Flatness and Localization

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In the previous post we examined several criteria for determining when an \(A\)-module \(M\) is flat, and in this post we turn to a criterion using localization. The following theorem shows that it suffices to verify §Flatness, ⁋Proposition 1 only for the maximal ideal.

Theorem 1 Fix a Noetherian local ring \((A, \mathfrak{m})\), and assume that \((E, \mathfrak{n})\) is a local Noetherian \(A\)-algebra satisfying \(\mathfrak{m}E\subseteq \mathfrak{n}\). Then for a finitely generated \(E\)-module \(M\), \(M\) is a flat \(A\)-module if and only if \(\Tor_1^A(A/\mathfrak{m}, M)=0\).

Proof

If \(M\) is a flat \(A\)-module then \(\Tor_1^A(A/\mathfrak{m}, M)=0\) is precisely the content of §Flatness, ⁋Proposition 1, so it suffices to prove the converse.

To prove the converse, we again invoke §Flatness, ⁋Proposition 1: assuming the given condition, we show that for any ideal \(\mathfrak{a}\) of \(A\) the multiplication map \(m:\mathfrak{a}\otimes_AM \rightarrow M\) is injective. To this end, suppose \(x\in \mathfrak{a}\otimes_AM\) lies in the kernel \(\ker m\), and let us show that \(x=0\). First, the \(E\)-module structure on \(M\) induces a natural \(E\)-module structure on \(\mathfrak{a}\otimes_AM\) as well, and from the assumption \(\mathfrak{m}E\subseteq \mathfrak{n}\) we know that for every \(n\),

\[\mathfrak{m}^n(\mathfrak{a}\otimes_AM )\subseteq \mathfrak{n}^n(\mathfrak{a}\otimes_AM)\]

holds. On the other hand, since these are finitely generated \(E\)-modules, §Blowup Algebras, ⁋Corollary 8 (Krull intersection theorem) yields

\[\bigcap \mathfrak{m}^n(\mathfrak{a}\otimes_AM)=\bigcap \mathfrak{n}^n(\mathfrak{a}\otimes_AM)=0.\]

Thus, to show \(x=0\) it suffices to show that \(x\in \mathfrak{m}^n(\mathfrak{a}\otimes_AM)\) for all \(n\). Now \(\mathfrak{m}^n(\mathfrak{a}\otimes_AM)\) can be identified with \((\mathfrak{m}^n \mathfrak{a})\otimes_AM\), and applying §Blowup Algebras, ⁋Lemma 7 (Artin–Rees) to the \(\mathfrak{m}\)-stable filtration

\[\mathfrak{m}\supseteq \mathfrak{m}^2\supseteq\cdots\]

and \(M'=\mathfrak{a}\), the induced filtration

\[\mathfrak{m}\cap \mathfrak{a}\supseteq \mathfrak{m}^2 \cap\mathfrak{a}\supseteq\cdots\]

is also \(\mathfrak{m}\)-stable; hence there exists \(N\) such that whenever \(m>N\),

\[\mathfrak{m}^{m+i}\cap \mathfrak{a}=\mathfrak{m}^i(\mathfrak{m}^m\cap \mathfrak{a})\]

holds for all \(i\). Thus, given any \(n\), choosing \(t>N+n\) gives

\[\mathfrak{m}^t\cap \mathfrak{a}=\mathfrak{m}^n(\mathfrak{m}^{t-n}\cap \mathfrak{a})\subseteq \mathfrak{m}^n \mathfrak{a},\]

and so instead of showing that \(x\in (\mathfrak{m}^n\mathfrak{a})\otimes_AM\) for arbitrary \(n\), we may show that \(x\in (\mathfrak{m}^t\cap \mathfrak{a})\otimes_AM\) for arbitrary \(t\).

