I'm stuck on this exercise.

$A$ is a commutative ring with unit. $N$ is an $A$-module. Then $N$ is flat $\Longleftrightarrow $ $\text{Tor}_{1}(A/a, N ) = 0 $ for every finitely generated ideal $a$ of $A$.

What I know: every module is isomorphic to the direct limit of its finitely generated submodules, I think I should use this information.

  • $\begingroup$ One direction is obvious, right? If $N$ is flat, then $\text{Tor}_1(M, N) = 0$ for any $A$-module $M$. $\endgroup$
    – Alex G.
    May 4, 2014 at 17:47
  • $\begingroup$ @AlexG.: yes, obviously $\endgroup$
    – WLOG
    May 4, 2014 at 17:47
  • $\begingroup$ There's a short proof of this fact in Lemma 2.1 in Hartshorne's "Deformation Theory" book on page 10. An early draft of the book is available for free: math.berkeley.edu/~robin/math274root.pdf‎ $\endgroup$ May 5, 2014 at 8:11
  • $\begingroup$ your link seems wrong $\endgroup$
    – WLOG
    May 5, 2014 at 9:19
  • 1
    $\begingroup$ @WLOG: Sorry, this link seems to work: math.berkeley.edu/~robin/math274root.pdf (anyway, just Google "hartshorne deformation theory pdf" if it doesn't). (also, you should add @ (my username), that way I get notified when you answer) $\endgroup$ May 5, 2014 at 11:28

1 Answer 1


Using the tensor-hom adjunction, one can see that $N$ is flat iff $N^c$ is injective, where $N^c=\mathrm{Hom}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ is the module of characters. By Baer's criterion, this happens if and only if $\mathrm{Ext}_A^1(A/a,N^c)=0$ for every ideal $a \subseteq A$. Using the isomorphism $\mathrm{Tor}_1^A(A/a,N)^c \simeq \mathrm{Ext}_A^1(A/a,N^c)$ and the fact that $M^c=0$ iff $M=0$ one can see that $N$ is flat if and only if $\mathrm{Tor}_1^A(A/a,N)=0$ for every ideal $a \subseteq A$. What remains is to reduce this to the finitely generated case.

Using the long exact sequence $$\dots \rightarrow \mathrm{Tor}_1^A(A/a,N) \rightarrow a \otimes_A N\rightarrow A \otimes_A N\rightarrow A/a \otimes_A N\rightarrow 0$$

one can see that $\mathrm{Tor}_1^A(A/a,N)=0$ is equivalent to the fact that the functor $-\otimes N$ preserves injectivity of the inclusion $i: a\subseteq A$.

So assume for contradiction that $\mathrm{Tor}_1^A(A/a,N)=0$ for every finitely generated ideal $a\subseteq A$ and that there is an ideal $b \subseteq A$ such that $\mathrm{Tor}_1^A(A/b,N)\neq 0$, i.e. the induced map $j\otimes_A N: b \rightarrow A \otimes_AN$ is not injective ($j$ denotes the inclusion $b \subseteq A$). That is, there is a nonzero element $\sum_{k=0}^n b_k \otimes n_k$ belonging to the kernel of $j\otimes_A N$.

Consider a finitely generated ideal $a:=\sum_{k=0}^n b_kA$. Then we have the inclusions $$a\stackrel{i}\hookrightarrow b \stackrel{j}\hookrightarrow A, \; \;a \stackrel{l}\hookrightarrow A, \;\;l=j\circ i $$

and hence $$a \otimes_A N \stackrel{i\otimes_A N}\rightarrow b\otimes_A N \stackrel{j\otimes_A N}\rightarrow A \otimes_A N, \;\; a \otimes_A N \stackrel{l\otimes_A N}\hookrightarrow A \otimes_A N, $$

and $l \otimes_A N= (j \otimes_A N) \circ( i \otimes_A N)$ (and the injectivity of $l \otimes_A N$ follows from $\mathrm{Tor}_1^A(A/a,N)=0$).

This is clearly a contradiction, since $\sum_{k=0}^n b_k \otimes n_k$ is a nonzero element of $a \otimes_A N$ (it is mapped to a nonzero element by $ i \otimes_A N$), which is sent by the injective map $ l \otimes_A N$ to $0$.

  • $\begingroup$ as soon as possible I'll check your answer :D $\endgroup$
    – WLOG
    May 4, 2014 at 19:25

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