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.