Now apply \(-\otimes_AM\) to the short exact sequence

\[0 \rightarrow \mathfrak{m}^t\cap \mathfrak{a} \rightarrow \mathfrak{a} \rightarrow \frac{\mathfrak{a}}{\mathfrak{m}^t\cap \mathfrak{a}} \rightarrow 0\]

to obtain the exact sequence

\[(\mathfrak{m}^t\cap \mathfrak{a})\otimes_AM \rightarrow \mathfrak{a}\otimes_AM \rightarrow \frac{\mathfrak{a}}{\mathfrak{m}^t\cap \mathfrak{a}}\otimes_AM \rightarrow 0;\]

in this situation it suffices to show that the image of \(x\) in \((\mathfrak{a}/\mathfrak{m}^t\cap \mathfrak{a})\otimes_AM\) is zero. On the other hand, consider the commutative diagram

and apply \(-\otimes_AM\) to obtain the commutative diagram

The left-hand map \(\mathfrak{a}\otimes_AM \rightarrow M\) is the multiplication map \(m\), so \(x\in\ker m\) is sent to \(0\) along the \(\llcorner\) composite. Hence it suffices to show that the right-hand map \((\mathfrak{a}/(\mathfrak{m}^t\cap \mathfrak{a}))\otimes_AM \rightarrow (A/\mathfrak{m}^t)\otimes_AM\) is injective. Via the isomorphism

\[\frac{\mathfrak{a}}{\mathfrak{m}^t\cap \mathfrak{a}}\cong \frac{\mathfrak{a}+\mathfrak{m}^t}{\mathfrak{m}^t}\]

the map \(\mathfrak{a}/(\mathfrak{m}^t\cap \mathfrak{a}) \rightarrow A/\mathfrak{m}^t\) inducing it is exactly the left-hand map in the short exact sequence

\[0 \rightarrow \frac{\mathfrak{a}+\mathfrak{m}^t}{\mathfrak{m}^t} \rightarrow \frac{A}{\mathfrak{m}^t}\rightarrow \frac{A}{\mathfrak{a}+\mathfrak{m}^t} \rightarrow 0.\]

Therefore, from the \(\Tor\) long exact sequence

\[\cdots \Tor_1^A(A/(\mathfrak{a}+\mathfrak{m}^t), M) \rightarrow \frac{\mathfrak{a}+\mathfrak{m}^t}{\mathfrak{m}^t}\otimes_AM \rightarrow \frac{A}{\mathfrak{m}^t}\otimes_AM \rightarrow\]

we see that what we must show is \(\Tor_1^A(A/(\mathfrak{a}+\mathfrak{m}^t), M)=0\).

Now \(A/(\mathfrak{a}+\mathfrak{m}^t)\) is annihilated by \(\mathfrak{m}^t\), and since \(\mathfrak{m}^t\) is finitely generated, it follows that \(A/(\mathfrak{a}+\mathfrak{m}^t)\) has finite length. Hence, if we show more generally that \(\Tor_1^A(N, M)=0\) for every \(A\)-module \(N\) of finite length, we obtain the desired conclusion.

We proceed by induction on length. If \(N\) has length \(1\), then by the discussion after §The Jordan-Hölder Theorem, ⁋Definition 1 we must have \(N=A/\mathfrak{m}\), and thus \(\Tor_1^A(N, M)=0\) is exactly the hypothesis of the theorem. Choose an \(A\)-module \(N\) of finite length and any proper submodule \(N'\) of \(N\). Applying the \(\Tor\) long exact sequence to

\[0 \rightarrow N' \rightarrow N \rightarrow N/N' \rightarrow 0\]

we obtain

\[\cdots \rightarrow\Tor_1^A(N', M) \rightarrow \Tor_1^A(N, M) \rightarrow \Tor_1^A(N/N', M) \rightarrow \cdots\]

By the inductive hypothesis \(\Tor_1^A(N', M)=\Tor_1^A(N/N',M)=0\), whence the desired result.

