Section 3.3 Flat Modules
Subsection Definition
“”―
Finally, we turn to the modules that make the tensor product exact.
Definition 3.35. Flat Module.
An \(R\)-module \(M\) is said to be flat if \(M \otimes_{R}-\) is an exact functor.
Remark 3.36.
Remark 4.35. By Theorem 2.63, \(M \otimes_{R}\) - is right exact. Therefore, \(M\) is flat if and only if for every injective \(R\)-module map \(i: A \longrightarrow B\text{,}\)
\begin{equation*}
M \otimes_{R} A \stackrel{1 \otimes i}{\longrightarrow} M \otimes_{R} B \quad \text { is injective. }
\end{equation*}
Lemma 3.37. Direct Summands of Flat Modules are Flat.
Given a family of \(R\)-modules \(\left\{M_{i}\right\}_{i \in I}\text{,}\) the direct sum \(\oplus_{i} M_{i}\) is flat if and only if every \(M_{i}\) is flat. In particular, direct summands of flat modules are flat.
Proof.
Given a family of \(R\)-module homomorphisms \(f_{i}: M_{i} \longrightarrow N_{i}\text{,}\) there is an \(R\)-module homomorphism
\begin{equation*}
\begin{array}{r}
\bigoplus_{i \in I} M_{i} \stackrel{\left(f_{i}\right)_{i \in I}}{\longmapsto} \bigoplus_{i \in I} N_{i} \\
\quad\left(m_{i}\right) \longmapsto
\end{array}
\end{equation*}
which is injective if and only if every \(f_{i}\) is injective.
Let \(f: A \longrightarrow B\) be an injective \(R\)-module homomorphism. There is a commutative diagram
\begin{equation*}
\begin{aligned}
& \left(\bigoplus_{i \in I} M_{i}\right) \otimes_{R} A \stackrel{\cong}{\longrightarrow} \bigoplus_{i \in I} M_{i} \otimes_{R} A \\
& \varphi:=1 \otimes f \downarrow \\
& \left(\bigoplus_{i \in I} M_{i}\right) \otimes_{R} B \stackrel{\downarrow}{\cong} \cong \bigoplus_{i \in I} M_{i} \otimes_{R} B
\end{aligned}
\end{equation*}
where the horizontal maps are the isomorphisms from Theorem 2.55. In particular, \(\varphi\) is injective if and only if \(\psi\) is injective. Moreover, \(\psi\) is injective if and only if each component is injective, meaning \(1 \otimes f: M_{i} \otimes A \longrightarrow M_{i} \otimes B\) is injective for all \(i\text{.}\)
On the one hand, \(\bigoplus_{i \in I} M_{i}\) is flat if and only if for every injective map \(f\text{,}\) the corresponding \(\phi\) is injective. On the other hand, all the \(M_{i}\) are flat if and only if for every injective map \(f, 1 \otimes f: M_{i} \otimes A \longrightarrow M_{i} \otimes B\) is injective for all \(i\text{,}\) or equivalently, as explained above, if \(\psi\) is injective for any given injective map \(f\text{.}\) This translates into the equivalence we want to show.
All projectives are flat.
Theorem 3.38. Every Projective Module is Flat.
Let \(R\) be any ring. Every projective \(R\)-module is flat.
Proof.
First, recall that \(R \otimes_{R}-\) is naturally isomorphic to the identity functor, by Lemma 3.40, and thus exact (see Remark 3.11). This shows that \(R\) is flat, and thus any free module, being a direct sum of copies of \(R\text{,}\) must also be flat by Lemma 3.37. Finally, every projective module is a direct summand of a free module, by Theorem 3.9. Direct summands of flat modules are flat, by Lemma 3.37, so every projective module is flat.
We can test whether a given module if flat by looking at the finitely generated submodules.
Theorem 3.39. Flatness and Finitely Generated Submodules.
If every finitely generated submodule of \(M\) is flat, then \(M\) is flat.
Proof.
Let \(i: A \longrightarrow B\) be an injective map of \(R\)-modules. We want to show that
\begin{equation*}
M \otimes_{R} A \stackrel{1 \otimes i}{\longrightarrow} M \otimes_{R} B
\end{equation*}
is injective. Suppose that \(u \in \operatorname{ker}\left(1_{M} \otimes i\right)\text{.}\) We are going to construct a finitely generated submodule \(N \subseteq M\text{,}\) with \(j: N \rightarrow M\) the inclusion, and an element \(v \in N \otimes_{R} A\) such that \(v \in \operatorname{ker}\left(1_{N} \otimes i\right)\) and \(u=\left(j \otimes 1_{A}\right)(v)\text{.}\) Once we do that, our submodule \(N\) is finitely generated, and thus flat by assumption, so \(1_{N} \otimes i\) is injective; therefore, \(v=0\) and thus we must have \(u=0\text{.}\) Therefore, \(1_{M} \otimes i\) is injective, and we conclude that \(M\) is flat.
