_{s}}] (except for the order in which they appear).

To prove the existence of this decomposition, use version II, factor each m_{i} as the product of powers of distinct primes, and apply the Chinese Remainder Theorem.

Conversely, the set of resulting prime powers (counting multiplicities!) determines the sequence m_{1}, …, m_{r}. That is, if m_{1} | ⋯ | m_{r}, we can recover m_{1}, …, m_{r} from the list of the prime powers appearing in the decompositions of m_{1}, …, m_{r}. Namely, for each prime which appears, the largest power of that prime divides some m_{i} and hence divides m_{r}. Hence m_{r} must be the product, over all (distinct) primes appearing, of the largest powers of those primes which occur. Removing these from consideration m_{r−1} must be the product, over all primes still remaining, of their largest powers still remaining, etc.

Thus Uniqueness in version III implies uniqueness in version II.

Therefore it remains only to show uniqueness in version III.

551, Fall '02                   B1h.    Uniqueness                   p. 23

Exercise. Show that the decomposition of T (G) into “primary” parts does not depend on the finite generation of G. More precisely, let G be a module over the PID R, and assume that G = T (G). For each association class [p] of primes in R define

G_{p} = {g ∈ G | p^{n}g = 0 for some positive integer n}.

Show that

a) Each G_{p} is an R-submodule of G.

b) G = ∐ · G_{p}, with one summand for each association class of primes in R.

1h. Uniqueness

We finally complete the proof of the fundamental theorem by proving uniqueness in version III. We use the invariants M[x] and xM of a torsion module M over a PID R.

Definition. Let M be a module over the commutative ring R. For each x ∈ R define M[x] = {m ∈ M | xm = 0} and xM = {xm | m ∈ M}.

It is obvious that each of these is a submodule of M for any x ∈ R. Furthermore, we may consider M[x] to be an R/xR-module by defining (r + xR)m = rm for each m ∈ M[x]. Moreover the following lemmas are easy to prove.

Lemma. Suppose that M and N are R-modules and M = M[x], N = N[x]. Then a mapping ϕ : M → N is an R-homomorphism if and only if it is an R/xR-homomorphism.

Lemma. If R is a commutative ring and M and M_{i} are R-modules such that M = ∐ M_{i}, then M[x] = ∐ M_{i}[x] and xM = ∐ xM_{i} for each x ∈ R.

Lemma. Let R be a PID, and M = R/p^{n}R, where p^{n} is a prime power in R, n ≥ 0. Then

a) If q ∈ R is a prime not associated with p, then M = qM and M[q] = 0.

b) M[p] ≅ R/pR (both as R-module and as R/pR-module).

c) If r ∈ Z^{+}, then

p^{r}M[p] ≅ \{ \begin{matrix}R/pR & if r < n \\ 0 & otherwise. \end{matrix}

d) If p′ is a prime associated with p, then M[p] = M[p′] and p^{r}M = (p′)^{r}M for any

r.

Proof. Under the canonical projection ϕ : R → R/p^{n}R = M, we have ϕ(qR) = qM. But gcd(q, p^{n}) = 1 so qR + p^{n}R = R. Hence M = ϕ(R) = ϕ(qR) = qM. Similarly, if m ∈ M[q], then m = r + p^{n}R fo some r ∈ R, and qr ∈ p^{n}R. Therefore p^{n}|qr, so p^{n}|r by unique factorization, so m = 0. This proves a).

Next, the preimage of M[p] is the ideal {r ∈ R | pr ∈ p^{n}R}, i.e., the ideal p^{n−1}R. So M[p] = p^{n−1}R/p^{n}R. Define ψ : R → M[p] by ψ(r) = p^{n−1}r + p^{n}R. This is a surjective module homomorphism, and ψ(r) = 0 if and only if p^{n−1}r ∈ p^{n}R, i.e., p|r (using unique factorization). This proves b).

Next, it is clear that p^{n}M = 0. If m < n then p^{m}M = p^{m}R/p^{n}R. But α : R → p^{m}R/p^{n}R defined by α(r) = p^{m}r +p^{n}R is a surjective homomorphism, with kernel p^{n−m}R by unique factorization. So p^{m}M ≅ R/p^{n−m}R and c) follows from b). The proof of d) is left to the reader. QED

Now suppose that

M ≅_{ϕ} ( ∐_{i} (∐_{j} R/p_{i}^{n_{ij}}R) ) ⊕ ∐_{j=1}^{m}_{R}R,  1D

where the p_{i} are pairwise non-associated primes, and the n_{ij} are positive integers. We must show that the isomorphism type of M determines the rank m, as well as the primes p_{i} (up to association and the order in which they occur) and the n_{ij} (up to order). This we do as follows. First, ϕ(T (M)) is the first direct sum, so M/T (M) ≅ ∐_{j=1}^{m} _{R}R. Therefore

m = rank(M/T (M)).

Second, for any fixed i, and any integer a, we can compute p_{i}^{a}T (M)[p_{i}], using the above lemmas. By the second lemma it is isomorphic to the direct sum of p_{i}^{a}N[p_{i}] as N ranges over the direct summands in (1D). But the third lemma implies that these terms are trivial a) for p_{k} not associated to p_{i}; and b) for those N = R/p_{i}^{n_{ij}} terms for which n_{ij} ≤ a. Moreover for the terms N = R/p_{i}^{n_{ij}} terms for which n_{ij} > a. we get N[p_{i}] ≅ R/p_{i}R. Therefore if we let c_{i,a} be the number of n_{ij} for which n_{ij} > a, we find that p_{i}^{a}T (M)[p_{i}] is isomorphic to the direct sum of c_{i,a} copies of R/p_{i}R. By the first lemma they are isomorphic as R/p_{i}R-modules. But R/p_{i}R is a field:

Lemma. In a PID, if p is a prime, then Rp is maximal.

(Proof: if Rp < Rx, then x is a proper divisor of p so is a unit.)

Therefore

c_{i,a} = dim_{R/p_{i}R}(p_{i}^{a}T (M)[p_{i}]).

Furthermore, the right side does not change if we replace p_{i} by an associate, by the second lemma (d).

Hence c_{i,a}, the is determined by the isomorphism type of M. (If ϕ : M → N is an isomorphism, then ϕ carries T (M) to T (N), so induces an isomorphism between the vector spaces, whence they have the same dimension.

Finally, for any given i and n, the number of terms p_{i}^{n} appearing in (1D) is c(i, a−1)−c(i, a) so is also determined by the isomorphism type of M.

551, Fall '02      B2a.    The module of a linear transformation      p. 1

2. Linear transformations; canonical forms.

2a. The module of a linear transformation

We shall be concerned here with a linear transformation T : V → V, where V is a finitedimensional vector space over a field F. For such a setup, the powers T^{n} (n ≥ 0) i.e., the iterated composites of T with itself, are defined, and they too are linear transformations from V to V. Considering T^{0} to be the identity transformation id_{V}, we can then make V into a module over the polynomial ring F [X] in one indeterminate X. Namely given a polynomial f ∈ F [X], say f(X) = a_{n}X^{n} + ⋯ + a_{1}X + a_{0}, and for any v ∈ V, define

fv = a_{n}T^{n}v + ⋯ + a_{1}T v + a_{0}v.

It is straightforward to check that the module axioms are satisfied. We shall call this module V_{T}.

Because F is a field, the ring F [X] is a Euclidean domain (relative to the degree function), so is a PID. Moreover:

Lemma. In this situation, V is a finitely generated torsion module for F [X].

