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Let $A$ be a commutative ring and $M$ an $A$-module. Let $I$ be any ideal of $A$. We have an epimorphism $M \otimes_A I \rightarrow IM$. It seems to me that this is not in general an isomorphism.

Q1: Any counterexample?

If $M$ is flat, then $M \otimes_A I \cong IM$. However, flatness seems to be a too strong condition for this equality to be true for any ideal $I$. I am interested in finding a characterization of $M$ such that $M \otimes_A I \cong IM$ for any ideal $I$ of $A$.

Q2: Any suggestions?

Manos
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3 Answers3

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An $A$-module $M$ is flat if and only if for every finitely generated ideal $I$ of $A$, $M\otimes_AI\rightarrow IM$ is an isomorphism. This is proved, e.g., as Theorem 1.2.4 of Liu's algebraic geometry textbook. He actually states it with all ideals $I$, but a direct limit argument reduces to the case of finitely generated ideals.

Keenan Kidwell
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  • Lam proves a noncommutative version of this in his book 'Lectures on Modules and Rings'. Indeed, Theorem 4.12 reads: "A right $R$-module $P$ is flat iff, for any (finitely generated) left ideal $\mathfrak{A}\subseteq R$, the natural map $P\otimes_R \mathfrak{A}\to P\mathfrak{A}$ is an isomorphism (of abelian groups)." –  Nov 10 '21 at 13:57
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Here is your counterexample: Let $A = k[x]/x^2$ and let $I = (x)$. Then $I$ is itself an $A$-module, $I^2 = 0$ but $I \otimes_A I \neq 0$.

Edit: To show that $I \otimes_A I \neq 0$. Intuitively the idea is that we cannot do $x \otimes x = 1 \otimes x^2$ because $1 \notin I$. To make this formal show that $I \times I \to I$ defined by $(ax, bx) \mapsto abx$ is a well-defined and nonzero bilinear map. It must then factor through $I \otimes_A I$.

Jim
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    More easy: $x : A \to I$ is an epimorphism with kernel $(x)$, so that $I = A/(x)$. Hence, $I \otimes_A I = A/(x) = k \neq 0$. – Martin Brandenburg Feb 08 '13 at 19:03
  • @MartinBrandenburg: Nice, but I think that should be $I \otimes_A I \simeq I$ and not $A/I$, as $M \otimes_A A/I \simeq M/IM$... although I guess $I$ and $A/I$ are isomorphic as $A$-modules... – Jim Feb 08 '13 at 20:36
  • Yes, $I \cong A/I$. I used the formula $A/I \otimes_A A/J = A/(I+J)$. – Martin Brandenburg Feb 08 '13 at 20:40
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It is easy to come up with a bunch of counterexampless, without knowing the notion of flatness, just by looking at simple special cases.

For $M=A/J$ we have $M \otimes_A I = I/IJ$, which injects to $M$ if and only if $IJ= I \cap J$. For $I=J$ this says $I^2=I$. But of course there are ideals with $I^2 \neq I$ (for example maximal ideals in a PID, or ideals with $I^2=0 \neq I$ as in Jim's example).

Martin Brandenburg
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