# THE DATA'S ALL GONE

Sun 07 Jul 2019 10:17:24 PM EDT

$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\Zi}{\mathbb{Z} [i]} \newcommand{\Zmp}{\mathbb{Z}_p [x]/(x^2+1)} \newcommand{\Rij}{R/(I \cap J)} \newcommand{\Rx}{R/I \times R/J}$$

# Problem 1

Let $$R$$ be a ring with identity and let $$I, J$$ be ideal of $$R$$ such that $$I + J = R$$. Show that there is an isomorphism $R/(I \cap J) \cong R/I \times R/J$

## Solution

Claim. The map $$\phi: \Rij \rightarrow \Rx$$ given by $a+ I\cap J \mapsto (a+I, a+J)$ is an isomorphism.

Proof. First, we must show that $$\phi$$ is well defined. Say that we have $$a+I\cap J = b+I\cap J$$. Then $$b=a+k$$, for some $$k \in I\cap J$$. (This is true because $$a-b \in I\cap J$$.) Then apply $$\phi$$ to $$a,b$$. We obtain $$(a+I, a+J)$$ and $$(a+k+I, a+k+J)$$. Since $$k \in I$$ and $$k \in J$$, $$(a+k+I, a+k+J) = (a+I, a+J)$$. So $$\phi(a+I\cap J) = \phi(b+I\cap J)$$. Thus, the map is well defined.

To show that $$\phi$$ is an isomorphism, we must show that it is a bijective homomorphism. First, we show that $$\phi$$ is a homomorphism.

Under addition: $\phi(a+I\cap J) + \phi(b+I\cap J)$ $= (a+I, a+J) + (b+I, b+J)$ $= (a+b+I, a+b+J)$ $= \phi(a+b+I\cap J)$ $= \phi(a+I\cap J + b+I\cap J)$

Under multiplication: $\phi(a+I\cap J) \phi(b+I\cap J)$ $= (a+I, a+J) (b+I, b+J)$ $= (ab+I, ab+J)$ $= \phi(ab + I\cap J)$ $= \phi( (a+I\cap J) (b+I\cap J) )$

We must now show that $$\phi$$ is surjective and injective.