Commutative Algebra/Torsion-free, flat, projective and free modules

Free modules
The following definitions are straightforward generalisations from linear algebra. We begin by repeating a definition we already saw in chapter 6.

We also have:

Theorem 11.3:

Let $$M_\alpha, \alpha \in A$$ be free modules. Then the direct sum
 * $$\bigoplus_{\alpha \in A} M_\alpha$$

is free.

Proof:

Let bases $$\{e_\beta\}_{\beta \in B_\alpha}$$ of the $$M_\alpha$$ be given. We claim that
 * $$\left\{ \left( 0, \ldots, 0, \overbrace{e_{\beta_\alpha}}^{\alpha\text{-th place}}, 0, \ldots, 0 \right) \big| \alpha \in A, \beta_\alpha \in B_\alpha \right\}$$

is a basis of
 * $$M := \bigoplus_{\alpha \in A} M_\alpha$$.

Indeed, let an arbitrary element $$(m_\alpha)_{\alpha \in A}$$ be given. Then by assumption, each of the $$m_\alpha$$ has a decomposition
 * $$m_\alpha = \sum_{j=1}^{n_\alpha} r_{j, \alpha} e_{\beta_{j, \alpha}}$$

for suitable $$e_{\beta_{j, \alpha}} \in \{e_\beta\}_{\beta \in B_\alpha}$$. By summing this, we get a decomposition of $$(m_\alpha)_{\alpha \in A}$$ in the aforementioned basis. Furthermore, this decomposition must be unique, for otherwise projecting gives a new composition of one of the particular $$m_\alpha$$.

The converse is not true in general!

Theorem 11.4:

Let $$M, N$$ be free $$R$$-modules, with bases $$\{e_\alpha\}_{\alpha \in A}$$ and $$\{f_\beta\}_{\beta \in B}$$ respectively. Then
 * $$M \otimes_R N$$

is a free module, with basis
 * $$\{e_\alpha \otimes f_\beta\}_{(\alpha, \beta) \in A \times B}$$,

where we wrote for short
 * $$e_\alpha \otimes f_\beta := [(e_\alpha, f_\beta)]$$

(note that it is quite customary to use this notation).

Proof:

We first prove that our supposed basis forms a generating system. Clearly, by summation it suffices to show that elements of the form
 * $$m \otimes n$$, $$m \in M, n \in N$$

can be written in terms of the $$e_\alpha \otimes f_\beta$$. Thus, write
 * $$m = \sum_{j=1}^\mu r_j e_{\alpha_j}$$ and $$n = \sum_{i=1}^\nu s_i b_{\beta_i}$$,

and obtain by the rules of computing within the tensor product, that
 * $$m \otimes n = \sum_{j=1}^\mu \sum_{i=1}^\nu r_j s_i e_{\alpha_j} \otimes b_{\beta_i}$$.

On the other hand, if
 * $$0 = \sum_{\alpha \in A, \beta \in B} t_{\alpha, \beta} e_\alpha \otimes f_\beta$$

is a linear combination (i.e. all but finitely many summands are zero), then all the $$t_{\alpha, \beta}$$ must be zero. The argument is this: Fix $$\alpha, \beta$$ and define a bilinear function
 * $$f: M \times N \to R, (m,n) \mapsto r_\alpha s_\beta$$,

where $$r_\alpha$$, $$s_\beta$$ are the coefficients of $$e_\alpha$$, $$f_\beta$$ in the decomposition of $$m$$ and $$n$$ respectively. According to the universal property of the tensor product, we obtain a linear map
 * $$g: M \otimes N \to R$$ with $$g \circ \pi = f$$,

where $$\pi: M \times N \to M \otimes N$$ is the canonical projection on the quotient space. We have the equations
 * $$g(e_{\alpha'} \otimes f_{\beta'}) = f(e_{\alpha'}, f_{\beta'}) = [\alpha = \alpha' \wedge \beta = \beta']$$,

and inserting the given linear combination into this map therefore yields the desired result.

Projective modules
The following is a generalisation of free modules:

Theorem 11.6:

Every free module is projective.

Proof:

Pick a basis $$\{m_j\}_{j \in J}$$ of $$M$$, let $$f: N \twoheadrightarrow M$$ be surjective and let $$g: K \to M$$ be some morphism. For each $$m_j$$ pick $$n_j \in N$$ with $$f(n_j) = m_j$$. Define
 * $$h:K \to N, h(k) = \sum_{i=1}^l r_i n_{j_i}$$ where $$g(k) = \sum_{i=1}^l r_i m_{j_i}$$.

