Flatness and Localization

In the previous post, we looked at several criteria for determining when an \(A\)-module \(M\) is flat; in this post, we specifically examine a criterion for determining this over a Noetherian local ring \((A, \mathfrak{m})\). The following theorem shows that, for a finitely generated module over a local Noetherian \(A\)-algebra \((E, \mathfrak{n})\) satisfying \(\mathfrak{m}E\subseteq \mathfrak{n}\), it suffices to check §Flatness, ⁋Proposition 1 for the maximal ideal alone.

Theorem 1 Fix a Noetherian local ring \((A, \mathfrak{m})\), and suppose that \((E, \mathfrak{n})\), satisfying \(\mathfrak{m}E\subseteq \mathfrak{n}\), is a local Noetherian \(A\)-algebra. 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 show the converse.

To show the converse as well, using §Flatness, ⁋Proposition 1, assuming the given conditions, it suffices to show that for any \(A\)-ideal \(\mathfrak{a}\), the multiplication map \(m:\mathfrak{a}\otimes_AM \rightarrow M\) is injective. To this end, assume that \(x\in \mathfrak{a}\otimes_AM\) belongs to the kernel \(\ker m\) of the multiplication map, and let us show that \(x=0\). First, from the structure defined on \(M\) as an \(E\)-module, there is naturally on \(\mathfrak{a}\otimes_AM\) an \(E\)-module structure as well, and from the assumption \(\mathfrak{m}E\subseteq \mathfrak{n}\), we know that for every \(n\), the relation

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

holds. Meanwhile, since these are finitely generated \(E\)-modules, part 1 of §Blowup Algebras, ⁋Corollary 8 (Krull intersection theorem) yields \((1-a)(\bigcap \mathfrak{n}^n(\mathfrak{a}\otimes_AM))=0\) for some \(a\in \mathfrak{n}\), and since \(E\) is a local ring, \(1-a\) is a unit, so

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

is satisfied. Therefore, to show that \(x=0\), it suffices to show that for every \(n\), \(x\in \mathfrak{m}^n(\mathfrak{a}\otimes_AM)\) holds. On the other hand, \(\mathfrak{m}^n(\mathfrak{a}\otimes_AM)\) can be identified with \((\mathfrak{m}^n \mathfrak{a})\otimes_AM\), and if we apply §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 filtration

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

is also \(\mathfrak{m}\)-stable, so there exists an \(N\) such that whenever \(m>N\), for all \(i\)

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

holds. Thus, whenever an arbitrary \(n\) is given, 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 instead of showing that \(x\in (\mathfrak{m}^n\mathfrak{a})\otimes_AM\) holds for every \(n\), we may show that for every \(t\), \(x\in (\mathfrak{m}^t\cap \mathfrak{a})\otimes_AM\) holds.

Now, taking the short exact sequence

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

and applying \(-\otimes_AM\) to it, we obtain the exact sequence

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

and in this situation, it suffices to show that \(x\), when mapped to \((\mathfrak{a}/\mathfrak{m}^t\cap \mathfrak{a})\otimes_AM\), becomes \(0\). Meanwhile, consider the commutative diagram

and the following commutative diagram obtained by applying \(-\otimes_AM\) to it:

Then the left map \(\mathfrak{a}\otimes_AM \rightarrow M\) is the multiplication map \(m\), and therefore \(x\in\ker m\) is sent via composition in the \(\llcorner\) direction to \(0\). Therefore, it suffices to show only that the right 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 \(\mathfrak{a}/(\mathfrak{m}^t\cap \mathfrak{a}) \rightarrow A/\mathfrak{m}^t\) giving this homomorphism is precisely the left-hand map in the following 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

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

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

However, \(A/(\mathfrak{a}+\mathfrak{m}^t)\) is annihilated by \(\mathfrak{m}^t\), and since \(\mathfrak{m}^t\) is finitely generated, we know from this that \(A/(\mathfrak{a}+\mathfrak{m}^t)\) has finite length. Therefore, we obtain the desired result if we show more generally that whenever an arbitrary \(A\)-module \(N\) of finite length is given, \(\Tor_1^A(N, M)=0\) holds.