Proof. A basis of V (as vector space) is finite by assumption and certainly generates V as an F [X]-module. So V is finitely generated. To see that it is a torsion module, either a) use the fundamental theorem to deduce that otherwise, it would have a free rank 1 submodule V_{0} ≅ F [X], leading to the contradiction dim V ≥ dim V_{0} = ∞; or b), set n = dim V and argue that given any v ∈ V, the vectors v, T v, …, T^{n}v form a linearly independent set, and then a dependence relation takes the form pv = 0 for some nonzero polynomial p ∈ F [X].

The fundamental theorem will therefore give a lot of information. In order to decode it, we first consider exactly what information is encoded in the module structure of V. This is the same as asking what it means for two such modules to be isomorphic.

Lemma. Let T : V → V and U : W → W be linear transformations, where V and W are finite-dimensional vector spaces over the same field F. Then the following conditions are equivalent:

a) V_{T} ≅ W_{U} as F [X] − modules;

b) There exists an isomorphism ϕ : V → W of vector spaces such that the diagram

\begin{aligned}T : & V & → & V \\ ↓ & ↓ \\ U : & W & → & W \end{aligned}

commutes;

c) There exist ordered bases B of V and C of W such that the matrices [T]_{B}^{B} and [U]_{C}^{C} coincide.

p. 2    B2.    Linear transformations; canonical forms.      551, Fall '02

Proof. If a) holds, then an isomorphism ϕ : V_{T} → W_{U} is in particular a bijective linear transformation and also satisfies ϕ(Xv) = Xϕ(v) for all v ∈ V. This means ϕ(T v) = Uϕ(v) so b) holds. If b) holds, then c) holds for an arbitrary basis B of V and its image C = ϕ(B), which is a basis of W. If c) holds, and B = {b_{1}, …, b_{n}} and C = {c_{1}, …, c_{n}}, then the mapping ϕ(∑ α_{i}b_{i}) = ∑ α_{i}c_{i} satisfies b). Finally if b) holds, then ϕ(T v) = Uϕ(v) for all v ∈ V, whence ϕ(T^{n}v) = U^{n}ϕ(v) follows by an inductive argument, and then since ϕ, T and U are all linear transformations, ϕ(fv) = fϕ(v) for all f ∈ F [X], and ϕ itself is the required isomorphism.

Two square matrices A, A′ of the same size with coefficients in F are called similar (over F) if and only if there is an invertible D of the same size with coefficients in F such that

DAD^{−1} = A′ or equivalently—as D is invertible—DA = A′D.

Given an ordered basis B of V, the invertible matrices of size dim V are just the matrices D = [id_{V}]_{B′}^{B} as B′ ranges over the set of all ordered bases of V, and so if A = [T]_{B}^{B}, then DAD^{−1} = [T]_{B′}^{B′}. Therefore

Lemma. The conditions of the above lemma are also equivalent to:

c’) For some bases B of V and C of W, the matrices [T]_{B}^{B} and [U]_{C}^{C} are similar;

c”) For any bases B of V and C of W, the matrices [T]_{B}^{B} and [U]_{C}^{C} are similar; The isomorphism type of V_{T} determines and is determined by the equivalence class of [T]_{B}^{B} under similarity.

We can therefore interpret the application of the fundamental theorem as a statement about similarity classes of matrices.

A submodule W ⊆ V_{T} is simply a subspace such that T (W) ⊆ W, since this condition immediately implies that fW ⊆ W for all f ∈ F [X]. Such a submodule W exists and has dimension m if and only if T, with respect to some basis, has matrix in upper-triangular “block” form

[ \begin{matrix}X & Y \\ 0 & Z \end{matrix} ]

where X is m × m and Z is (n − m) × (n − m). For if T is similar to such a matrix, then that matrix equals [T]_{B}^{B} for some ordered basis B, and the zeroes in the lower left imply that the first m elements of B span a subspace W such that T (W) ⊆ W. Conversely, if W exists, then an ordered basis of W followed by further vectors constitutes an ordered basis B of V for which [T]_{B}^{B} is in upper-triangular block form.

Similarly, to say that V_{T} has a direct sum decomposition V_{T} = V_{1} ⊕ V_{2} as F [X]-modules, where V_{1} and V_{2} have dimension m and n − m, respectively, is equivalent to saying that for some B, [T]_{B}^{B} is in block-diagonal form

[ \begin{matrix}X_{1} & 0 \\ 0 & X_{2} \end{matrix} ] .

If the decomposition exists, one can take B to consist of a basis of V_{1} followed by a basis of V_{2}; and conversely, if the matrix is in this form, V_{1} can be recovered as the span of the first m elements in the ordered basis B, and V_{2} as the span of the last n − m elements.

551, Fall '02             B2b.    Rational canonical form             p. 3

Let R = F [X]. What does it mean to say that V_{T} ≅ R/fR as R-modules, where 0 ≠ f ∈ R? For simplicity let us assume that f is monic, i.e. has leading coefficient 1. (There is no loss of generality in this assumption since f has a (unique) associate f^{∗} which is monic, and fR = f^{∗}R.) Write f(X) = X^{n} + a_{n−1}X^{n−1} + ⋯ + a_{1}X + a_{0}. Then V_{T} ≅ R/fR if and only if for some basis B,

[T]_{B}^{B} = [ \begin{matrix}0 & 0 & 0 & 0 & ⋯ & 0 & −a_{0} \\ 1 & 0 & 0 & 0 & ⋯ & 0 & −a_{1} \\ 0 & 1 & 0 & 0 & ⋯ & 0 & −a_{2} \\ 0 & 0 & 1 & 0 & ⋯ & 0 & −a_{3} \\ & & & & ⋯ \\ 0 & 0 & 0 & 0 & ⋯ & 1 & −a_{n−1} \end{matrix} ] .b

The matrix on the right is called the companion matrix of the monic polynomial f, written C(f). Thus C(f) is a certain n × n matrix, where n = deg f.

To see this, notice that in \overline{R} = R/fR, every coset \overline{g} has a unique representative g which is a polynomial of degree strictly less than n = deg f, by the division algorithm. Let us write \overline{h} = h + fR. Then the images \overline{1}, \overline{X}, \overline{X^{2}}, …, \overline{X^{n−1}} form a basis of \overline{R} = R/fR, corresponding to a basis B of V. Multiplication by X transforms this to the basis \overline{X}, \overline{X^{2}}, …, \overline{X^{n}}. Therefore the matrix of T with respect to B has as its columns the coordinates of \overline{X}, \overline{X^{2}}, …, \overline{X^{n}} with respect to the basis \overline{1}, \overline{X}, \overline{X^{2}}, …, \overline{X^{n−1}}. Since \overline{f} = 0 we have f(\overline{X}) = 0, so \overline{X^{n}} = −a_{0} − a_{1}\overline{X} − ⋯, and we obtain the companion matrix.

2b. Rational canonical form

With this interpretation, we immediately get “rational canonical form” theorems. For example, from version II we get:

Theorem. Let V be a finite-dimensional vector space and T : V → V a linear transformation. Then there exists a unique r ≥ 0 and unique monic polynomials f_{1}, …, f_{r} such that f_{1} | f_{2} | ⋯ | f_{r} and [T]_{B}^{B} is the block diagonal matrix

[ \begin{matrix}C(f_{1}) & 0 & ⋯ & 0 \\ 0 & C(f_{2}) & ⋯ & 0 \\ & & ⋯ \\ 0 & 0 & ⋯ & C(f_{r}) \end{matrix} ] .

and its matrix version:

Theorem. Let F be a field and A a square matrix over F. Then there exist unique r and monic polynomials f_{1}, …, f_{r} ∈ F [X] such that f_{1} | f_{2} | ⋯ | f_{r} and A is similar to the block-diagonal matrix whose blocks are the companion matrices C(f_{1}), …, C(f_{r}).

