Free Modules

All modules are over a fixed ring A.

We already mentioned finite free modules earlier. Here we will consider general free modules.

Definition.

Let $I$ be any set. The free A-module on I is a direct sum of copies of A, indexed by $i\in I$ :

$A^{\oplus I} := \oplus_{i\in I} A.$

More generally, an A-module M is free if $M \cong A^{\oplus I}$ for some I. Note that M is finite free if we can pick I to be finite.

Exercise A

Prove that if A is a non-trivial ring and $A^n \cong A^m$ , then $n = m$ . [Hint: pick a maximal ideal of A; reduce this to a statement in linear algebra.]

For each $i\in I$ , let

$e_i := ( \ldots, 0, 0, \overbrace{1_A}^{i\text{-th coord}}, 0, 0, \ldots)\in A^{\oplus I}$

which takes zeros at all entries except i, where it takes $1_A$ .

Why do we use the direct sum and not the product? Well, the direct sum allows us to have the following nice properties. Following the case of linear algebra, we define:

Definition.

Let M be an A-module. A collection of elements $(m_i)_{i\in I}$ of M is said to be linearly independent over A, if for any distinct $i_1, \ldots, i_n \in I$ ,

$a_1, \ldots, a_n \in A, \ a_1 m_{i_1} + a_2 m_{i_2} + \ldots + a_n m_{i_n} = 0 \implies a_1 = a_2 = \ldots = a_n.$

If $(m_i)$ is linearly independent and generates the whole module, we call it a basis of M.

The following result comes at no surprise.

Proposition 1.

The $(e_i)$ form a basis of $A^{\oplus I}$ , called the standard basis.

M is free if and only if it has a basis indexed by I.

Proof

Easy exercise. ♦

The free module also satisfies the following universal property.

Universal Property of Free Module.

Let M be an A-module, and $\nu : I \to M$ be a set of elements in M, indexed by I; we write $m_i := \nu(i) \in M$ . Then there is a unique A-linear map

$\psi : A^{\oplus I} \to M, \quad \psi(e_i) = m_i.$

Note

In other words, we have a bijection of sets:

$\begin{aligned} \mathrm{hom}_{A\text{-}\mathbf{Mod}}(A^{\oplus I}, M) &\stackrel \cong\to \mathrm{hom}_{\mathbf{Set}}(I, M), \\ \psi &\mapsto (i \mapsto \psi(e_i)).\end{aligned}$

Exercise B

Prove the universal property.
Prove that $\psi$ is surjective if and only if the $(m_i)_{i\in I}$ generate M.

This universal property is a special case of adjoint functors in category theory, which we will see later.

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Projective Modules

One can imagine projective modules as a generalization of free modules. Geometrically, if we think of modules over a coordinate ring $k[V]$ as vector bundles over V, then projective modules correspond to locally trivial bundles. To get acquainted with this concept, first we give the abstract definition before going through its implications.

Recall that for any A-module M, the functor $\mathrm{Hom}_A(M, -)$ is left-exact.

Definition.

The module M is said to be projective if $\mathrm{Hom}_A(M, -)$ is exact.

Thus, M is projective if and only if: for any surjective $f:N_1 \to N_2$ , the map $f_* : \mathrm{Hom}_A(M, N_1) \to \mathrm{Hom}_A(M, N_2)$ is surjective. Equivalently, any $g : M\to N_2$ lifts to some $h : M\to N_1$ such that $f\circ h = g$ :

projective_module_cd

Note

We only demand existence, not uniqueness, of h, i.e. this is not a universal property.

Lemma 1.

A free module is projective.

Proof

Without loss of generality let $M = A^{\oplus I}$ and $g : M \to N_2$ be an A-linear map. Let $n_i' = g(e_i) \in N_2$ , where $e_i \in M$ are elements of the standard basis. Since $f$ is surjective, for each $i\in I$ there exists $n_i \in N_1$ such that $f(n_i) = n_i'$ . By the universal property of the free module, there is a unique $h:M\to N_1$ which maps $e_i\mapsto n_i$ for each $i\in I$ . Then $(f\circ h)(e_i) = f(n_i) = n_i' = g(e_i)$ . Hence $f\circ h = g$ . ♦

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Splitting Lemma

Recall that for a submodule $N\subseteq M$ we do not have $M \cong N \oplus (M/N)$ in general. However, suppose we can find another $P\subseteq M$ such that:

$N + P = M, \quad N \cap P = 0,$

then we get an isomorphism $N \oplus P \cong M$ by mapping $(x,y) \mapsto x+y$ . The proof for this is a straight-forward exercise.

Splitting Lemma.

Suppose we have the following short exact sequence of A-modules.

$0 \longrightarrow N \stackrel f\longrightarrow M \stackrel g\longrightarrow P \longrightarrow 0.$

If there is an A-linear map $i : P\to M$ such that $g\circ i = 1_P$ , then $N \oplus P \cong M$ by mapping $(x,y) \mapsto f(x) + i(y)$ .

