Section 3.4 Commutative Local Rings
We have shown that
\begin{equation*}
\text { Free } \Longrightarrow \text { projective } \Longrightarrow \text { flat. }
\end{equation*}
Over a local ring, these three notions actually coincide. To show this, we need a little bit of commutative algebra. First, some notation: when \(R\) is a local ring, meaning \(R\) has a unique maximal ideal \(\mathfrak{m}\text{,}\) we write \((R, \mathfrak{m})\) to denote the ring \(R\) and its maximal ideal. Now note that for any \(R\)-module \(M\text{,}\) the module \(M / \mathfrak{m} M\) is annihilated by \(\mathfrak{m}\text{,}\) so it is also a module over a ring \(R / \mathfrak{m}\text{,}\) which is a field.
The following is a classical result in commutative algebra, known by some as Nakayama’s Lemma. As noted in [Mat89, page 8], Nakayama himself claimed that this should be attributed to Krull and Azumaya, but it’s not clear which of the three actually had the commutative ring statement first. So some authors (eg, Matsumura) prefer to refer to it as NAK. There are actually a range of statements, rather than just one, that go under the banner of Nakayama’s Lemma a.k.a. NAK.
Theorem \(4.45(\mathrm{NAK})\text{.}\) Let \((R, \mathfrak{m}, k)\) be a local ring, and \(M\) be a finitely generated module. If \(M=\mathfrak{m} M\text{,}\) then \(M=0\text{.}\)
The theorem above is the theorem most commonly referred to as NAK. The proof involves only elementary tools, and a fun linear algebra-inspired trick called the Determinantal Trick. While we will not include the details here, they can be found in any standard Commutative Algebra book. We will however use this result to prove another statement that is also commonly referred to as NAK, which allows us to talk about minimal generating sets for finitely generated modules over local rings.
Remark 4.46. Let \(R\) be any commutative ring, and consider an \(R\)-module \(M\) and an ideal \(I\text{.}\) If \(I M=0\text{,}\) meaning that \(a m=0\) for all \(a \in I\) and all \(m \in M\text{,}\) then \(M\) can be given the structure of an \(R / I\)-module, as follows: for any \(m \in M\) and any \(r \in R\text{,}\)
\begin{equation*}
(r+I) m=r m
\end{equation*}
The fact that \(I\) kills \(M\) is what makes this action well-defined. The fact that \(M\) is actually an \(R\)-module under this action is a consequence of the fact that \(M\) is an \(R\)-module; checking these details is routine, and we leave them as an exercise.
Notice that the structure of \(M\) as an \(R / I\)-module is essentially the same as its structure as an \(R\)-module. There are many properties of \(M\) as an \(R\)-module that pass onto its \(R / I-\) module structure, and typically such results are easy to check.
Here is a special case of this: if \((R, \mathfrak{m})\) is a commutative local ring, and \(M\) is an \(R\)-module, then the module \(M / \mathfrak{m} M\) is killed by \(\mathfrak{m}\text{,}\) and thus it is also a module over \(R / \mathfrak{m}\text{.}\) Now notice that \(R / \mathfrak{m}\) is a field, so \(M / \mathfrak{m} M\) is actually a vector space over the field \(R / \mathfrak{m}\text{.}\)
Theorem 4.47. Let \((R, \mathfrak{m})\) be a commutative local ring, and \(M\) be a finitely generated module. For \(m_{1}, \ldots, m_{s} \in M\text{,}\)
\begin{equation*}
m_{1}, \ldots, m_{s} \text { generate } M \Longleftrightarrow \overline{m_{1}}, \ldots, \overline{m_{s}} \text { generate } M / \mathfrak{m} M
\end{equation*}
Thus, any generating set for \(M\) consists of at least \(\operatorname{dim}_{k}(M / \mathfrak{m} M)\) elements.