This is obtained by taking V = F^{n} to be the space of column vectors, and T : V → V to be defined by T (v) = Av, and applying the previous theorem.

The polynomials f_{1}, …, f_{r} are called the invariant factors of T or A, as the case may be.

Similarly, there is an “elementary divisor” version, corresponding to version III of the fundamental theorem, giving a canonical form for the similarity class of any square matrix over F as a block-diagonal matrix with the blocks being companion matrices of powers of monic irreducible polynomials (unique up to the order of blocks); these prime powers are the “elementary divisors” of the transformation or matrix.

Theorem. Let A be a square matrix over a field F. Then there exist irreducible polynomials p_{1}, …, p_{r} and positive integers n_{1}, …, n_{r} such that A is similar to the block-diagonal matrix whose blocks are C(p_{1}^{n_{1}}), …, C(p_{r}^{n_{r}}). The prime powers p_{1}^{n_{1}}, …, p_{r}^{n_{r}} are uniquely determined up to order.

Example. The invariant factor version implies that every 2×2 real matrix is similar (over R) to exactly one of the following:

[ \begin{matrix}α & 0 \\ 0 & α \end{matrix} ] , α ∈ R or [ \begin{matrix}0 & γ \\ 1 & β \end{matrix} ] , β, γ ∈ R

the first having invariant factors X − α, X − α and the second having the single invariant factor X^{2} − βX − γ. The elementary divisor version implies that every real matrix is similar (over R) to exactly one of the following:

[ \begin{matrix}α & 0 \\ 0 & β \end{matrix} ] , α, β ∈ R or [ \begin{matrix}0 & γ \\ 1 & β \end{matrix} ] , β, γ ∈ R, β^{2} + 4γ ≤ 0.

This is because the monic irreducible polynomials over R are the polynomials X−α, α ∈ R and the polynomials X^{2} +βX +γ with β^{2} < 4γ. (Notice that the companion matrix of this last polynomials has entries −β and −γ in the last column.) Now the elementary divisors of a 2 × 2 matrix either are a) two irreducible polynomials of degree 1; b) an irreducible polynomial of degree 2; or c) the square of an irreducible polynomial of degree 1. It is the third possibility which is covered above by the case β^{2} + 4γ = 0.

The rational forms for 2 × 2 matrices over C are similar, except that the quadratic irreducible possibility does not exist.

Those of the above forms with nonzero determinant therefore are a set of representatives for the conjugacy classes of the group GL_{2}(R).

Caution: the term “rational canonical form” is not universally interpreted this way.

2c. Jordan canonical form

Definition. The field F is algebraically closed if and only for every nonconstant polynomial f ∈ F [X] there exists a zero, i.e., an element z ∈ F such that f(z) = 0.

This immediately implies that X − z divides f. (Write f(X) = q(X)(X − z) + y, with y ∈ F. Then 0 = f(z) = y.) Consequently over an algebraically closed field, the monic

551, Fall '02              B2c.    Jordan canonical form              p. 5

irreducible elements of F [X] are just the linear polynomials X − α, α ∈ F. Conversely, if these are all the irreducibles, then the fact that F [X] is a UFD (uniqueness doesn’t matter) implies that every polynomial has a linear factor, hence a zero, so F is algebraically closed.

The most famous example of an algebraically closed field is C. But in fact every field can be embedded in an algebraically closed field (proof next semester).

To analyze linear transformations and matrices over an algebraically closed field, there is a more useful canonical form than rational canonical form. It uses the elementary divisor version of the fundamental theorem, and “Jordan blocks” instead of companion matrices. Every monic prime power in F [X], in this case, is of the form f(X) = (X −α)^{n}. Instead of the basis \overline{1}, \overline{X}, …, \overline{X^{n−1}} of R/fR which we used for companion matrices, we can use the basis \overline{1}, \overline{X} − α, (\overline{X} − α)^{2}, …, (\overline{X} − α)^{n−1}. We have \overline{X}(\overline{X} − α)^{i} = (α + \overline{X} − α)(\overline{X} − α)^{i} = α(\overline{X} − α)^{i} + (\overline{X} − α)^{i+1}. As (\overline{X} − α)^{n} = 0 we get the lower triangular matrix

J(α, n) = [ \begin{matrix}α & 0 & 0 & ⋯ & 0 & 0 \\ 1 & α & 0 & ⋯ & 0 & 0 \\ 0 & 1 & α & ⋯ & 0 & 0 \\ & & & ⋯ \\ 0 & 0 & 0 & ⋯ & 1 & α \end{matrix} ] ,

all diagonal entries being α, all sub-diagonal entries being 1, and all other entries being 0. Version III of the fundamental theorem then implies (we state only the matrix version):

A matrix will be said to be in Jordan canonical form (over F) if and only if it is in block-diagonal form

[ \begin{matrix}J(α_{1}, n_{1}) & 0 & ⋯ & 0 \\ 0 & J(α_{2}, n_{2}) & ⋯ & 0 \\ & & ⋯ \\ 0 & 0 & ⋯ & J(α_{r}, n_{r}) \end{matrix} ]

for some α_{i} ∈ F and some positive integers n_{1}, …, n_{r}.

Version III of the fundamental theorem then implies:

Theorem. Let F be an algebraically closed field. Then any square matrix over F is similar to a matrix in Jordan canonical form. This canonical form is unique except for the order in which the diagonal blocks appear.

For the above canonical forms to be useful it is important that they can be calculated, or at least partial information about them can be calculated. The important invariants μ_{A} and χ_{A} (minimal and characteristic polynomials) will be discussed below. For Jordan canonical form, there is a nice characterization of the number of blocks J(α, n) for any α ∈ F and any positive integer n, although it requires fairly detailed information about the matrix in question. We define the nullspace N of an n × n matrix A to be the space of all vectors v such that Av = 0, and the nullity ν(A) to be dim N(A). If A = [T]_{B}^{B} (or even [T]_{B′}^{B}!) then ν(A) = dim ker(T). Therefore if A and A′ are similar, then ν(A) = ν(A′).

Exercise. If A is in block diagonal form,

A = [ \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ] ,

then ν(A) = ν(A_{1}) + ν(A_{2}).

Proposition. Let F be an algebraically closed field and A a square matrix (or T : V → V a linear transformation). Let α ∈ F and let n be a positive integer. Then the number of Jordan blocks J(α, m) such that m > n in the Jordan canonical form of A (or T) is

ν((A − αI)^{n+1}) − ν((A − αI)^{n}), or dim ker((T − α)^{n+1}) − dim ker((T − α)^{n}).

Here we write α for the linear transformation v ↦ αv of V.

Once we know how many blocks J(α, m) there are satisfying m > n, and we know this for all n, subtraction gives us the number of J(α, n) blocks.

Proof. If A and A′ are similar, then so are A−αI and A′ −αI and hence so are (A−αI)^{n} and (A′ − αI)^{n} for every n. Hence the nullities of these two matrices coincide. The upshot of this remark is that in proving our result, we are free to replace A by any matrix similar to it, and so we may assume that A is already in Jordan canonical form. Using the exercise above, we reduce to the case of a single Jordan block A = J(α, n). But then A − β is invertible for all β ≠ α, so the nullities ν((A − β)^{n}) are all 0 and our formula predicts no Jordan blocks J(β, m), which is correct. For α itself, if A = J(α, m), then A−αI = J(0, m), and as one takes the powers of J(0, m), the 1’s slide down one diagonal at a time, until J(0, m)^{m} = 0. Therefore ν(J(0, m)^{i}) = i for 0 ≤ i ≤ m, and ν(J(0, m)^{i}) = m for i > m. Hence

ν(J(0, m)^{n+1}) − ν(J(0, m)^{n}) = \{ \begin{matrix}1 & if 0 ≤ n < m \\ 0 & if n ≥ m \end{matrix} ,

so our formula counts blocks as claimed.

