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Given a non zero element $x \in M$ and the nonzero submodule $M' = Ax$ of a commutative ring with unity $A$, and B a flat $A$-module, I'm trying to see that as I can show that $$ (\operatorname{Ann}(x))^e \subset m^e \subsetneq B $$ where $m$ is a maximal ideal in $A$, then $$ M'_B \simeq A/\operatorname{Ann}(x) \otimes_A B \simeq B/ (\operatorname{Ann}(x))^e \neq 0 $$

But although I can intuitively see why this would be true I can't show the last isomorphism, is it something trivial? I'm studying through Atiyah and Macdonald but I don't recall seeing such a property.

mark
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Daniel Moraes
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    Does $\operatorname{Ann}(x)^e$ here indicate the tensor product of the annihilator with $B$? Or did you possibly mean that $B$ is a ring which is flat as an $A$-module and this is the usual extension of scalars? – Elie Belkin Jul 27 '21 at 20:50
  • In the case $B$ is an $A$ algebra, so i guess this is the case of the tensor product – Daniel Moraes Jul 28 '21 at 12:56

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Throughout this answer I am assuming that by $\operatorname{Ann}(x)^e$ you mean the product of $\operatorname{Ann}(x)$ and $B$. If so, then this is true for an arbitrary ideal and module. See Atiyah-Macdonald, Chapter 2, Exercise 2 if you would like to prove it for yourself. I will construct an alternative proof from the usual one utilizing a nice fact about flat modules.

To show this, we can take the exact sequence $$0 \to \operatorname{Ann}(x) \to A \to A/\operatorname{Ann}(x) \to 0$$ and tensor it with $B$ to get an exact sequence (by flatness) $$0 \to \operatorname{Ann}(x) \otimes_A B \to B \to A/\operatorname{Ann}(x) \otimes_A B \to 0$$

Thus $B/(\operatorname{Ann}(x) \otimes_A B) \cong A/\operatorname{Ann}(x)\otimes_A B$, and it suffices to construct an isomorphism of $\operatorname{Ann}(x) \otimes_A B$ and $\operatorname{Ann}(x)B$. We can do this by taking the inclusion map of our ideal into $A$ and tensoring by $B$, to get a map $\operatorname{Ann}(x) \otimes_A B \to B$ sending $a \otimes_A b$ to $ab$. It is injective because $B$ is flat, and the image of this map is precisely $\operatorname{Ann}(x) B$, giving the desired isomorphism.

Elie Belkin
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  • Oh i missed that it was direct, i was trying to think on it as a ring, but your comment already made me realize it was indeed direct. Anyways this was still a very nice proof thank you! – Daniel Moraes Jul 28 '21 at 12:58