We proceed by induction. If \(N\) has length \(1\), then from the argument following §The Jordan-Hölder Theorem, ⁋Definition 1, we must have \(N=A/\mathfrak{m}\), and so \(\Tor_1^A(N, M)=0\) coincides exactly with the hypothesis of the theorem. Choose an \(A\)-module \(N\) of finite length and an arbitrary proper submodule of \(N\) other than \(0\), denoted \(N'\). Then, to the exact sequence

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

applying the \(\Tor\) long exact sequence yields

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

Now by the inductive hypothesis, \(\Tor_1^A(N', M)=\Tor_1^A(N/N',M)=0\), so we obtain the desired result.

Meanwhile, 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 conditions. In Corollary 3, we assume the conditions of Theorem 1 and show the converse of this claim. For this, we first need the following lemma.

Lemma 2 Let an \(A\)-module \(M\) be given, and suppose \(a\in A\) is 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

If we consider a free resolution

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

of the \(A\)-module \(M\), by definition the \(i\)-th homology of the following chain complex

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

is \(\Tor_i^A(M,N)\). Meanwhile, consider the following complex

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

obtained by applying \(A/(a)\otimes_A-\) to (1). Then, since the homology of this complex is given by

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

this becomes a free resolution of \(M/aM\). Therefore, using (2) to compute \(\Tor_i^{A/(a)}(N, M/aM)\), through the following isomorphism

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

we obtain the desired result.

Using this, we can show the following.

Corollary 3 Fix a Noetherian local ring \((A, \mathfrak{m})\), and suppose that \((E, \mathfrak{n})\) is a local Noetherian \(A\)-algebra satisfying \(\mathfrak{m}E\subseteq \mathfrak{n}\). If \(a\in \mathfrak{m}\) is simultaneously a non-zerodivisor on \(A\) and a non-zerodivisor on a 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

Suppose \(M/aM\) is a flat \(A/(a)\)-module. Since the residue field of \(A/(a)\) is \(A/\mathfrak{m}\), from the assumption

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

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

While the theorem above was basically Corollary 3 cutting by an element on the base side, conversely, the following form, which cuts by an element on the fiber side, is used to lift flatness from the fiber toward the base.

Corollary 4 Fix a Noetherian local ring \((A, \mathfrak{m})\), and suppose that \((E, \mathfrak{n})\), satisfying \(\mathfrak{m}E\subseteq \mathfrak{n}\), is a local Noetherian \(A\)-algebra. For a flat \(A\)-module that is a finitely generated \(E\)-module \(M\) and an element \(f\in \mathfrak{n}\), if the image of \(f\) is a non-zerodivisor on \(M/\mathfrak{m}M\), then \(f\) is a non-zerodivisor on \(M\) and \(M/fM\) is a flat \(A\)-module. In particular, if the images of \(f_1,\ldots, f_r\in \mathfrak{n}\) form a regular sequence on \(M/\mathfrak{m}M\), then \(f_1,\ldots, f_r\) form a regular sequence on \(M\) and \(M/(f_1,\ldots, f_r)M\) is a flat \(A\)-module.

Proof

For convenience of the proof, write \(\kappa=A/\mathfrak{m}\), and let the kernel of \(\times f:M \rightarrow M\) be \(K\) and the cokernel \(C=M/fM\).

First, for the exact sequence

\[0 \rightarrow K \rightarrow M \rightarrow fM \rightarrow 0\]

applying \(-\otimes_A\kappa\), \(M\otimes_A\kappa \rightarrow fM\otimes_A\kappa\) is surjective. If we compose this with the map induced by the inclusion \(fM\hookrightarrow M\), \(fM\otimes_A\kappa \rightarrow M\otimes_A\kappa\), this is precisely, on \(M\otimes_A\kappa=M/\mathfrak{m}M\), multiplication by \(f\), and since this is injective by assumption, \(M\otimes_A\kappa \rightarrow fM\otimes_A\kappa\) is also injective, giving an isomorphism. In particular, \(fM\otimes_A\kappa \rightarrow M\otimes_A\kappa\) is also injective.