2d. Minimal polynomial

Two of the most important of the invariants of a linear transformation or square matrix (over a field F) are its minimal polynomial and characteristic polynomial, both elements of F [X].

The minimal polynomial is defined as follows. Let V be a finite dimensional vector space over the field F. The space A = Hom_{F}(V, V), in addition to being a vector space over F, carries a (noncommutative) ring structure, with multiplication given by composition of functions. Such a ring/vector space A is called an F -algebra if scalar multiplication and multiplication obey the axiom:

α(xy) = (αx)y = x(αy)

551, Fall '02                B2d.    Minimal polynomial                p. 7

for all α ∈ F and all x, y ∈ A (it is assumed also that the addition in the ring and the addition in the vector space are the same operation). This axiom does hold in the present case. Another algebra, isomorphic in fact to this one, is the algebra of n × n matrices over F (where n = dim V). Another one is F [X] itself.

Given T ∈ Hom_{F} (V, V), consider the mapping

ϕ_{T} : F [X] → Hom_{F}(V, V), ϕ_{T} (f) = f(T).

This is an F -algebra homomorphism, i.e., preserves both the vector space and ring structures. The kernel of this ring homomorphism, like that of any ring homomorphism, is an ideal, so as F [X] is a PID, ker ϕ_{T} = F [X]μ_{T} for a unique monic polynomial μ_{T}, called by definition the minimal polynomial of T. An equivalent definition would be: μ_{T} ∈ F [X] is the minimal polynomial of T if and only if μ_{T} (T) = 0 and μ_{T} is the unique monic polynomial of least degree satisfying this equation. A further property is that if f ∈ F [X] is any polynomial such that f(T) = 0, then μ_{T} | f (in F [X]). Similarly we can define the minimal polynomial of a square matrix A, or of any element x of an F -algebra (in the last case, there is the possibility that μ_{x} = 0, i.e., f(x) ≠ 0 for every nonzero polynomial f ∈ F [X]).

The minimal polynomial of a linear transformation is of course equal to the minimal polynomial of any matrix representing it. This implies that similar matrices have the same minimal polynomial (a fact which is not too hard to prove directly anyway).

The relationship between minimal polynomial and the invariants of the last sections is:

Proposition. Let A be a square matrix over a field F. The minimal polynomial of A is the invariant factor of A of largest degree.

The identical statement is also valid for linear transformations: V → V, V a finitedimensional vector space over F.

Proof. We give the proof for a linear transformation T : V → V. Let f_{1}, …, f_{r} be the invariant factors of T. Thus by definition each f_{i} is monic and

V_{T} ≅ R/f_{1}R ⊕ ⋯ ⊕ R/f_{r}R and f_{1} | f_{2} | ⋯ | f_{r}.

Also by definition, μ_{T} is the unique monic polynomial of least degree such that μ_{T} V_{T} = 0, this equation being equivalent to μ_{T} (T) = 0. Since each f_{i} divides f_{r}, we have f_{r}R ⊆ f_{i}R for each i and so f_{r}V_{T} = 0. Therefore μ_{T} | f_{r}. Conversely μ_{T} (R/f_{r}R) = 0 and so μ_{T} · 1 ≡ 0 mod f_{r}, that is, f_{r} | μ_{T}. Therefore f_{r} = μ_{T}.

Thus the minimal polynomial of T is analogous to the exponent of a finite abelian group.

The following observations are easy to check:

Proposition.

a) If A is in block diagonal form,

A = [ \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ] ,

then μ_{A} = lcm(μ_{A_{1}}, μ_{A_{2}}).

b) μ_{C(f)} = f if f is monic. (The previous proposition has actually verified this.)

c) μ_{J(α,n)} = (X − α)^{n}. (Again, this follows from the previous proposition, or more messily from a matrix calculation.)

Consequently, for a matrix A, if the Jordan canonical form of A has Jordan blocks J(α_{1}, n_{1}), …, J(α_{r}, n_{r}), then

μ_{A} = lcm((X − α_{1})_{1}^{n}, …, (X − α_{r})_{r}^{n} = ∏_{α}(X − α)^{n_{α}}

where the product is over the distinct diagonal entries α of the Jordan form of A, and for each α, n_{α} is the size of the largest Jordan block appearing for α.

Thus for example, a complex matrix with minimal polynomial (X^{2} − 1)(X^{3} − 1) = (X − 1)^{2}(X + 1)(X − ω)(X − ω^{2}) has Jordan blocks J(−1, 1), J(ω, 1), J(ω^{2}, 1), at least one of each; at least one Jordan block J(1, 2), and possibly some J(1, 1)’s.

This also gives the following characterization of diagonalizability:

Proposition. A square matrix A over an algebraically closed field is diagonalizable, i.e., similar to a diagonal matrix, if and only if its minimal polynomial is square-free, i.e., has no multiple roots (i.e. no repeated linear factors).

The same can be said for a linear transformation T : V → V; we define diagonalizability of such a transformation to mean that there exists a basis v_{1}, …, v_{n} of V and scalars α_{i} ∈ F such that T (v_{i}) = α_{i}v_{i} for each i = 1, …, n.

2e. Characteristic polynomial; eigenvalues

We assume familiarity with the rudimentary theory of determinants; every square matrix A over a commutative ring R has a determinant det A ∈ R, which can be obtained by row or column expansion. Moreover,

a) det(AB) = det(A) det(B),

b) det(I) = 1, and

c) A adj A = adj AA = det AI, where adj A is the transpose of the matrix of cofactors: (adj A)_{ij} = (−1)^{i+j} det M_{ji}, where M_{ji} is obtained from A by crossing out the i-th row and j-th column.

d) The matrix A is invertible if and only if det A ≠ 0. (In one direction compute det(AA^{−1}), using a) and b); in the other direction use c).)

e) det [ \begin{matrix}A & 0 \\ B & C \end{matrix} ] = det A det C.

Using determinants we get another useful invariant of a square matrix or linear transformation (maybe the most useful), the characteristic polynomial.

551, Fall '02      B2e.    Characteristic polynomial; eigenvalues      p. 9

Definition. Let A be an n × n matrix over a commutative ring F. The characteristic polynomial of A is defined to be

χ_{A}(X) = det(XI − A).

The determinant on the right is evaluated in the polynomial ring F [X]. Some authors define it as det(A − XI). This is an inconsequential difference; what matters is just the association class of χ_{A}, and that one sticks consistently to one definition or the other.

Salient properties of characteristic polynomials:

a) If A and A′ are similar then χ_{A} = χ_{A′}.

b) If A is n × n, then χ_{A}(X) = X^{n} − (Tr A)X^{n−1} + ⋯ + (−1)^{n} det A. Here Tr A, the trace of A, is defined as ∑_{i=1}^{n} A_{ii}, the sum of the entries of A on the main diagonal.

c) A is invertible if and only if χ_{A}(0) ≠ 0. (This is immediate since χ_{A}(0) = ± det A.)

d) μ_{A} | χ_{A}. That is, χ_{A}(A) = 0. This is the Cayley-Hamilton Theorem.

Proof. (of Cayley-Hamilton) (proof number 1, 066) Let V = F^{n} be the space of column vectors, with entries in F; let v_{1}, …, v_{n} be the standard basis of V (i.e., the columns of the identity matrix). Make V an F [X]-module in the usual way: for an arbitrary f(X) = a_{n}X^{n} + ⋯ + a_{1}X + a_{0} ∈ F [X], define fv = f(A)v = a_{n}A^{n}v + ⋯ + a_{1}Av + a_{0}v.

