TITLE: Show that $R/I\otimes_R R/J\cong R/(I+J)$
QUESTION [15 upvotes]: Let $R$ be a commutative ring with identity and $I$ and $J$ be ideals in $R$.

Show that $R/I\otimes_R R/J\cong R/(I+J)$ as $R$-modules.

This is what I tried.
Define a map $f:R/I\times R/J\to R/(I+J)$ as $f(\bar r, \bar s)=rs+(I+J)$.
This is a well-defined bilinear map.
By the universal property of tensor products, we get an $R$-linear map $F:R/I\otimes_R R/J\to R/(I+J)$ which sends $\bar r\otimes \bar s$ to $rs+(I+J)$.
This map is surjective.
So all we need to show is that the kernel of $F$ is $0$.
But I am unable to do this.

REPLY [5 votes]: You can also show directly that this map is injective.  First, you show that every element of $R/I \otimes_R R/J$ can be written in the form $\bar{r} \otimes \bar{1}$.  Then, suppose such an element is in the kernel of $F$.  This implies that $r \in I+J$, so that $r=x+y$ for some $x \in I$ and $y \in J$.  Then $$\bar{r} \otimes \bar{1} = \bar{x} \otimes \bar{1} + \bar{y} \otimes \bar{1} = 0 + \bar{1} \otimes \bar{y} = 0.$$