Show that $R/I\otimes_R R/J\cong R/(I+J)$

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.

Solution 1:

Consider the map $g: R/(I+J)\to R/I \otimes_R R/J$ given by $\overline{a} \mapsto \overline{a} \otimes 1$. If you confirm for yourself that this is a well-defined function, it's obvious that it's an inverse to your map $f$.

Solution 2:

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 $$\bar{r} \otimes \bar{1} = \bar{x} \otimes \bar{1} + \bar{y} \otimes \bar{1} = 0 + \bar{1} \otimes \bar{y} = 0.$$