Given any matrix C = [c_{ij}(X)] over F [X] (we shall presently take C = XI − A), there is an F -homomorphism ψ_{C} : V → V defined by ψ_{C}(v_{j}) = ∑_{i} c_{ij}(A)v_{i}, j = 1, …, n. (To define homomorphisms out of free modules, precisely what is required is to specify the images of the basis elements.) The following facts follow directly from the definition:

1) If C = fI for some polynomial f ∈ F [X] then ψ_{C} is left multiplication by f(A). (ψ_{C}(v_{j}) = f(A)v_{j}.)

2) ψ_{CD} = ψ_{C} ∘ ψ_{D} for any n × n matrices C and D over F [X].

3) If C = XI − A, then ψ_{C} = 0. (Namely, ψ_{C}(v_{j}) = ∑_{i} c_{ij}(A)v_{i} = Av_{j} − ( ∑_{i} a_{ij}v_{i}) = 0.)

Now let D = adj(XI − A), another n × n matrix over F [X]. Then CD = det(X − IA)I = χ_{A}I, all being matrices over F [X]. By a), ψ_{CD} is left multiplication by χ_{A}(A). But by b) and c), ψ_{CD} = ψ_{C}ψ_{D} = 0. Therefore χ_{A}(A) = 0. QED

Exercise. If A is an n × n matrix over F and A^{m} = 0 for some m, then A^{n} = 0.

Compare the following properties of χ_{A} with those of μ_{A} in the previous section.

Proposition. If

a) If A is in block diagonal form,

A = [ \begin{matrix}A_{1} & 0 \\ 0 & A_{2} \end{matrix} ] ,

then χ_{A} = χ_{A_{1}}χ_{A_{2}}.

b) χ_{C(f)} = f if f is monic.

c) χ_{J(α,n)} = (X − α)^{n}.

Part a) follows from property (e) of determinants, above. For b) and c), you can calculate directly, or be sneaky: we know that these are the minimal polynomials, and since they have the right degree and divide the characteristic polynomials, and every polynomial in sight is monic, they must be the characteristic polynomials. Consequently:

Proposition. χ_{A} is

a) the product of all the invariant factors of A;

b) the product ∏_{α}(X − α)^{m_{a}}, the product being taken over all distinct diagonal entries of the Jordan form of A, with m_{α} being the number of times α appears there (in the case that F is algebraically closed).

Corollary. Let T be a linear transformation V → V. Then μ_{T} = χ_{T} if and only if V_{T} is “cyclic”, i.e., generated as F [X]-module by a single element.

Exercise. Let T : V → V and let W be a T -invariant subspace of V. If there is no w ∈ W such that the images v, T v, T^{2}v, … generate W, then there is no v ∈ V such that the images v, T v, T^{2}v, … generate V. (Hint. Show that the negations of these assertions are that V_{T} and W_{T |W} are cyclic R-modules in the sense just defined.)

Corollary. Any irreducible factor of χ_{A} also divides μ_{A}.

Proof. χ_{A} = f_{1} ⋯ f_{r} with f_{1} | f_{2} | ⋯ | f_{r} = μ_{A}.

Corollary. If A (or T) has at most two invariant factors, then μ_{A} and χ_{A} determine the similarity class of A (or the isomorphism class of V_{T}).

Namely, the invariant factors are μ_{A} and χ_{A}/μ_{A} (the latter existing only if μ_{A} ≠ χ_{A}).

QED

Exercise. Show that χ_{A} and μ_{A} determine the similarity class of A if A is 3 × 3, but not in general if A is 4 × 4 or larger.

Exercise. If A is a complex 4 × 4 matrix and μ_{A} = (X^{2} + 1)(X − i), then what are the possibilities for χ_{A}? Give an example to show that each possibility occurs.

The equation χ_{A}(X) = 0 is the characteristic equation and its roots are the eigenvalues of A. When each eigenvalue α is counted with its multiplicity as a root of the characteristic equation (i.e., the number of linear factors (X − α) of χ_{A}) there are n eigenvalues in all (in an algebraically closed field containing F). The eigenvalues of T are defined similarly.

551, Fall '02                 B2f.    Bilinear forms                 p. 11

Proposition. α is an eigenvalue of A (or T) if and only if there is a nonzero column vector v such that Av = αv (or a nonzero v ∈ V such that T (v) = αv).

Proof. χ_{A}(α) = 0 if and only if det(αI − A) = 0 if and only if αI − A is not invertible if and only if there exists v such that (αI − A)v = 0. QED

Such vectors are called eigenvectors; for a fixed α, the eigenvectors together with 0 form a subspace (“eigenspace”).

Exercise. If F is algebraically closed, then the multiplicity of an eigenvalue α is at least the dimension of the eigenspace corresponding to α. The multiplicity of α is exactly equal to the dimension of the generalized eigenspace

V_{α} = ∪_{n=1}^{∞} ker(T − α)^{n}.

A square matrix is nilpotent if and only if its only eigenvalue is 0.

As an example, if A is a 5 × 5 complex matrix which is nilpotent (A^{n} = 0 for some n), then χ_{A} = X^{5}. The largest Jordan block has size m, where μ_{A} = X^{m}. For each of the values m = 5, 4 and 1 there is only one similarity class possible for A: J(0, 5); block(J(0, 4), J(0, 1)), and 0, respectively. For each of the values m = 3, 2 there are two similarity classes possible: block(J(0, 3), J(0, 2)) and block(J(0, 3), J(0, 1), J(0, 1)) for m = 3, and for m = 2 there are either one or two blocks of size 2, and the rest are of size 1.

Exercise. The number of isomorphism classes of abelian groups of order 1024 is the same as the number of similarity classes of 10 × 10 complex matrices. Please do not calculate this number!

Exercise. If A is a nilpotent matrix, then Tr A^{n} = 0 for all n ≥ 0. Show that the converse is false. (Look around: it’s true over some fields!)

Exercise. If T : V → V and W is a T -invariant subspace of V, then μ_{T |W} | μ_{T} and χ_{T |W} | χ_{T}.

Exercise. If A and B are diagonalizable matrices and AB = BA, then they are “simultaneously” diagonalizable, that is, there exists D such that both DAD^{−1} and DBD^{−1} are diagonal.

2f. Bilinear forms

An important function of matrices is that they are the concrete versions not only of linear transformations, but also, in the square case, of bilinear forms. This leads to a rather different equivalence relation on the set of all n × n matrices over a field F. We briefly indicate the flavor here.

Definition. Let V be a finite-dimensional vector space over a field F. A bilinear form on V is a function B : V × V → F such that

\begin{aligned}B(v, α_{1}w_{1} + α_{2}w_{2}) & = α_{1}B(v, w_{1}) + α_{2}B(v, w_{2}) and \\ B(α_{1}v_{1} + α_{2}v_{2}, w) & = α_{1}B(v_{1}, w) + α_{2}B(v_{2}, w) \end{aligned}

for all v, v_{1}, v_{2}, w, w_{1}, w_{2} ∈ V; that is, for each v ∈ V the functions B(·, v) and B(v, ·) are linear functions V → F.

Given a bilinear form B, if we choose a basis C = {v_{1}, …, v_{n}} of V, we obtain the matrix of B with respect to the basis, namely

[B]_{C} = [B(v_{i}, v_{j})]_{i,j=1}^{n} .

This matrix completely determines B, since B(∑_{i} α_{i}v_{i}, ∑_{j} β_{j}v_{j}) = ∑_{i,j} α_{i}β_{j}(v_{i}, v_{j}) by the bilinearity of B. Furthermore, starting with any matrix A, one can define

B(∑_{i} α_{i}v_{i}, ∑_{j} β_{j}v_{j}) = ∑_{i,j} α_{i}β_{j}A_{ij}

and obtain a bilinear form with matrix A. Thus as with linear transformations, the choice of a basis of V determines a bijective correspondence between the set of square matrices of size dim V and the set of bilinear forms on V.

