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.

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