Let’s say that \(u=m_{1} \otimes a_{1}+\cdots+m_{n} \otimes a_{n}\text{.}\) In Theorem 2.33, we constructed the tensor product \(M \otimes_{R} B\) as a quotient of the free module \(F\) on \(M \times B\) by the submodule \(S\) with all the necessary relations we need to impose. This gives us a short exact sequence
\begin{equation*}
0 \longrightarrow S \longrightarrow F \stackrel{\pi}{\longrightarrow} M \otimes_{R} B \longrightarrow 0
\end{equation*}
The fact that \(m_{1} \otimes i\left(a_{1}\right)+\cdots+m_{n} \otimes i\left(a_{n}\right)=0\) means we can rewrite this element as \(\pi(s)\) for some \(s \in S\text{.}\) This element \(s\) is a linear combination of elements of finitely many \((m, b) \in M \times B\text{.}\) Let \(c_{1}, \ldots, c_{t}\) be all the \(M\)-coordinates of those elements.
Now we take \(N\) to be the finitely generated submodule of \(M\) generated by \(m_{1}, \ldots, m_{n}\) and \(c_{1}, \ldots, c_{t}\text{,}\) and \(v=m_{1} \otimes a_{1}+\cdots+m_{n} \otimes a_{n} \in N \otimes A\text{.}\) Now
\begin{equation*}
\left(j \otimes 1_{A}\right)(v)=\left(j \otimes 1_{A}\right)\left(m_{1} \otimes a_{1}+\cdots+m_{n} \otimes a_{n}\right)=m_{1} \otimes a_{1}+\cdots+m_{n} \otimes a_{n} \in M \otimes_{R} A,
\end{equation*}
and
\begin{equation*}
\left(1_{N} \otimes i\right)(v)=\left(1_{N} \otimes i\right)\left(m_{1} \otimes a_{1}+\cdots+m_{n} \otimes a_{n}\right)=m_{1} \otimes i\left(a_{1}\right)+\cdots+m_{n} \otimes i\left(a_{n}\right)=0
\end{equation*}
as desired.
The reason we needed to add in these extra elements \(n_{i}\) is that a priori \(N \otimes B\) is not necessarily a submodule of \(M \otimes B\text{,}\) so we do not necessarily have \(m_{1} \otimes i\left(a_{1}\right)+\cdots+m_{n} \otimes i\left(a_{n}\right)=0\) in \(\left(R m_{1}+\cdots+R m_{n}\right) \otimes B\) without adding in all relations that make it true.
Subsection Torsion Submodules
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Definition 3.40. Torsion Submodule.
Let \(R\) be a domain and \(M\) be an \(R\)-module. The torsion submodule of \(M\) is
\begin{equation*}
T(M):=\{m \in M \mid r m=0 \text { for some regular element } r \in R\}
\end{equation*}
The elements of \(T(M)\) are called torsion elements, and we say that \(M\) is torsion if \(T(M)=M\text{.}\) Finally, \(M\) is torsion free if \(T(M)=0\text{.}\)
Theorem 3.41. Closure Properties of Torsion Free Modules.
Torsion free modules are closed under taking submodules, quotients, direct sums, and localizations.
Lemma 3.42. Flat Implies Torision Free in Domains.
If \(R\) is a domain and \(M\) is a flat \(R\)-module, then \(M\) is torsion free.
Proof.
Let \(Q=\operatorname{frac}(R)\) be the fraction field of \(R\text{,}\) which is a torsion free \(R\)-module. Now \(M \otimes_{R} Q\) is a \(Q\)-vector space, so isomorphic to a direct sum of copies of \(Q\text{.}\) In particular, \(M \otimes_{R} Q\) is torsion free as an \(R\)-module. Since \(M\) is flat, the inclusion \(R \subseteq Q\) induces an injective \(R\)-module map
\begin{equation*}
0 \longrightarrow M \otimes_{R} R \longrightarrow M \otimes_{R} Q
\end{equation*}
and since \(M \cong M \otimes_{R} R\text{,}\) by Lemma 2.54, we conclude that \(M\) is isomorphic to a submodule of \(M \otimes_{R} Q\text{.}\) By Theorem 3.41, submodules of torsion free modules are also torsion free, so \(M\) is torsion free.
