Quotienting by an ideal in tensor product of algebras

Question

All rings are commutative ring with unity.

Let $A$ and $B$ are two $R$-algebras and $I$ and $J$ are two ideals of $A$ and $B$ respectively. I want to show that $A\otimes_R B/(I\otimes_R B+A\otimes_R J)\cong (A/I)\otimes_R (B/J)$.

Theres is a map from $A\otimes_R B/(I\otimes_R B+A\otimes_R J)$ to $(A/I)\otimes_R (B/J)$ (as $I\otimes_R B+A\otimes_R J$ is in the kernel of the map $A\otimes_R B\rightarrow (A/I)\otimes_R (B/J))$ and this map is also surjective. But how would I show that this is also an injective map?

Thank you.

Answer 1

Well, I'd consider the canonical epimorphism

$$A\otimes_R B\rightarrow (A/I)\otimes_R (B/J): a\otimes b\rightarrow (a+I)\otimes (b+J)$$

and show that the kernel is

$$I\otimes_R B + A\otimes_R J.$$

Then the result follows from the first isomorphism theorem.

It is clear that $I\otimes_R B + A\otimes_R J$ lies in the kernel. Let $a\otimes b$ be in the kernel, i.e., $(a+I)\otimes (b+J) = 0$. Please consider When can we say elements of tensor product are equal to $0$?

Answer 2

$\require{AMScd}\DeclareMathOperator\Ker{Ker}$ Let $\alpha:A\to A/I$ and $\beta:B\to B/J$ be the canonical projections. Then $\alpha\otimes_B\beta:A\otimes_RB\to(A/I)\otimes_R(B/J)$ is surjective as composition of surjective homomorphisms. A diagram chasing on the diagram below shows $\Ker(\alpha\otimes_R\beta)=\Ker(\alpha\otimes_R1_B)+\Ker(1_A\otimes_R\beta)$. $\begin{CD} @. A\otimes_RJ\\ @. @VVV\\ I\otimes_RB@>>>A\otimes_RB@>>>(A/I)\otimes_RB@>>>\{0\}\\ @VVV @VVV @VVV\\ I\otimes_R(B/J)@>>>A\otimes_R(B/J) @>>>(A/I)\otimes_R(B/J)@>>>\{0\}\\ @VVV @VVV @VVV\\ \{0\} @. \{0\} @. \{0\}\\ \end{CD}$