Note

For any A-modules N and P, we always have the canonical sequence

$0\longrightarrow N \stackrel f\longrightarrow N\oplus P \stackrel g \longrightarrow P \longrightarrow 0, \quad f(n) = (n, 0),\ g(n, p) = p.$

The map $i(p) = (0, p)$ then satisfies $g\circ i = 1_P$ . The splitting lemma effectively states the converse: existence of such an i means the exact sequence is isomorphic to this form.

Proof

Note that since $g\circ i$ is injective so is $i$ . Now we prove the following:

$M = f(N) + i(P), \quad f(N) \cap i(P) = 0.$

For the first claim, let $m\in M$ and $p = g(m) \in P$ . Let $m' = i(p)$ . Note that $g(m - m') = p - gi(p) = 0$ so we have $m-m' = f(n)$ for some $n\in N$ . This gives $m = f(n) + m' = f(n) + i(p)$ .

For the seecond claim, suppose $f(n) = i(p)$ for some $n\in N$ and $p\in P$ . Then $p = g(i(p)) = g(f(n)) = 0$ . ♦

Exercise C

Prove the other splitting lemma: if there is an A-linear map $j : M\to N$ such that $j\circ f = 1_N$ , then $N\oplus P \cong M$ .

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Projective vs Free

The splitting lemma gives us the following classification of projective modules.

Theorem 1.

An A-module M is projective if and only if there exists an A-module N such that $M\oplus N$ is free.

Furthermore, if M is finitely generated and projective, we can pick N such that $M\oplus N$ is finite free.

Proof

Consider the equivalence in the first statement.

(⇒) Let M be projective. Pick any generating subset $S\subseteq M$ (e.g. the whole module) which gives a surjective $f: A^{\oplus S} \to M$ mapping $e_s \mapsto s$ . If M is finitely generated, we pick S to be finite. Since M is projective, the identity map $1_M : M\to M$ lifts to a map $h: M \to A^{\oplus S}$ such that $f\circ h = 1_M$ . By the splitting lemma we have $A^{\oplus S} \cong M \oplus N$ for some N. Note that this also proves the second statement.

(⇐) Suppose $M \oplus N$ is free, and hence projective. It suffices to show: if $M\oplus N$ is projective so is M. But we have a natural isomorphism

$\mathrm{Hom}_A(M\oplus N, P) \cong \mathrm{Hom}_A(M, P) \oplus \mathrm{Hom}_A(N, P)$

of functors in P. Since $M\oplus N$ is projective, the LHS is an exact functor, hence so is the RHS, and it follows that $\mathrm{Hom}_A(M, -)$ is exact. ♦

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Examples and Consequences

At first glance, it is not entirely clear there are projective modules that are not free. Here are two examples.

Example 1

Let $B = A\times A'$ . Then $M = A\times \{0\}$ and $N=\{0\} \times A'$ are B-modules such that $M \oplus N \cong B$ . Hence M and N are projective even though they are clearly not free.

Example 2

A more interesting example occurs for $A = \mathbb Z[\sqrt{-5}]$ , a non-PID. Recall that the prime ideals

$\mathfrak p = (1+\sqrt{-5}, 2), \quad \mathfrak q = (1 - \sqrt{-5}, 2)$

are non-principal. However, they are projective because of the following.

Exercise D

Prove that we have an isomorphism of A-modules:

$A \oplus A \longrightarrow \mathfrak p \oplus \mathfrak q, \quad (x, y) \mapsto ((1+\sqrt{-5})x + 2y, ?\, x + ?\, y).$

Example 3

Let $A = \mathbb C[X, Y]/(Y^2 - X^3 + X)$ and $\mathfrak m = (X, Y)$ be a maximal ideal of A. Then $\mathfrak m$ is a projective module because of the following.

Exercise E

Prove that we have an isomorphism of A-modules

$A \oplus A \longrightarrow \mathfrak m \oplus \mathfrak m, \quad (f, g) \mapsto (X f + Y g, ?\, f + ?\, g).$

Now we discuss some consequences of the above results.

Corollary 1.

If $(M_i)$ is a collection of projective A-modules, then $\oplus_i M_i$ is projective.

Proof

For each i let $N_i$ be an A-module such that $M_i \oplus N_i$ is free. Then $(\oplus_i M_i) \oplus (\oplus_i N_i) \cong \oplus_i (M_i \oplus N_i)$ is also free. Hence $\oplus_i M_i$ is projective. ♦

Corollary 2.

If M is a projective A-module, then $S^{-1}M$ is a projective $S^{-1}A$ -module.

Proof

Indeed, if $M \oplus N \cong A^{\oplus I}$ then

$(S^{-1}M) \oplus (S^{-1}N) \cong S^{-1}(M\oplus N) \cong S^{-1}(A^{\oplus I}) \cong (S^{-1}A)^{\oplus I}$

because localization commutes with taking direct sums. Hence $S^{-1}M$ is also projective over $S^{-1}A$ . ♦

Is projective a local property then? In other words, suppose M is an A-module such that $M_{\mathfrak m}$ is projective over $A_{\mathfrak m}$ for all maximal $\mathfrak m$ , must M be projective? We will answer this in a later article (spoiler: the answer is a conditional ‘yes’).

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Commutative Algebra 27

Free Modules

Projective Modules

Splitting Lemma

Projective vs Free