Proof. The implication \((\Rightarrow)\) is clear. For \((\Leftarrow)\text{,}\) given \(m_{1}, \ldots, m_{s} \in M\) such that \(\overline{m_{1}}, \ldots, \overline{m_{s}}\) generate \(M / \mathfrak{m} M\text{,}\) consider
\begin{equation*}
N:=R m_{1}+\cdots+R m_{s} \subseteq M
\end{equation*}
Since \(M / \mathfrak{m} M\) is generated by the image of \(N\text{,}\) we have \(M=N+\mathfrak{m} M\text{.}\) By taking the quotient by \(N\text{,}\) we see that
\begin{equation*}
M / N=(N+\mathfrak{m} M) / N=\mathfrak{m}(M / N)
\end{equation*}
By Theorem 4.45, \(M / N=0\) and thus \(M=N\text{.}\)
As we mentioned above, this allows us to talk about minimal generating sets.
Definition 4.48. Let \((R, \mathfrak{m})\) be a local ring, and \(M\) a finitely generated module. A set of elements \(\left\{m_{1}, \ldots, m_{t}\right\}\) is a minimal generating set of \(M\) if the images of \(m_{1}, \ldots, m_{t}\) form a basis for the \(R / \mathfrak{m}\) vector space \(M / \mathfrak{m} M\text{.}\)
Note that every finitely generated module over a local ring has a minimal generating set, that every minimal generating set has the same number of elements, and that any set of generators for \(M\) contains a minimal generating set, all thanks to plain old linear algebra. In particular, we can now define the following:
Definition 4.49. Let \(M\) be a finitely generated module over a commutative local ring \((R, \mathfrak{m})\text{.}\) The minimal number of generators of \(M\text{,}\) denoted \(\mu(M)\text{,,}\) is the number of elements in any minimal generating set for \(M\text{.}\)
We now have all the key commutative algebra ingredients needed to show that for finitely generated modules over a noetherian local ring, projective \(=\) free. However, the proof follows more easily with one more homological tool we haven’t developed yet, so we will hold off on proving this for now - in fact, you will soon be able to prove it easily, so this will be in the next problem set.
Exercise 58. Let \((R, \mathfrak{m})\) be a commutative local ring, and let \(M\) be a finitely presented module. Then
\begin{equation*}
M \text { is flat } \Longleftrightarrow M \text { is projective } \Longleftrightarrow M \text { is free. }
\end{equation*}
Kaplansky [Kap58] showed that this holds even for modules that are not necessarily finitely presented, but generated by countably many elements.
Definition 4.50. An \(R\)-module \(M\) is locally free if \(M_{P}\) is free for every prime ideal \(P\text{.}\)
Exercise 59. Let \(R\) be a commutative ring, \(M\) and \(N\) be \(R\)-modules, and \(P\) be a prime ideal. Show that
\begin{equation*}
\left(M \otimes_{R} N\right)_{P} \cong M_{P} \otimes_{R_{P}} N_{P}
\end{equation*}
Exercise 60. Let \(R\) be a commutative ring, \(P\) be a prime ideal, and \(M\) be an \(R_{P}\)-module. Let \(N\) be \(M\) as an \(R\)-module via restriction of scalars. Then as \(R_{P}\)-modules, we have an isomorphism
\begin{equation*}
N_{P} \cong M
\end{equation*}
Exercise 61. Let \(R\) be a commutative ring. Show that a homomorphism of \(R\)-modules \(f: M \rightarrow N\) is surjective if and only if \(f_{P}\) is surjective for all primes \(P\text{.}\)
Exercise 3.48.
Let \(R\) be a noetherian ring, \(W\) be a multiplicative set, \(M\) be a finitely generated \(R\)-module, and \(N\) an arbitrary \(R\)-module. Show that
\begin{equation*}
\operatorname{Hom}_{W^{-1} R}\left(W^{-1} M, W^{-1} N\right) \cong W^{-1} \operatorname{Hom}_{R}(M, N) \text {. }
\end{equation*}
In particular, if \(P\) is prime,
\begin{equation*}
\operatorname{Hom}_{R_{P}}\left(M_{P}, N_{P}\right) \cong \operatorname{Hom}_{R}(M, N)_{P}
\end{equation*}
Theorem 4.51. Let \(R\) be a commutative noetherian ring and let \(M\) be a finitely presented \(R\)-module. Then
\begin{equation*}
M \text { is projective } \Longleftrightarrow M \text { is flat } \Longleftrightarrow M \text { is locally free. }
\end{equation*}
Proof. We already know that projectives are flat, by Theorem 4.37.