In general, the converse does not hold.
Example 3.43. Torsion Free but not Flat Module.
Example 4.41. Let \(k\) be a field and \(R=k[x, y]\text{.}\) Consider the ideal \(\mathfrak{m}=(x, y)\text{.}\) This is a submodule of the torsion free module \(R\text{,}\) and thus torsion free by Theorem 3.41. However, it is not flat. For example, when we apply \(R / \mathfrak{m} \otimes_{R}-\) to the inclusion \(\mathfrak{m} \subseteq R\) we obtain a map of \(R / \mathfrak{m}\)-vector spaces
\begin{equation*}
\mathfrak{m} / \mathfrak{m}^{2} \longrightarrow R / \mathfrak{m}
\end{equation*}
This map cannot possibly be injective: \(\mathfrak{m} / \mathfrak{m}^{2}\) is a \(2\)-dimensional \(R / \mathfrak{m}\)-vector space, while \(R / \mathfrak{m}\) is \(1\)-dimensional.
The converse does hold over a PID.
Theorem 3.44. Flat iff Torsion Free in PID.
If \(R\) is a principal ideal domain, an \(R\)-module \(M\) is flat if and only if it is torsion free.
Proof.
\((\Rightarrow)\) This is just a special case of Lemma 3.42.
\((\Leftarrow)\) Suppose \(M\) is a torsion free finitely generated \(R\)-module. The structure theorem for PIDs says that \(M\) must be isomorphic to a direct sum of copies of cyclic modules. The cyclic module \(R / I\) has torsion (all the elements are killed by \(I\) ) unless \(I=0\text{.}\) Therefore, \(M\) must be isomorphic to a direct sum of copies of \(R\text{,}\) and thus free. By Theorem 3.3, \(M\) is projective, and by Theorem 3.38 projectives are flat, so \(M\) is flat.
Now let \(M\) be any torsion free \(R\)-module. All of the finitely generated submodules of \(R\) are also torsion free by Theorem 3.41, and thus flat by what we have shown so far. By Theorem 3.39, \(M\) must be flat.
Not all flat modules are projective.
Example 3.45. Flat but not Projective Module.
Example 4.43. The \(\mathbb{Z}\)-module \(\mathbb{Q}\) is torsion free and thus flat, by Theorem 3.44. However, \(\mathbb{Q}\) is not a projective \(\mathbb{Z}\)-module. Suppose by contradiction, that \(\mathbb{Q}\) is a projective \(\mathbb{Z}\)-module. By Theorem 3.9, \(\mathbb{Q}\) must be a direct summand of a free module, say \(F=\bigoplus_{I} \mathbb{Z}\text{.}\) Consider the inclusion \(\iota: \mathbb{Q} \hookrightarrow F\text{,}\) and pick \(i \in I\) such that the image of \(\mathbb{Q}\) contains some element with a nonzero entry in the \(i\) component. Now consider the projection \(\pi: F \longrightarrow \mathbb{Z}\) onto the \(i\) th factor. By assumption, the composition \(\pi i: \mathbb{Q} \longrightarrow \mathbb{Z}\) is nonzero. However, there are no nontrivial abelian group homomorphisms \(\mathbb{Q} \longrightarrow \mathbb{Z}\text{,}\) contradicting the fact that \(\pi i\) is nonzero. We conclude that \(\mathbb{Q}\) is not projective.
However, for finitely generated modules over a commutative noetherian local ring, every flat module is free, and thus flat, projective, and free all coincide. However, to prove that we need a little bit of commutative algebra, which we introduce in the next section.
Finally, here is a very important example: localization is flat.
Theorem 3.46. Flatness of Localization.
Let \(R\) be a commutative ring, and \(W \ni 1\) a multiplicative subset of \(R\text{.}\) Then \(W^{-1} R\) is flat over \(R\text{.}\)
Proof.
By Theorem 2.75, tensoring with \(W^{-1} R\) is localizing at \(W\text{.}\) But localization is exact, so tensoring with \(W^{-1} R\) is exact, and thus \(W^{-1} R\) is a flat \(R\)-module.
Example 3.47.
If \(R\) is a domain then its fraction field \(Q\) is a flat module.