However, if we want “equivalence” of matrices then to mean “same bilinear form, different basis”, we do not come out with similarity but rather the following equivalence relation on matrices:

A ∼ A′ ⟺ ∃ an invertible D such that A′ = D^{T} AD,

the “T” indicating transpose.

For if w_{j} = ∑_{i} d_{ij}v_{i}, j = 1, …, n, then

(w_{i}, w_{j}) = (∑_{k} d_{ki}v_{k}, ∑_{ℓ} d_{ℓj}v_{ℓ}) = ∑_{k,ℓ} d_{ki}(v_{k}, v_{ℓ})d_{ellj}.

The “canonical form” problem for bilinear forms — i.e., the problem of classifying them — is thus completely different from the classification of linear transformations we have developed in the previous sections.

There is some connection, which can be formulated in terms of the dual space V^{∗}.

Definition. If V is a finite-dimensional vector space over F, then V^{∗} = Hom_{F} (V, F).

The elements of Hom_{F} (V, F) are called linear functionals on V. When V is infinitedimensional, V^{∗} can get out of hand and it is normally defined to be a subspace of Hom_{F} (V, F), e.g. consisting in the case of Banach spaces, for example, of all bounded linear functionals.

551, Fall '02             B2g.    Orthogonal complements             p. 13

For finite dimensions, however, the fact that direct sums are coproducts means that

Hom_{F}(V_{1} ⊕ V_{2}, F) ≅_{ϕ} Hom_{F}(V_{1}, F) × Hom_{F}(V_{2}, F)

under the mapping ϕ(f) = (f|V_{1}, f|V_{2}). Moreover, if V is 1-dimensional, and 0 ≠ v_{1} ∈ V, then Hom_{F} (V, F) ≅ F, via f ↦ f(v_{1}). Note however this isomorphism is not “natural” but depends on the choice of v_{1}. The displayed isomorphism is “natural”, however.

Choosing a basis v_{1}, …, v_{n} of V, in general, we therefore get an isomorphism

V^{∗} = Hom(V, F) = Hom(F v_{1}⊕⋯⊕F v_{n}, F) ≅ Hom(F v_{1}, F)×⋯×Hom(F v_{n}, F) ≅ F ⊕⋯⊕F ≅ V.

However the isomorphism we get here depends on the choice of basis. Following the progress of v_{i} from right to left, we see that under our isomorphism v_{i} corresponds to the functional v_{i}^{∗} defined by

v_{i}^{∗}(v_{j}) = δ_{ij} (= 1 or 0 according as i = j or i ≠ j).

The basis v_{1}^{∗}, …, v_{n}^{∗} of V^{∗} is called the dual basis to v_{1}, …, v_{n}. We have

v = ∑_{j} v_{j}^{∗}(v)v_{j}

for each v ∈ V; this is obvious for v = v_{i} and then extends to all v by linearity.

Now a bilinear form B on V determines a linear transformation β_{B} : V → V^{∗} by β_{B}(v) = B(·, v), that is,

β_{B}(v)(w) = B(v, w)

for all v, w ∈ V. Conversely a linear transformation β_{B} : V → V^{∗} determines a bilinear form B by the same equation. Then for a basis C = {v_{1}, …, v_{n}} and its dual basis C^{∗} = {v_{1}^{∗}, …, v_{n}^{∗}}, we have B(v_{i}, v_{j}) = β_{B}(v_{i})(v_{j}), so

β_{B}(v_{i}) = ∑ β_{B}(v_{i})(v_{j})v_{j}^{∗} = ∑ B(v_{i}, v_{j})v_{j}^{∗}.

Thus

[B]_{C} = [β_{B}]_{C^{∗}}^{C}.

Exercise. A bilinear form B is called nondegenerate (on the left) if and only if the only v ∈ V such that B(v, w) = 0 for all w ∈ V is v = 0. Show that B is nonsingular if and only if β_{B} is injective (hence an isomorphism in the finite-dimensional case).

2g. Orthogonal complements

For simplicity let us write (v, w) for B(v, w). In order for orthogonality to be an easy concept, we shall assume from now on that our bilinear forms B satisfy the condition:

if (v, w) = 0, then (w, v) = 0.  2A

This condition is satisfied, for example, if either

a) B is symmetric: (v, w) = (w, v) for all v, w ∈ V, or

b) B is alternating: (v, v) = 0 for all v ∈ V; this implies that (v, w) + (w, v) = (v + w, v + w) − (v, v) − (w, w) = 0 so (v, w) = −(w, v) for all v, w ∈ V, i.e., B is antisymmetric. (Conversely, antisymmetry implies the alternating condition as long as 1 ≠ −1, i.e, as long as the underlying field is not Z_{2} and is not any other field of “characteristic 2”.)

Under the hypothesis (2A), we can then define v ⊥ w, for v, w ∈ V, to mean (v, w) = 0, and have that ⊥ is a symmetric relation. Then for any subset (usually a subspace) W ⊆ V, we define

W^{⊥} = {v ∈ V | v ⊥ w for all w ∈ W }.

The bilinearity of B implies immediately that W^{⊥} is a subspace of V. Notice the following properties:

a) If W_{1} ⊆ W_{2}, then W_{2}^{⊥} ⊆ W_{1}^{⊥};

b) W ⊆ W^{⊥}^{⊥}.

We shall write V = W_{1} ⊥ W_{2} if V = W_{1} ⊕ W_{2} and in addition, W_{1} ⊆ W_{2}^{⊥}.

For the moment assume that B is nondegenerate. This means that the mapping β_{B} : V → V^{∗} described in the previous section is an isomorphism of vector spaces. For any subspace W ⊆ V, the restriction mapping gives a homomorphism

ϕ : V^{∗} → W^{∗},

which is surjective; choosing any complement X to W in V (i.e. writing V = W ⊕ X) we may extend any linear functional W → V to a linear functional on V by prescribing that X go to 0. Let γ be the composite

ϕ ∘ β_{B} : V → V^{∗} → W^{∗},

which is still surjective. The kernel of ϕ ∘ β_{B} is precisely the set of those v ∈ V such that (v, w) = 0 for all w ∈ W, i.e., W^{⊥}. Therefore V/W^{⊥} ≅ W^{∗}, and we have proved:

Proposition. If B is a nondegenerate symmetric or alternating bilinear form on the finitedimensional vector space V, then for any subspace W ⊆ V, we have

dim V = dim W + dim W^{⊥}.

Corollary. If B is a nondegenerate symmetric or alternating bilinear form on the finitedimensional vector space V, then for any subspace W ⊆ V, we have

W = W^{⊥}^{⊥}.

The proposition gives the equality of the dimensions of these subspaces, and we already know that W ⊆ W^{⊥}^{⊥}.

Corollary. Suppose that B is a nondegenerate symmetric or alternating bilinear form on the f.d.v.s. V. If B|W is nondegenerate, then V = W ⊥ W^{⊥}, and W^{⊥} is also nondegenerate.

The first statement holds since the nondegeneracy of B|W means exactly that W ∩W^{⊥} = 0. By the corollary, W ⊥ ∩W^{⊥}^{⊥}, proving the last statement.

Examples.

