## The tensor product of the quotients of a commutative ring by two ideals is isomorphic to the quotient by their sum

Let $R$ be a commutative ring with ideals $I$ and $J$. Let $R/I$ and $R/J$ be $R$-modules (in fact $(R,R)$-bimodules) in the usual way.

1. Prove that every element of $R/I \otimes_R R/J$ can be written as a simple tensor of the form $(1 + I) \otimes (r + J)$.
2. Prove that $R/I \otimes_R R/J \cong_R R/(I+J)$.

We will prove the first result first for simple tensors; the extension to arbitrary sums of tensors follows by tensor distributivity. Let $(a+I) \otimes (b+J)$ be an arbitrary simple tensor in $R/I \otimes_R R/J$. Now $(a+I) \otimes (b+J) = (1+I)a \otimes (b+J)$ $= (1+I) \otimes a(b+J)$ $= (1+I) \otimes (ab+J)$, as desired.

Now define $\varphi : R/I \times R/J \rightarrow R/(I+J)$ by $(a+I, b+J) \mapsto (a+b) + (I+J)$. First, suppose $a_1-a_2 \in I$ and $b_1-b_2 \in J$. Then $b_1(a_1-a_2) \in I$ and $a_2(b_1-b_2) \in J$, so that $a_1b_1 - a_2b_2 \in I+J$. Thus $\varphi$ is well-defined. It is clear that $\varphi$ is $R$-balanced, and so induces an $R$-module homomorphism $\Phi : R/I \otimes_R R/J \rightarrow R/(I+J)$. Since $\varphi((1+I) \otimes (r+J)) = r + (I+J)$, $\Phi$ is surjective. Now suppose $(1+I) \otimes (r+J)$ is in the kernel of $\Phi$; then $r \in I+J$. Say $r = a+b$ where $a \in I$ and $b \in J$. Then $(1 + I) \otimes (r + J) = (1+I) \otimes (a+J) + (1+I) \otimes (b+J)$ $= (a+I) \otimes (1+J) + (1+I) \otimes (b+J)$ $= 0+0 = 0$, and hence $\Phi$ is an isomorphism.