This is well-defined since the linear combination describing $$g(k)$$ is unique. Furthermore, it is linear, since we have
 * $$g(k + r k') = \sum_{i=1}^l r_i m_{j_i} + r \sum_{i'=1}^{l'} r_i' m'_{j_i'}$$,

where the right hand side is the sum of the linear combinations coinciding with $$g(k)$$ and $$g(k')$$ respectively, which is why $$h(k + rk') = h(k) + r h(k')$$. By linearity of $$f$$ and definition of the $$n_j$$, it has the desired property.

There are a couple equivalent definitions of projective modules.

Theorem 11.7:

A module $$M$$ is projective if and only if there exists a module $$N$$ such that $$K := M \oplus N$$ is free.

Proof:

$$\Rightarrow$$: Define the module
 * $$L := \bigoplus_{m \in M} R$$

(this obviously is a free module) and the function
 * $$f: L \to M, (0, \ldots, 0, \overbrace{r}^{m\text{-th place}}, 0, \ldots, 0) \mapsto rm$$.

$$f$$ is a surjective morphism, whence we obtain a commutative diagram


 * [[File:Diagram used in the proof of the free summand characterisation of projective modules.svg|130px]];

that is, $$f \circ h = \operatorname{Id}_M$$.

We claim that the map
 * $$\varphi: M \oplus \ker f \to L, \varphi(m,k) := h(m) + k$$

is an isomorphism. Indeed, if $$h(m) + k = 0$$, then $$f(h(m) + k) = f(h(m)) = m = 0$$ and thus also $$k=0$$ (injectivity) and further $$\varphi((rm, 0)) = (0, \ldots, 0, \overbrace{r}^{m\text{-th place}}, 0, \ldots, 0) + k$$, where $$k \in \ker f$$, which is why
 * $$\varphi((rm, -k)) = (0, \ldots, 0, \overbrace{r}^{m\text{-th place}}, 0, \ldots, 0) = (0, \ldots, 0, \overbrace{r}^{m\text{-th place}}, 0, \ldots, 0)$$

(surjectivity).

$$\Leftarrow$$: Assume $$M \oplus L$$ is a free module. Assume $$f: N \twoheadrightarrow M$$ is a surjective morphism, and let $$g: K \to M$$ be any morphism. We extend $$g$$ to $$\tilde g: K \to M \oplus L$$ via
 * $$\tilde g(k) := (g(k), 0)$$.

This is still linear as the composition of the linear map $$g$$ and the linear inclusion $$M \hookrightarrow M \oplus L$$. Now $$M \oplus L$$ is projective since it's free. Hence, we get a commutative diagram


 * Projective-proof2.svg

where $$\tilde h$$ satisfies $$(f \times \operatorname{Id}_L) \circ \tilde h = \tilde g$$. Projecting $$\tilde h$$ to $$N$$ gives the desired diagram for $$M$$.

Definition 11.8:

An exact sequence of modules
 * $$0 \rightarrow K \rightarrow N \rightarrow M \rightarrow 0$$

is called split exact iff we can augment it by three isomorphisms such that
 * Split_exact_sequence.svg

commutes.

Theorem 11.9:

A module $$M$$ is projective iff every exact sequence
 * $$0 \rightarrow K \rightarrow N \rightarrow M \rightarrow 0$$

is split exact.

Proof:

$$\Rightarrow$$: The morphism $$N \rightarrow M$$ is surjective, and thus every other morphism with codomain $$M$$ lifts to $$N$$. In particular, so does the projection $$\pi: K \oplus M \to M$$. Thus, we obtain a commutative diagram


 * Projective-proof3.svg

where we don't know yet whether $$h$$ is an isomorphism, but we can use $$h$$ to define the function
 * $$\tilde h: K \otimes M \to N, \tilde h(k, m) := g(k) + h(0, m)$$,

which is an isomorphism due to injectivity:

Let $$\tilde h(k,m) = 0$$, that is $$h(0,m) + g(k) = 0$$. Then first
 * $$m = f(h(0,m)) = f(h(0,m) + g(k)) = f(0) = 0$$

and therefore second
 * $$g(k) = h(0,m) + g(k) = 0 \Rightarrow k = 0$$.

And surjectivity:

Let $$n \in N$$. Set $$m := f(n)$$. Then
 * $$h(0,m) - n \in \ker f = \operatorname{im} h$$

and hence $$g(k) = h(0,m) - n$$ for a suitable $$k \in K$$, thus
 * $$n = \tilde h(-k, m)$$.

We thus obtain the commutative diagram


 * Projective-proof4.svg

and have proven what we wanted.

$$\Leftarrow$$: We prove that $$M \oplus N$$ is free for a suitable $$N$$.

We set
 * $$K := \bigoplus_{m \in M} R$$, $$f: K \to M$$

where $$f$$ is defined as in the proof of theorem 11.7 $$\Rightarrow$$. We obtain an exact sequence
 * $$0 \rightarrow \ker f \overset{\iota}{\hookrightarrow} K \overset{f}{\rightarrow} M \rightarrow 0$$

which by assumption splits as


 * Projective-proof5.svg

which is why $$\ker f \oplus M$$ is isomorphic to the free module $$K$$ and hence itself free.

