# 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.

• Construct instead an inverse. Aug 7, 2015 at 23:45

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 $f$.
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 $$\bar{r} \otimes \bar{1} = \bar{x} \otimes \bar{1} + \bar{y} \otimes \bar{1} = 0 + \bar{1} \otimes \bar{y} = 0.$$
• Why should every element be of the form $\bar{r}\otimes \bar{1}$? Apr 21, 2019 at 6:18
• Write down an arbitrary element, which is of the form $\sum_{i=1}^n \overline{r_i} \otimes \overline{s_i}$, then use the properties of the tensor product to reduce it to the form $\bar{r} \otimes \bar{1}$. Apr 21, 2019 at 12:13