# 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.

Definition 6.1 (generators of modules):

Let ${\displaystyle M}$ be a module over the ring ${\displaystyle R}$. A generating set of ${\displaystyle M}$ is a subset ${\displaystyle \{m_{j}\}_{j\in J}\subseteq M}$ such that

${\displaystyle \forall n\in M:\exists j_{1},\ldots ,j_{k}\in J,r_{1},\ldots ,r_{k}\in R:n=\sum _{l=1}^{k}r_{l}m_{j_{l}}}$.

We also have:

Definition 11.1:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module. A subset ${\displaystyle \{m_{j}\}_{j\in J}}$ of ${\displaystyle M}$ is called linearly independent if and only if, whenever ${\displaystyle j_{1},\ldots ,j_{n}\in J}$, we have

${\displaystyle r_{1}m_{j_{1}}+\cdots +r_{n}m_{j_{n}}=0\Rightarrow r_{1},\ldots ,r_{n}=0}$.

Definition 11.2:

A free ${\displaystyle R}$-module is a module ${\displaystyle M}$ over ${\displaystyle R}$ where there exists a basis, that is, a subset ${\displaystyle \{m_{j}\}_{j\in J}}$ of ${\displaystyle M}$ that is a linearly independent generating set.

Theorem 11.3:

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

${\displaystyle \bigoplus _{\alpha \in A}M_{\alpha }}$

is free.

Proof:

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

${\displaystyle \left\{\left(0,\ldots ,0,\overbrace {e_{\beta _{\alpha }}} ^{\alpha {\text{-th place}}},0,\ldots ,0\right){\big |}\alpha \in A,\beta _{\alpha }\in B_{\alpha }\right\}}$

is a basis of

${\displaystyle M:=\bigoplus _{\alpha \in A}M_{\alpha }}$.

Indeed, let an arbitrary element ${\displaystyle (m_{\alpha })_{\alpha \in A}}$ be given. Then by assumption, each of the ${\displaystyle m_{\alpha }}$ has a decomposition

${\displaystyle m_{\alpha }=\sum _{j=1}^{n_{\alpha }}r_{j,\alpha }e_{\beta _{j,\alpha }}}$

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

The converse is not true in general!

Theorem 11.4:

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

${\displaystyle M\otimes _{R}N}$

is a free module, with basis

${\displaystyle \{e_{\alpha }\otimes f_{\beta }\}_{(\alpha ,\beta )\in A\times B}}$,

where we wrote for short

${\displaystyle e_{\alpha }\otimes f_{\beta }:=[(e_{\alpha },f_{\beta })]}$

(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

${\displaystyle m\otimes n}$, ${\displaystyle m\in M,n\in N}$

can be written in terms of the ${\displaystyle e_{\alpha }\otimes f_{\beta }}$. Thus, write

${\displaystyle m=\sum _{j=1}^{\mu }r_{j}e_{\alpha _{j}}}$ and ${\displaystyle n=\sum _{i=1}^{\nu }s_{i}b_{\beta _{i}}}$,

and obtain by the rules of computing within the tensor product, that

${\displaystyle m\otimes n=\sum _{j=1}^{\mu }\sum _{i=1}^{\nu }r_{j}s_{i}e_{\alpha _{j}}\otimes b_{\beta _{i}}}$.

On the other hand, if

${\displaystyle 0=\sum _{\alpha \in A,\beta \in B}t_{\alpha ,\beta }e_{\alpha }\otimes f_{\beta }}$

is a linear combination (i.e. all but finitely many summands are zero), then all the ${\displaystyle t_{\alpha ,\beta }}$ must be zero. The argument is this: Fix ${\displaystyle \alpha ,\beta }$ and define a bilinear function

${\displaystyle f:M\times N\to R,(m,n)\mapsto r_{\alpha }s_{\beta }}$,

where ${\displaystyle r_{\alpha }}$, ${\displaystyle s_{\beta }}$ are the coefficients of ${\displaystyle e_{\alpha }}$, ${\displaystyle f_{\beta }}$ in the decomposition of ${\displaystyle m}$ and ${\displaystyle n}$ respectively. According to the universal property of the tensor product, we obtain a linear map

${\displaystyle g:M\otimes N\to R}$ with ${\displaystyle g\circ \pi =f}$,

where ${\displaystyle \pi :M\times N\to M\otimes N}$ is the canonical projection on the quotient space. We have the equations

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

and inserting the given linear combination into this map therefore yields the desired result.${\displaystyle \Box }$

## Projective modules

The following is a generalisation of free modules:

Definition 11.5:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module. ${\displaystyle M}$ is called projective if and only if for a fixed module ${\displaystyle N}$ and a fixed surjection ${\displaystyle f:N\twoheadrightarrow M}$ every other module morphism with codomain ${\displaystyle M}$ (call ${\displaystyle g:K\to M}$) has a factorisation

.