Theorem 11.10:

Let $$M$$ and $$N$$ be projective $$R$$-modules. Then $$M \otimes N$$ is projective.

Proof:

We choose $$L, K$$ $$R$$-modules such that $$M \oplus L$$ and $$N \oplus K$$ are free. Since the tensor product of free modules is free, $$(M \oplus L) \otimes (N \oplus K)$$ is free. But
 * $$(M \oplus L) \otimes (N \oplus K) \cong (M \otimes N) \oplus (M \otimes K) \oplus (L \otimes N) \oplus (L \otimes K)$$,

and thus $$M \otimes N$$ occurs as the summand of a free module and is thus projective.

Theorem 11.11:

Let $$M_\alpha, \alpha \in A$$ be $$R$$-modules. Then $$\bigoplus_{\alpha \in A} M_\alpha$$ is projective if and only if each $$M_\alpha$$ is projective.

Proof:

Let first each of the $$M_\alpha$$ be projective. Then each of the $$M_\alpha$$ occurs as the direct summand of a free module, and summing all these free modules proves that $$\bigoplus_{\alpha \in A} M_\alpha$$ is the direct summand of free modules.

On the other hand, if $$\bigoplus_{\alpha \in A} M_\alpha$$ is the summand of a free module, then so are all the $$M_\alpha$$s.

Flat modules
The following is a generalisation of projective modules:

The morphisms in the right sequence induced by any morphism $$f$$ are given by the bilinear map
 * $$(x, m) \mapsto f(x) \otimes m$$.

Theorem 11.13:

The module $$S^{-1}R$$ is a flat $$R$$-module.

Proof: This follows from theorems 9.10 and 10.?.

Theorem 11.14:

Flatness is a local property.

Proof: Exactness is a local property. Furthermore, for any multiplicatively closed $$S \subseteq R$$
 * $$S^{-1} (M \otimes_R N) \cong S^{-1}M \otimes_{S^{-1}R} S^{-1}N$$

by theorem 9.11. Since every $$S^{-1}R$$-module is the localisation of an $$R$$-module (for instance itself as an $$R$$-module via $$rn = r/1 n$$), the theorem follows.

Theorem 11.15:

A projective module is flat.

Proof:

We first prove that every free module is flat. This will enable us to prove that every projective module is flat.

Indeed, if $$M$$ is a free module and $$\{e_\alpha\}, \alpha \in A$$ a basis of $$M$$, we have
 * $$M \cong \bigoplus_{\alpha \in A} R$$

via
 * $$\sum_{\alpha \in A} r_\alpha e_\alpha \mapsto (r_\alpha)_{\alpha \in A}$$,

where all but finitely many of the summands on the left are nonzero. Hence, by distributivity of direct sum over tensor product, if we are given any exact sequence
 * $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$,

to show that the sequence
 * $$0 \rightarrow A \otimes M \rightarrow B \otimes M \rightarrow C \otimes M \rightarrow 0$$

is exact, all we have to do is to prove that
 * $$0 \rightarrow A \otimes M \rightarrow B \otimes M \rightarrow C \otimes M \rightarrow 0$$

is exact, since we may then augment the latter sequence by suitable isomorphisms

Theorem 11.16:

direct sum flat iff all summands are

Theorem 11.17:

If $$M, N$$ are flat $$R$$-modules, then $$M \otimes_R N$$ is as well.

Proof:

Let
 * $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$

be an exact sequence of modules.

Torsion-free modules
The following is a generalisation of flat modules:

Lemma 11.19:

The torsion of a module is a submodule of that module.

Proof:

Let $$m, n \in T(M)$$, $$r \in R$$. Obviously $$m - n \in T(M)$$ (just multiply the two annihilating elements together), and further $$s(rm) = r(sm) = 0$$ if $$sm = 0$$ (we used commutativity here).

We may now define torsion-free modules. They are exactly what you think they are.

Theorem 11.21:

A flat module is torsion-free.

To get a feeling for the theory, we define $$S$$-torsion for a multiplicatively closed subset $$S \subseteq R$$.

Definition 11.22:

Let $$S \subseteq R$$ be a multiplicatively closed subset of a ring $$R$$, and let $$M$$ be an $$R$$-module. Then the $$S$$-torsion of $$M$$ is defined to be
 * $$T_S(M) := \{m \in M | \exists s \in S: sm = 0\}$$.

Theorem 11.23:

Let $$S \subseteq R$$ be a multiplicatively closed subset of a ring $$R$$, and let $$M$$ be an $$R$$-module. Then the $$S$$-torsion of $$M$$ is precisely the kernel of the canonical map $$\pi_S: M \to S^{-1}M$$.