Suppose \(M\) is flat. We claim that \(M_{P}\) is flat for every prime ideal \(P\text{.}\) First, note that \(M_{P} \cong R_{P} \otimes_{R} M\text{,}\) by Theorem 3.57; moreover, \(R_{P}\) is a flat \(R\)-module by Exercise 53 . Note moreover that any \(R_{P}\)-module can also be viewed as an \(R\)-module by extension of scalars along the canonical localization map. Now given any short exact sequence of \(R_{P}\)-modules, say
\begin{equation*}
0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0
\end{equation*}
tensoring with \(M_{P}\) over \(R_{P}\) can be done in two steps: first we view this as a short exact sequence of \(R\)-modules, and tensor with \(M\text{,}\) but \(M\) is a flat \(R\)-module, so
\begin{equation*}
0 \longrightarrow A \otimes_{R} M \longrightarrow B \otimes_{R} M \longrightarrow C \otimes_{R} M \longrightarrow 0
\end{equation*}
is exact. Then we tensor with \(R_{P}\text{,}\) but this is also flat \(R\)-module, so we get a short exact sequence again:
\begin{equation*}
0 \longrightarrow\left(A \otimes_{R} M\right) \otimes_{R} R_{P} \longrightarrow\left(B \otimes_{R} M\right) \otimes_{R} R_{P} \longrightarrow\left(C \otimes_{R} M\right) \otimes_{R} R_{P} \longrightarrow 0
\end{equation*}
By Theorem 3.57 and Exercise 59, for each \(R_{P}\)-module \(X\) we have
\begin{equation*}
\left(X \otimes_{R} M\right) \otimes_{R} R_{P} \cong\left(X \otimes_{R} M\right)_{P} \cong X_{P} \otimes_{R_{P}} M_{P}
\end{equation*}
But \(X_{P} \cong X\text{,}\) by ??, so we conclude that
\begin{equation*}
\left(X \otimes_{R} M\right) \otimes_{R} R_{P} \cong X \otimes_{R_{P}} M_{P}
\end{equation*}
Thus
\begin{equation*}
0 \longrightarrow A \otimes_{R_{P}} M_{P} \longrightarrow B \otimes_{R_{P}} M_{P} \longrightarrow C \otimes_{R_{P}} M_{P} \longrightarrow 0
\end{equation*}
is exact, and \(M_{P}\) is a flat \(R_{P}\)-module.
So whenever \(M\) is flat, \(M_{P}\) is a flat \(R_{P}\)-module for all primes \(P\text{.}\) By Exercise \(58, M_{P}\) must be free over \(R_{P}\) for all primes \(P\text{,}\) that is, \(M\) is locally free.
Finally, suppose that \(M\) is locally free. We want to show that \(M\) is projective. So by Theorem 4.4, we need to show that for all surjective \(R\)-module maps \(f: A \rightarrow B\text{,}\) the map \(f_{*}: \operatorname{Hom}_{R}(M, A) \rightarrow \operatorname{Hom}_{R}(M, B)\) is surjective. By Exercise 61 , it is enough to show that \(f_{P}\) is surjective for all primes \(P\text{.}\) By Exercise 62,
\begin{equation*}
\operatorname{Hom}_{R_{P}}\left(M_{P}, A_{P}\right) \cong \operatorname{Hom}_{R}(M, A)_{P} \quad \text { and } \quad \operatorname{Hom}_{R_{P}}\left(M_{P}, B_{P}\right) \cong \operatorname{Hom}_{R}(M, B)_{P}
\end{equation*}
and
\begin{equation*}
\left(f_{*}\right)_{P}=\left(f_{P}\right)_{*}: \operatorname{Hom}_{R_{P}}\left(M_{P}, A_{P}\right) \rightarrow \operatorname{Hom}_{R_{P}}\left(M_{P}, B_{P}\right) .
\end{equation*}
But \(M_{P}\) i free, and thus projective by Theorem 4.3, so \(\left(f_{P}\right)_{*}\) is surjective. Since this holds for all \(P\text{,}\) by Exercise 61 we conclude that \(f_{*}\) is surjective, and thus \(M\) is projective.
Note that the noetherianity assumption is just so that finitely generated implies finitely presented; the statement is also true for a general commutative ring if instead of finitely generated modules we take finitely presented.
Summary