Theorem 11.6:

Every free module is projective.

Proof:

Pick a basis ${\displaystyle \{m_{j}\}_{j\in J}}$ of ${\displaystyle M}$, let ${\displaystyle f:N\twoheadrightarrow M}$ be surjective and let ${\displaystyle g:K\to M}$ be some morphism. For each ${\displaystyle m_{j}}$ pick ${\displaystyle n_{j}\in N}$ with ${\displaystyle f(n_{j})=m_{j}}$. Define

${\displaystyle h:K\to N,h(k)=\sum _{i=1}^{l}r_{i}n_{j_{i}}}$ where ${\displaystyle g(k)=\sum _{i=1}^{l}r_{i}m_{j_{i}}}$.

This is well-defined since the linear combination describing ${\displaystyle g(k)}$ is unique. Furthermore, it is linear, since we have

${\displaystyle g(k+rk')=\sum _{i=1}^{l}r_{i}m_{j_{i}}+r\sum _{i'=1}^{l'}r_{i}'m'_{j_{i}'}}$,

where the right hand side is the sum of the linear combinations coinciding with ${\displaystyle g(k)}$ and ${\displaystyle g(k')}$ respectively, which is why ${\displaystyle h(k+rk')=h(k)+rh(k')}$. By linearity of ${\displaystyle f}$ and definition of the ${\displaystyle n_{j}}$, it has the desired property.${\displaystyle \Box }$

There are a couple equivalent definitions of projective modules.

Theorem 11.7:

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

Proof:

${\displaystyle \Rightarrow }$: Define the module

${\displaystyle L:=\bigoplus _{m\in M}R}$

(this obviously is a free module) and the function

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

${\displaystyle f}$ is a surjective morphism, whence we obtain a commutative diagram

;

that is, ${\displaystyle f\circ h=\operatorname {Id} _{M}}$.

We claim that the map

${\displaystyle \varphi :M\oplus \ker f\to L,\varphi (m,k):=h(m)+k}$

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

${\displaystyle \varphi ((rm,-k))=(0,\ldots ,0,\overbrace {r} ^{m{\text{-th place}}},0,\ldots ,0)=(0,\ldots ,0,\overbrace {r} ^{m{\text{-th place}}},0,\ldots ,0)}$

(surjectivity).

${\displaystyle \Leftarrow }$: Assume ${\displaystyle M\oplus L}$ is a free module. Assume ${\displaystyle f:N\twoheadrightarrow M}$ is a surjective morphism, and let ${\displaystyle g:K\to M}$ be any morphism. We extend ${\displaystyle g}$ to ${\displaystyle {\tilde {g}}:K\to M\oplus L}$ via

${\displaystyle {\tilde {g}}(k):=(g(k),0)}$.

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

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

Definition 11.8:

An exact sequence of modules

${\displaystyle 0\rightarrow K\rightarrow N\rightarrow M\rightarrow 0}$

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

commutes.

Theorem 11.9:

A module ${\displaystyle M}$ is projective iff every exact sequence

${\displaystyle 0\rightarrow K\rightarrow N\rightarrow M\rightarrow 0}$

is split exact.

Proof:

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

where we don't know yet whether ${\displaystyle h}$ is an isomorphism, but we can use ${\displaystyle h}$ to define the function

${\displaystyle {\tilde {h}}:K\otimes M\to N,{\tilde {h}}(k,m):=g(k)+h(0,m)}$,

which is an isomorphism due to injectivity:

Let ${\displaystyle {\tilde {h}}(k,m)=0}$, that is ${\displaystyle h(0,m)+g(k)=0}$. Then first

${\displaystyle m=f(h(0,m))=f(h(0,m)+g(k))=f(0)=0}$

and therefore second

${\displaystyle g(k)=h(0,m)+g(k)=0\Rightarrow k=0}$.

