Let $I,J$ be two ideals of Noetherian ring $R$. How to compute kernel of following homomorphism directly: $$\phi: R/I\oplus R/J\to R/(I+J) $$ $$(a+I,b+J)\to (a-b)+I+J $$
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Hint: the sequence
$$0 \to R/(I \cap J) \xrightarrow{\Delta} R/I \oplus R/J \xrightarrow{\phi} R/(I+J) \to 0$$
where $\Delta(a + (I \cap J)) = (a + I, a + J)$, is exact.
Edit: To see exactness, let $(a+I,b+J) \in \ker \phi$. Then $a - b \in I + J$, so $a - b = i + j$ for some $i \in I, j \in J$. Then $a - i = b + j$, so $\Delta((a-i)+(I \cap J)) = ((a-i)+I, (b+j)+J) = (a+I,b+J)$.
zcn
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I know this SES but I wanna compute kernel of $\phi$ directly(by definition). – Stella Feb 01 '14 at 08:25
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Are you trying to show that the sequence is exact? – zcn Feb 01 '14 at 08:32
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@MartinBrandenburg: It's not hard to check that this is exact. Was that worth a downvote? – zcn Feb 02 '14 at 21:25
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Apparently we have posted the proof simultaneously. Sorry for that. – Martin Brandenburg Feb 02 '14 at 21:30
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The homomorphism $$i : R/(I \cap J) \to R/I \times R/J, ~ a+(I \cap J) \mapsto (a+I,a+J)$$ is clearly injective and factors through the kernel of $\phi$. Conversely, let $(a+I,b+J)$ be in the kernel of $\phi$, i.e. $a+I+J=b+I+J$. Choose $i \in I$, $j \in J$ such that $a=b+i+j$, or $c:=a-i=b+j$. Then $(a+I,b+J)=(c+I,c+J)$ lies in the image of $i$.
This shows that $i$ is a kernel of $\phi$.
By the way, this also says that we have an isomorphism of $R$-algebras $$R/(I \cap J) \cong R/I \times_{R/(I+J)} R/J.$$ For $I+J=R$ we recover the Chinese Remainder Theorem.
Martin Brandenburg
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