Quotienting by an ideal in tensor product of algebras 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.  
 A: 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 $0$?
A: $\require{AMScd}\DeclareMathOperator\Ker{Ker}$
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 $\Ker(\alpha\otimes_R\beta)=\Ker(\alpha\otimes_R1_B)+\Ker(1_A\otimes_R\beta)$.
$\begin{CD}
@. A\otimes_RJ\\
@. @VVV\\
I\otimes_RB@>>>A\otimes_RB@>>>(A/I)\otimes_RB@>>>\{0\}\\
@VVV @VVV @VVV\\
I\otimes_R(B/J)@>>>A\otimes_R(B/J) @>>>(A/I)\otimes_R(B/J)@>>>\{0\}\\
@VVV @VVV @VVV\\
\{0\} @. \{0\} @. \{0\}\\
\end{CD}$
