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For commutative ring with unit $A$, ideals $I, J$ it holds

$$A/ I \otimes_A A/J \cong A/(I+J).$$

A proof can be found here (Problem 10.4.16) for example.

However, I'd be interested in a less explicit proof of this fact. Does anybody know a nice way to do this, for example with the universal property or isomorphism theorems?

Thank you.

So (after some minutes of thinking by myself) here would be my approach:

$A/ I \otimes_A A/J \cong (A/ J) / ( I \cdot A/J) \cong (A/J) / ((I+J)/J) \cong A/(I+J)$

Everything okay? (using isomorphism theorem and $A/I \otimes M \cong M/IM$)

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Yes, that's true of course. Just a little stupidity when writing it down, as the iso. $I \cdot A/J \cong I \cap J / J$ somehow seemed more obvious to me. –  Louis Jul 15 '13 at 21:58
    
Your proof is correct. –  Martin Brandenburg Jul 16 '13 at 6:31
    
Note that this is a special case of math.stackexchange.com/questions/123543 –  Martin Brandenburg Jul 16 '13 at 11:16

2 Answers 2

up vote 2 down vote accepted

I'm assuming you mean this isomorphism as an isomorphism of $A$-algebras (I'm going to assume $A$ is commutative so this makes sense). Then $A/I$ has the following universal property: for any $A$-algebra $B$ such that the elements of $I$ map to zero under the structural map $A\rightarrow B$, there is a unique $A$-algebra map $A/I\rightarrow B$. Similarly for $A/J$. The tensor product $(A/I)\otimes_A(A/J)$ is the coproduct in the category of $A$-algebras. This means that an $A$-algebra map out of the tensor product to an $A$-algebra $B$ is the same as a pair $(\varphi,\psi)$ with $\varphi:A/I\rightarrow B$ and $\psi:A/J\rightarrow B$, $A$-algebra maps. Notice that for an $A$-algebra $B$, there is at most one $A$-algebra map $A/I\rightarrow B$, and likewise for $A/J$. So such a pair, if it exists, is unique, and it exists if and only if the structural map $A\rightarrow B$ kills $I$ and $J$, i.e., if and only if it kills $I+J$. This is the condition that characterizes the $A$-algebra $A/(I+J)$ (i.e. its universal property). So $A$-algebra maps $(A/I)\otimes_A(A/J)\rightarrow B$ and $A$-algebra maps $A/(I+J)\rightarrow B$ are ``the same" for all $A$-algebras $B$. Then there is a unique $A$-algebra isomorphism $(A/I)\otimes_A(A/J)\cong A/(I+J)$ compatible with the natural maps from $A/I$ and $A/J$ to source and target. The uniqueness implies that this isomorphism (and its inverse) are the natural maps one would write down if one wanted to prove these two $A$-algebras were isomorphic.

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A proof in 4 hints:

Define a map $f : \left(A/I\right) \times \left(A/J\right) \to A/\left(I+J\right)$ by

$f\left(a+I, b+J\right) = ab+I+J$ for all $a\in A$ and $b\in A$.

1) Prove that this $f$ is well-defined and $A$-bilinear.

Hence, by the universal property of the tensor product, this $f$ induces a linear map $f' : \left(A/I\right) \otimes_A \left(A/J\right) \to A/\left(I+J\right)$ satisfying

$f'\left(\left(a+I\right) \otimes_A \left(b+J\right)\right) = ab+I+J$ for all $a\in A$ and $b\in A$.

Define a map $g : A/\left(I+J\right) \to \left(A/I\right) \otimes_A \left(A/J\right)$ by

$g\left(a+I+J\right) = \left(a+I\right) \otimes_A \left(a+J\right)$ for all $a\in A$.

2) Prove that this map $g$ is well-defined and linear.

3) Prove that $f'\circ g = \operatorname*{id}$. (This is very easy.)

4) Prove that $g\circ f' = \operatorname*{id}$. (Here it is best to use the universal property of the tensor product again, this time its uniqueness part. Alternatively, recall that a tensor product is generated by pure tensors, so that two linear maps from a tensor product must be identical if they are equal on every pure tensor. And we can compute the value of $g\circ f'$ on pure tensors directly.)

From 3) and 4), it follows that the maps $f'$ and $g$ are mutually inverse, so that $\left(A/I\right) \times \left(A/J\right) \cong A/\left(I+J\right)$ as $A$-modules.

Note that $A$ does not have to be commutative as long as $I$ and $J$ are two-sided ideals.

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1  
Thanks for your answer, but please notice that I already knew this proof (it was linked in the question) and asked for a less explicit approach. –  Louis Jul 15 '13 at 21:33
    
Oh, I thought that by "explicit" you meant the simple-tensor argument in part (a) of the problem. (That's avoided by my proof, so it's not the one you linked.) –  darij grinberg Jul 15 '13 at 22:13

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