Now, to the exact sequence

\[0 \rightarrow fM \rightarrow M \rightarrow C \rightarrow 0\]

we apply the long exact sequence of \(\Tor^A_\bullet(\kappa, -)\). Since \(M\) is flat, \(\Tor_1^A(\kappa, M)=0\) (§Flatness, ⁋Proposition 1), and therefore

\[0 \rightarrow \Tor_1^A(\kappa, C) \rightarrow fM\otimes_A\kappa \rightarrow M\otimes_A\kappa\]

is exact. Since we saw above that the right-hand morphism is injective, \(\Tor_1^A(\kappa, C)=0\), and since \(C\) is a finitely generated \(E\)-module, by Theorem 1, \(C=M/fM\) is a flat \(A\)-module.

We must now show that \(K=0\). Since \(C\) is flat, if for any \(A\)-module \(N\) we compute \(\Tor_i^A(N, C)\) via a projective resolution of \(N\), then since \(-\otimes_AC\) is exact, it vanishes for \(i\geq 1\); likewise, \(\Tor_1^A(N,M)=0\). Then from the long exact sequence of \(0 \rightarrow fM \rightarrow M \rightarrow C \rightarrow 0\),

\[0=\Tor_2^A(N,C) \rightarrow \Tor_1^A(N, fM) \rightarrow \Tor_1^A(N,M)=0\]

we have \(\Tor_1^A(N, fM)=0\), and by §Flatness, ⁋Proposition 1, \(fM\) is also a flat \(A\)-module. Therefore, the sequence obtained from \(0 \rightarrow K \rightarrow M \rightarrow fM \rightarrow 0\) by applying \(-\otimes_A\kappa\) is exact on the left as well, so \(K\otimes_A\kappa \rightarrow M\otimes_A\kappa\) is injective; but its image is the kernel of \(M\otimes_A\kappa \rightarrow fM\otimes_A\kappa\), and since this morphism is an isomorphism, the image is zero. That is, \(K\otimes_A\kappa=K/\mathfrak{m}K=0\), and since \(K\) is a finitely generated \(E\)-module and \(\mathfrak{m}K\subseteq \mathfrak{n}K\), by Nakayama’s lemma (§Integral Extensions, ⁋Lemma 8 (Nakayama)), \(K=0\).

The final claim is obtained by induction on \(r\). We saw above that \(M/f_1M\) is a flat \(A\)-module, and since

\[(M/f_1M)/\mathfrak{m}(M/f_1M)=(M/\mathfrak{m}M)/f_1(M/\mathfrak{m}M)\]

the images of \(f_2,\ldots, f_r\) form a regular sequence on this quotient, so the same argument can be repeated.

Rees algebra

Definition 5 For a ring \(A\) and an ideal \(\mathfrak{a}\), the Rees algebra is

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

Also, in the same setting, we define the extended Rees algebra by

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

Here, for \(n\leq 0\), we adopt the convention that \(\mathfrak{a}^n=A\).

If \(A\) is an algebra over a field \(\mathbb{K}\), then the extended Rees algebra contains \(t^{-1}\), so it is a \(\mathbb{K}[t^{-1}]\)-algebra, and in this case the following holds.

Proposition 6 Fix a field \(\mathbb{K}\), a \(\mathbb{K}\)-algebra \(A\), and, in \(A\), an ideal \(\mathfrak{a}\), and write \(R=A[\mathfrak{a}t, t^{-1}]\). Then the extended Rees algebra \(R\) is a flat \(\mathbb{K}[t^{-1}]\)-module. Moreover, if \(\bigcap \mathfrak{a}^i=0\), then all elements of the form \(1-t^{-1}s\) (\(s\in R\)) are non-zerodivisors in \(R\).

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.