Tensor of quotients

Question

I would like to prove the following: let $A$ be a ring and $I,J\subset A$ two ideals. Then: $$A/I\otimes_AA/J\cong A/(I+J)$$ I have seen a proof using the Yoneda lemma (even though I haven't understood it fully yet), but I've been told there's also a proof using the short exact sequence: $$0\to I\to A\to A/I\to0$$ The only obvious thing to do here, is to apply the functor $-\otimes_AA/J$ to the sequence. This functor is right exact, and thus we get the exact sequence: $$I\otimes_AA/J\to A\otimes_AA/J\cong A/J\to A/I\otimes_AA/J\to0$$ Now my idea would be to try to show that the kernel of the homomorphism on the left is (isomorphic to) $(I+J)/J$, and then I would be done. However, I don't know how to show it.

Any advice? Alternatively, is there any other simple way to work this proof?

Generally, $A/I \otimes_A M \cong M/IM$ for every left $A$-module $M$ and every right ideal $I$ of $A$. — darij grinberg, Jan 08 '14 at 22:34

score 1 · Accepted Answer · answered Jan 08 '14 at 21:56

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The image of $I \otimes_A A/J \to A/J$ is obviously $(I+J)/J$, hence $A/I \otimes_A A/J = (A/J)/((I+J)/J) = A/(I+J)$.

answered Jan 08 '14 at 21:56

Martin Brandenburg

163,620

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Ah, hell, you're right. I have to take the image, not the kernel. Thanks. – Daniel Robert-Nicoud Jan 08 '14 at 21:58
Can you explain the obvious part to me? – T.J. Gaffney Jan 08 '14 at 21:59
Since $A \to A/J$ is surjective, also $I \to I \otimes_A A/J$ is surjective. Hence, the image equals the image of $I \to A/J$. Now it is clear, right? – Martin Brandenburg Jan 08 '14 at 22:02

Tensor of quotients

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