On the other hand, if \(M\) is a flat \(A\)-module then for any \(A/(a)\)-module \(N\),

\[(M/aM)\otimes_{A/(a)}N=(A/(a)\otimes_A M)\otimes_{A/(a)} N\cong M\otimes_AN\]

so \(M/aM\) is a flat \(A/(a)\)-module without any additional hypotheses. In Corollary 3 we establish the converse of this assertion under the hypotheses of Theorem 1. For this we require the following lemma.

Lemma 2 Let \(M\) be an \(A\)-module, and let \(a\in A\) be a non-zerodivisor on both \(A\) and \(M\). Then for any \(A/(a)\)-module \(N\),

\[\Tor_i^{A/(a)}(N, M/aM)=\Tor_i^A(N,M)\]

holds.

Proof

Consider a free resolution of the \(A\)-module \(M\)

\[\cdots \rightarrow F_2 \rightarrow F_1 \rightarrow F_0\tag{1}\]

By definition, the \(i\)-th homology of the chain complex

\[\cdots \rightarrow N\otimes_A F_2 \rightarrow N\otimes_AF_1 \rightarrow N\otimes_A F_0\]

is \(\Tor_i^A(M,N)\). On the other hand, consider the complex obtained by applying \(A/(a)\otimes_A-\) to (1):

\[\cdots \rightarrow F_2/aF_2 \rightarrow F_1/aF_1 \rightarrow F_0/aF_0 \rightarrow M/aM \rightarrow 0\tag{2}\]

The homology of this complex is given by

\[\Tor_i^A(A/(a), M)=\begin{cases} M/aM&\text{if $i=0$}\\ 0&\text{otherwise}\end{cases}\]

so (2) is a free resolution of \(M/aM\). Therefore, computing \(\Tor_i^{A/(a)}(N, M/aM)\) using (2), we obtain the desired result from the isomorphism

\[N\otimes_{A/(a)} F_i/aF_i=N\otimes_{A/(a)} ((A/(a))\otimes_A F_i)\cong N\otimes_A F_i.\]

Using this we can prove the following.

Corollary 3 Fix a Noetherian local ring \((A, \mathfrak{m})\), and assume that \((E, \mathfrak{n})\) is a local Noetherian \(A\)-algebra satisfying \(\mathfrak{m}E\subseteq \mathfrak{n}\). If \(a\in \mathfrak{m}\) is a non-zerodivisor in \(A\) and simultaneously a zerodivisor on the finitely generated \(E\)-module \(M\), then \(M\) is a flat \(A\)-module if and only if \(M/aM\) is a flat \(A/(a)\)-module.

Proof

Assume that \(M/aM\) is a flat \(A/(a)\)-module. For the residue field \(A/\mathfrak{m}\) of \(A\), the assumption gives

\[\Tor_1^{A/(a)}(A/\mathfrak{m}, M/aM)=0,\]

and applying Lemma 2 we deduce that \(\Tor_1^A(A/\mathfrak{m}, M)=0\). Hence by Theorem 1, \(M\) is a flat \(A\)-module.

Rees algebra

Definition 4 For a ring \(A\) and an ideal \(\mathfrak{a}\), the Rees algebra is

\[A[\mathfrak{a}t]=\bigoplus_{n=0}^\infty \mathfrak{a}^n t^n\subseteq A[t].\]

In the same setting, the extended Rees algebra is defined by

\[A[\mathfrak{a}t, t^{-1}]=\bigoplus_{n=-\infty}^\infty \mathfrak{a}^nt^n\subseteq A[t, t^{-1}].\]

Then the following corollary is almost obvious.

Proposition 5 Fix a field \(\mathbb{K}\) and a \(\mathbb{K}\)-algebra \(A\). Then the Rees algebra \(A[\mathfrak{a}t, t^{-1}]\) is a flat \(\mathbb{K}[t]\)-module. Moreover, if \(\bigcap \mathfrak{a}^i=0\), then every element of the form \(1-t s\) (\(s\in S\)) is a non-zerodivisor in \(A[\mathfrak{a}t, t^{-1}]\).