Ex. A. The usual Euclidean inner product E on R^{n} is a bilinear form; it is symmetric and nondegenerate. It is also positive definite ((v, v) > 0 for all v ≠ 0), which is a stronger condition than nondegeneracy. There exist orthonormal bases C (i.e., [E]_{C} = I). Any positive definite symmetric bilinear form B on R^{n} is equivalent to E, i.e., there exists an orthonormal basis for B. Namely, choose any v ∈ V, v ≠ 0. Then B(v, v) > 0 and we let c = B(v, v)^{−1/2} and w_{1} = cv. Then B(w_{1}, w_{1}) = 1, and in particular B is nondegenerate on Rw_{1}. Let V′ = Rw_{1}^{⊥}. Then V = Rw_{1} ⊥ V′ by the proposition, and our assertion follows by induction.

Ex. B. Let us consider an arbitrary bilinear symmetric form B on V = R^{n}, dropping the positive definiteness and nondegeneracy conditions. First choose any vector-space complement W to V^{⊥} and write

V = V^{⊥} ⊕ W = V^{⊥} ⊥ W;

observe also that W is nondegenerate, because W^{⊥} is orthogonal to both V^{⊥} and W, hence W^{⊥} ⊆ V^{⊥}, whence W^{⊥} ∩W ⊆ V^{⊥} ∩W = 0. This means that to obtain our form B, we first may study the nondegenerate case and then attach an arbitrary number of basis vectors orthogonal to everything.

In the nondegenerate case, we must have (v, v) ≠ 0 for some v. For otherwise,

2(v, w) = (v + w, v + w) − (v, v) − (w, w) = 0

for all v, w ∈ V, by bilinearity and symmetry. Choosing any v with (v, v) ≠ 0 and setting w_{1} = |(v, v)|^{−1/2}v, we have (w_{1}, w_{1}) = ±1. Then V′ = Rw_{1} is nondegenerate, so V = Rw_{1} ⊥ V′ and we may continue. In this way we have proved the existence part of Sylvester’s Theorem:

Theorem. Let B be a symmetric bilinear form on the real finite-dimensional vector space V. Then V has an orthogonal basis consisting of vectors v for which (v, v) = 0 or ±1. Moreover, the number of basis vectors for which (v, v) takes on each of these three values is uniquely determined by B, i.e., independent of the choice of basis.

Proof. It remains to prove the uniqueness statement. Choose a basis C consisting of mutually orthogonal vectors c_{1}, …, c_{m}, d_{1}, …, d_{n} and e_{1}, …, e_{p}, with (c_{i}, c_{i}) = 1, (d_{i}, d_{i}) = 0 and (e_{i}, e_{i}) = −1 for all i. Let V_{C}^{+,0} be the span of the c’s and d’s; then (v, v) ≥ 0 for all v ∈ C^{+,0}. Likewise if V_{C}^{−} is the span of the e′s, then (v, v) < 0 for all 0 ≠ v ∈ V_{C}^{−}. We can make the same construction for any basis as in the theorem. We must prove that the dimensions m, n, p are unique; but they are determined by the four dimensions V_{C}^{±,0} and V_{C}^{±}, so it is enough to prove the equality of these. As V_{C}^{±,0} and V_{C}^{∓} are complementary, it is enough to show the uniqueness of dim V_{C}^{±,0}. Let D be another such basis. Then V_{C}^{+,0} ∩ V_{D}^{−} = 0, since its nonzero vectors v satisfy (v, v) ≥ 0 and (v, v) < 0 simultaneously. Therefore dim V_{C}^{+,0} ≤ dim V − dim V_{D}^{−} = dim V_{D}^{+,0}. By symmetry, we have equality. QED

Ex. C. Suppose that B is a nondegenerate alternating form (here F can be any field). Let v ∈ V, v ≠ 0. Since B is nondegenerate, (v, v′) ≠ 0 for some v′ ∈ V. Replacing v′ by a scalar multiple we may arrange that (v, v′) = 1. Such a pair v, v′ is called a hyperbolic pair, and satisfies

(v, v) = (v′, v′) = 0, (v, v′) = 1, (v′, v) = −1.

Obviously v′ ∉ F v. Let W = F v + F v′. Then dim W = 2 (W is a “hyperbolic plane”), and the matrix of the form B with respect to {v, v′} is

[ \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ] .

In particular B|W is nondegenerate, so B|W^{⊥} is too by a corollary in the previous section. By induction we have proved:

Theorem. Let B be a nondegenerate alternating bilinear form on a f.d.v.s. V. Then V is the orthogonal sum of hyperbolic planes for B. In particular, dim V is even.

Corollary. Any two nondegenerate alternating bilinear forms on a f.d.v.s. V are equivalent.

2h. Sesquilinear and Hermitian Forms.

The results of this section are specific to vector spaces over C. For them, since z\overline{z} ≥ 0 for all z ∈ C, it is more favorable to consider “hermitian” forms. These are an example of sesquilinear forms (“\frac{11}{2}-linear”).

A mapping ϕ : V → W of complex vector spaces is called conjugate-linear if and only if ϕ(v + v′) = ϕ(v) + ϕ(v′) (ϕ is “bi-additive”) and ϕ(αv) = \overline{α}ϕ(v) for all v, v′ ∈ v and α ∈ C. Here \overline{α} is the complex conjugate of α.

Exercise. For a conjugate-linear mapping ϕ : V → W, ker ϕ and im ϕ are subspaces and dim V = dim ker ϕ + dim im ϕ.

Definition. A hermitian form on a complex vector space V is a form B : V × V → V which is linear in the first argument (for each fixed value of the second) and conjugate linear in the second (for each fixed value of the first), and which is hermitian-symmetric:

(w, v) = \overline{(v, w)}

for all v, w ∈ V.

Of these three conditions, either of the first two follows from the others. Without the hermitian-symmetry, the form still could be called sesquilinear.

For such a form, (v, w) = 0 ⟺ (w, v) = 0, so our general results on orthogonality from the previous section still hold. In particular, the definition of nondegeneracy is the same, and if V is nondegenerate, then V = W ⊥ W^{⊥} for any nondegenerate subspace W ⊆ V.

We have (v, v) = \overline{(v, v)} ∈ R by the hermitian-symmetry, for all v ∈ V. We again call the form positive-definite if and only if

(v, v) > 0 for all 0 ≠ v ∈ V.

If a basis C = {v_{1}, …, v_{n}} of V is given, then a hermitian form on V is determined by its values (v_{i}, v_{j}):

(∑_{i} α_{i}v_{i}, ∑_{j} β_{j}v_{j}) = ∑_{i,j} α_{i}\overline{β}_{j}(v_{i}, v_{j}).

So we can again speak of the matrix [B]_{C} of B with respect to C.

Proposition. The matrix of a hermitian form is hermitian-symmetric.

p. 2       B2h.    Sesquilinear and Hermitian Forms.         551, Fall '02

Proof. By definition A is hermitian-symmetric if and only if A^{T} = \overline{A}, the matrix obtained by replacing each entry by its complex conjugate. In our situation this is just (v_{i}, v_{j}) = \overline{(v_{j}, v_{i})}. QED Exercise. If V is a f.d.v.s over C with a positive-definite hermitian form, then V has an orthonormal basis, i.e., its matrix with respect to some basis is I. Any two positive-definite hermitian forms on V are equivalent.

A hermitian form B gives rise to a conjugate-linear mapping β_{B} : V → V^{∗}, by β_{B}(v)(w) = (w, v). For each fixed v, this expression is linear in w, so ϕ is defined. The axioms imply that β_{B} is conjugate-linear. The form is nondegenerate if and only if this mapping is injective (equivalently, surjective; see the exercise above).

Notice that if V_{0} is a real vector space with a symmetric form, with matrix A (which is then symmetric) relative to a basis C = {v_{1}, …, v_{n}}, then we may form a complex vector space V with the same basis, containing V_{0} as an R-subspace, and extend the given form on V_{0} to a hermitian form on V, defined with respect to C by the same matrix (which is hermitian-symmetric, being real symmetric).

