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Given a commutative ring $A$ (with identity) and two cyclic $A$-modules, $M$ and $N$ with generators $x$ and $y$, respectively.

How do you show that $\mathrm{Ann}(x\otimes_A y) = \mathrm{Ann}(x) + \mathrm{Ann}(y)$?


It is simple to show that $\mathrm{Ann}(x\otimes y) \subseteq \mathrm{Ann}(x) + \mathrm{Ann}(y)$ by taking the acting element inside the tensor.

My attempt:

For any pure tensor $(r_1x)\otimes (r_2 y) = (r_1r_2)(x\otimes y)$, so that any general tensor can be written as $r (x\otimes y)$ and thus $M \otimes N$ is cyclic with generator $x\otimes y$. Therefore $M \otimes N \cong A / \mathrm{Ann}(x\otimes y)$

Let $I = \mathrm{Ann}(x) + \mathrm{Ann}(y)$. (They're both ideals)

I defined a homomorphism $\psi: A/I \rightarrow M\otimes N$; $r + I \mapsto r (x \otimes y)$. It is well-defined and surjective. If it were injective, then we'd have $A/I \cong M\otimes N \cong A/\mathrm{Ann}(x\otimes y)$ and because of the first inclusion the two ideals would be equal (I think).

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1 Answer 1

up vote 1 down vote accepted

The question is equivalent to the isomorphism $A/I \otimes_A A/J \cong A/(I+J)$ of $A$-modules. Actually they are also isomorphic as $A$-algebras. Beyond the usual proof which you can read everywhere, there is a one-line proof with the Yoneda-Lemma.

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Thanks! For anyone interested, there's a proof of that isomorphism here –  Felipe Apr 29 '13 at 21:17
    
$A/I$ is the initial $A$-algebra whose structural morphism kills $I$. It follows that $A/I \otimes A/J$ is the initial $A$-algebra whose structural morphism kills $I$ and $J$, i.e. $I+J$. Hence this is $A/(I+J)$. –  Martin Brandenburg Apr 30 '13 at 8:51

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