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

$\underset{\alpha \in A}{⨁}{M}_{\alpha }$

is free.

Proof:

Let bases $\{e_{\beta }{\}}_{\beta \in B_{\alpha }}$ of the $M_{\alpha }$ be given. We claim that

$\left\{(0,\dots ,0,\stackrel{\alpha \text{-th place}}{\overbrace{e_{{\beta }_{\alpha }}}},0,\dots ,0)\right|\alpha \in A,{\beta }_{\alpha }\in B_{\alpha }\}$

is a basis of

$M:=\underset{\alpha \in A}{⨁}{M}_{\alpha }$.

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 $m_{\alpha }$.$◻$

Template:DABE 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

$g(e_{{\alpha }^{\prime }}\otimes f_{{\beta }^{\prime }})=f(e_{{\alpha }^{\prime }},f_{{\beta }^{\prime }})=[\alpha ={\alpha }^{\prime }\wedge \beta ={\beta }^{\prime }]$,

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

$g(k+r{k}^{\prime })=\sum _{i=1}^l{r}_i{m}_{j_i}+r\sum _{i^{\prime }=1}^{l^{\prime }}{r_i}^{\prime }m{\text{'}}_{{j_i}^{\prime }}$,

where the right hand side is the sum of the linear combinations coinciding with $g(k)$ and $g(k^{\prime })$ respectively, which is why $h(k+r{k}^{\prime })=h(k)+rh(k^{\prime })$. 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

$L:=\underset{m\in M}{⨁}R$

(this obviously is a free module) and the function

$f:L\to M,(0,\dots ,0,\stackrel{m\text{-th place}}{\overbrace{r}},0,\dots ,0)\mapsto rm$.

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

;

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

We claim that the map

$\phi :M\oplus \ker f\to L,\phi (m,k):=h(m)+k$

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 $\phi ((rm,0))=(0,\dots ,0,\stackrel{m\text{-th place}}{\overbrace{r}},0,\dots ,0)+k$, where $k\in \ker f$, which is why

$\phi ((rm,-k))=(0,\dots ,0,\stackrel{m\text{-th place}}{\overbrace{r}},0,\dots ,0)=(0,\dots ,0,\stackrel{m\text{-th place}}{\overbrace{r}},0,\dots ,0)$

(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

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

Definition 11.8:

An exact sequence of modules

$0\to K\to N\to M\to 0$

is called split exact iff we can augment it by three isomorphisms such that

commutes.

Theorem 11.9:

A module $M$ is projective iff every exact sequence

$0\to K\to N\to M\to 0$

is split exact.

Proof:

$\Rightarrow$: The morphism $N\to M$ 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

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

$h(0,m)-n\in \ker f=\mathrm{im}h$

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

and have proven what we wanted.

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

We set

$K:=\underset{m\in M}{⨁}R$, $f:K\to M$

where $f$ is defined as in the proof of theorem 11.7 $\Rightarrow$. We obtain an exact sequence

$0\to \ker f\stackrel{\iota }{\hookrightarrow }K\stackrel{f}{\to }M\to 0$

which by assumption splits as

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 $\underset{\alpha \in A}{⨁}{M}_{\alpha }$ 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 $\underset{\alpha \in A}{⨁}{M}_{\alpha }$ is the direct summand of free modules.

On the other hand, if $\underset{\alpha \in A}{⨁}{M}_{\alpha }$ 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 $rn=r/1n$), 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

$M\cong \underset{\alpha \in A}{⨁}R$

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

$0\to A\to B\to C\to 0$,

to show that the sequence

$0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$

is exact, all we have to do is to prove that

$0\to A\otimes M\to B\otimes M\to C\otimes M\to 0$

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

$0\to A\to B\to C\to 0$

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

$T_S(M):=\{m\in M|\mathrm{\exists }s\in S:sm=0\}$.

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 ${\pi }_S:M\to S^{-1}M$.