2i. The Spectral Theorem

We have earlier derived an algebraic necessary and sufficient condition for a matrix, or a linear transformation V → V, to be diagonalizable: its minimal polynomial is squarefree. When V comes equipped with a bilinear form (i.e., some geometry), there may be geometric sufficient conditions for diagonalizability. We give an important one here: the spectral theorem.

Suppose then that T : V → V is a linear transformation, and V is a complex vector space with a nondegenerate hermitian form B. The mapping β_{B} is then a bijection: V → V^{∗}. We use β_{B} to define the “adjoint” T^{∗} of T.

Lemma. There exists a unique linear transformation T^{∗} : V → V such that

(T v, w) = (v, T^{∗}w)

for all v, w ∈ V.

Proof. Our mapping β_{B}(v)(w) = (w, v) is a bijection β_{B} : V → V^{∗}. We therefore want β_{B}(T^{∗}w)(v) = (T v, w). For each w ∈ V, the mapping γ_{w} : v ↦ (T v, w) is linear in v, so lies in V^{∗}. Thus we want β_{B}(T^{∗}(w)) = γ_{w} for each w. As β_{B} is bijective, there is a unique function T^{∗} satisfying the desired equation. Then (v, T^{∗}(αw)) = (T v, αw) = \overline{α}(T v, w) = \overline{α}(v, T^{∗}(w)) = (v, αT^{∗}(w)) for all v, w ∈ V. The uniqueness of T^{∗} shows T^{∗}(αw) = αT^{∗}(w). The additivity of T^{∗} follows similarly. QED

It is easy to check that (T_{1} + T_{2})^{∗} = T_{1}^{∗} + T + 2^{∗}, (αT)^{∗} = \overline{α}T^{∗} and T^{∗∗} = T. E.g., (v, T^{∗∗}w) = (T^{∗}v, w) = \overline{(w, T^{∗}v)} = \overline{(}T w, v) = (v, T w) for all v, w ∈ V, so T^{∗∗}w = T w for all w.

The following elementary fact is important:

Proposition. ker T = (im T^{∗})^{⊥} and (ker T)^{⊥} = im T^{∗}.

Proof. Since dim V is finite and the form is nondegenerate, the second statement follows from the first by taking orthogonal complements. For any v ∈ V, the following statements

551, Fall '02               B2i.    The Spectral Theorem               p. 3

are equivalent: v ∈ ker T; T v = 0; (T v, w) = 0 for all w ∈ W; (v, T^{∗}w) = 0 for all w ∈ W; v ∈ (im T^{∗})^{⊥}.

Definition. In the above situation, T is called

a) self-adjoint if and only if T^{∗} = T;

b) unitary if and only if T^{∗} = T^{−1};

c) normal if and only if T T^{∗} = T^{∗}T.

Thus self-adjoint and unitary transformations are normal.

Exercise. With respect to an orthonormal basis C of V, the matrices of T and T^{∗} satisfy

[T^{∗}]_{C}^{C} = \overline{[T]_{C}^{C}^{T}} .

Moreover, if A = [T]_{C}^{C}, then T is self-adjoint, unitary or normal according as A^{T} = \overline{A}, A^{T} \overline{A} = I, or A^{T} \overline{A} = \overline{A}A^{T}.

Spectral Theorem. Let V be a finite-dimensional complex vector space with a nondegenerate hermitian bilinear form. Let T : V → V be a linear transformation. Then T is normal if and only if V has an orthonormal basis consisting of eigenvectors of T. Moreover, if T is self-adjoint, all its eigenvalues are real; if T is unitary, then all its eigenvalues are on the unit circle.

We prove this as the culmination of several simple remarks.

1. If W is a subspace of V and T (W) ⊆ W, then T^{∗}(W^{⊥}) ⊆ W^{⊥}.

Proof. For any u ∈ W^{⊥} and v ∈ W, (v, T^{∗}(u)) = (T v, u) = 0.

2. If W is a subspace of V, and both W and W^{⊥} are T -invariant, then W is T^{∗}-invariant and (T |W)^{∗} = T^{∗}|W.

Proof. By 1, W is T^{∗}-invariant. For any u, v ∈ W, (u, (T |W)^{∗}v) = ((T |W)u, v) = (T u, v) = (u, T^{∗}v). Hence by the uniqueness of (T |W)^{∗}, T^{∗}v = (T |W)^{∗}v.

3. If T is normal and W is a subspace of V such that both W and W^{⊥} are T -invariant, then T |W is normal.

Proof. Use 2 and the definition of normality.

4. If T is normal then so is T − λ id_{V} for any λ ∈ C.

Proof. (T − λ id_{V})^{∗} = T^{∗} − \overline{λ} id_{V} commutes with T and hence with T − λ id_{V}.

5. If T U = UT then T leaves ker U and im U invariant.

Proof. T (Uv) = U(T v) for all v ∈ V. Let v range over ker U to get T (ker U) ⊆ ker U. Let v range over V to get T (im U) ⊆ im U.

Now we prove the “hard” direction of the spectral theorem. Suppose that T is normal. Using the fundamental theorem of algebra, take an eigenvalue λ of T. By the proposition,

V = ker(T − λ id_{V}) ⊥ im((T − λ id_{V})^{∗}) = ker(T − λ id_{V}) ⊥ im(T^{∗} − \overline{λ} id_{V}).

Since T is normal, it commutes with T^{∗} − \overline{λ} id_{V}, so both the displayed subspaces are T -invariant by 5. By 3, the restriction of T to each is normal. By choice the first is nontrivial, so by induction on dim V the second has an orthonormal basis of eigenvectors of T. Obviously ker(T − λ id_{V}) does too, and assembling these two bases we obtain the desired orthonormal basis of V.

If T is self-adjoint, then for any eigenvector v and corresponding eigenvalue λ,

λ(v, v) = (T v, v) = (v, T v) = \overline{(T v, v)}

is real. Also (v, v) = \overline{(v, v)} is real, so λ is as well. Likewise if T is unitary, then T is invertible by definition, and the equation T v = λv implies that T^{−1}v = λ^{−1}v. Now λ(v, v) = (T v, v) = (v, T^{−1}v) = \overline{λ}^{−1}(v, v), so λ = \overline{λ}^{−1} is on the unit circle.

Conversely, if there exists an orthonormal basis v_{1}, …, v_{n} of eigenvectors of T, say T v_{i} = λ_{i}v_{i}, then we put Uv_{i} = \overline{λ}_{i}v_{i} for each i and extend U to a linear transformation on V; then

(T v_{i}, v_{j}) = δ_{ij}λ_{i} = δ_{ij}λ_{j} = (v_{i}, Uv_{j})

for all i and j, and so U = T^{∗}. But obviously T U = UT since the matrices of both T and U with respect to C are diagonal. Hence T is normal. QED

In matrix terms, if we fix an orthonormal basis C and set A = [T]_{C}^{C}, then T is self-adjoint (unitary, normal) iff A^{T} = \overline{A} (A^{T} = \overline{A}^{−1}, A^{T} \overline{A} = \overline{A}A^{T}). The “transition” matrix from C to another basis is easily seen to be unitary if and only if the other basis is orthonormal. Thus in matrix terms we have:

Matrix Spectral Theorem. Let A be a square complex matrix. Then A is normal if and only if there is a unitary matrix U such that UAU^{−1} is diagonal. Moreover, if A is self-adjoint (resp. unitary), then the eigenvalues of A are real (resp. on the unit circle).

Exercise. Adapt the theorem and proof to the case of a real vector space with a positive definite symmetric bilinear (real-valued) form, and a self-adjoint linear transformation T. (Hint. The matrix of T being symmetric, all eigenvalues of T are known a priori to be real, by the theorem already proved.)