And surjectivity:

Let ${\displaystyle n\in N}$. Set ${\displaystyle m:=f(n)}$. Then

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

and hence ${\displaystyle g(k)=h(0,m)-n}$ for a suitable ${\displaystyle k\in K}$, thus

${\displaystyle n={\tilde {h}}(-k,m)}$.

We thus obtain the commutative diagram

and have proven what we wanted.

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

We set

${\displaystyle K:=\bigoplus _{m\in M}R}$, ${\displaystyle f:K\to M}$

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

${\displaystyle 0\rightarrow \ker f{\overset {\iota }{\hookrightarrow }}K{\overset {f}{\rightarrow }}M\rightarrow 0}$

which by assumption splits as

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

Theorem 11.10:

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

Proof:

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

${\displaystyle (M\oplus L)\otimes (N\oplus K)\cong (M\otimes N)\oplus (M\otimes K)\oplus (L\otimes N)\oplus (L\otimes K)}$,

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

Theorem 11.11:

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

Proof:

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

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

## Flat modules

The following is a generalisation of projective modules:

Definition 11.12:

An ${\displaystyle R}$-module ${\displaystyle M}$ is called flat if and only if tensoring by it preserves exactness:

${\displaystyle 0\rightarrow N\rightarrow L\rightarrow K\rightarrow 0}$ exact implies ${\displaystyle 0\rightarrow N\otimes _{R}M\rightarrow L\otimes _{R}M\rightarrow K\otimes _{R}M\rightarrow 0}$ exact.

The morphisms in the right sequence induced by any morphism ${\displaystyle f}$ are given by the bilinear map

${\displaystyle (x,m)\mapsto f(x)\otimes m}$.

Theorem 11.13:

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

Proof: This follows from theorems 9.10 and 10.?.${\displaystyle \Box }$

Theorem 11.14:

Flatness is a local property.

Proof: Exactness is a local property. Furthermore, for any multiplicatively closed ${\displaystyle S\subseteq R}$

${\displaystyle S^{-1}(M\otimes _{R}N)\cong S^{-1}M\otimes _{S^{-1}R}S^{-1}N}$

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

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 ${\displaystyle M}$ is a free module and ${\displaystyle \{e_{\alpha }\},\alpha \in A}$ a basis of ${\displaystyle M}$, we have

${\displaystyle M\cong \bigoplus _{\alpha \in A}R}$

via

${\displaystyle \sum _{\alpha \in A}r_{\alpha }e_{\alpha }\mapsto (r_{\alpha })_{\alpha \in A}}$,

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

${\displaystyle 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0}$,

to show that the sequence

${\displaystyle 0\rightarrow A\otimes M\rightarrow B\otimes M\rightarrow C\otimes M\rightarrow 0}$

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

${\displaystyle 0\rightarrow A\otimes M\rightarrow B\otimes M\rightarrow C\otimes M\rightarrow 0}$

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 ${\displaystyle M,N}$ are flat ${\displaystyle R}$-modules, then ${\displaystyle M\otimes _{R}N}$ is as well.

Proof:

Let

${\displaystyle 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0}$

be an exact sequence of modules.

## Torsion-free modules

The following is a generalisation of flat modules:

Definition 11.18:

Let ${\displaystyle M}$ be an ${\displaystyle R}$-module. The torsion of ${\displaystyle M}$ is defined to be the set

${\displaystyle T(M):=\{m\in M|\exists r\in R:rm=0\}}$.

Lemma 11.19:

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

Proof:

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

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

Definition 11.20:

Let ${\displaystyle M}$ be a module. ${\displaystyle M}$ is called torsion-free if and only if

${\displaystyle T(M)=\{0\}}$.

Theorem 11.21:

A flat module is torsion-free.

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

Definition 11.22:

Let ${\displaystyle S\subseteq R}$ be a multiplicatively closed subset of a ring ${\displaystyle R}$, and let ${\displaystyle M}$ be an ${\displaystyle R}$-module. Then the ${\displaystyle S}$-torsion of ${\displaystyle M}$ is defined to be

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

Theorem 11.23:

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