How can I prove that $\ cong(R)‎\cong‎ Id(R)$?

Question

If $R$ is an arbitrary ring, $\ cong(R)$ is the set of all congruence of $R$ and $Id(R)$ is the set of all Ideals of $R$, How can I prove that $\ cong(R)‎\cong‎ Id(R)$?

Answer 1

Suppose I have a congruence $\sigma$ on $R$. Let $I_\sigma$ be the $\sigma$-class of $0$; it is not hard to see that $I_\sigma$ is an ideal (exercise).

Now, suppose $\sigma, \sigma'$ have the same $0$-class, that is, $I_\sigma=I_{\sigma'}$. Then for $(a, b)\in\sigma$, we have $(a-b, 0)\in\sigma$, so $(a-b, 0)\in\sigma'$ since $I_\sigma=I_{\sigma'}$ - but that means $(a, b)\in\sigma$! So $\sigma'\subseteq\sigma$, and similarly $\sigma\subseteq\sigma'$; that is, $\sigma=\sigma'$.

This means the map $\sigma\mapsto I_\sigma$ is injective. All that remains to do is show that it is surjective. To do this, I need to associate to each ideal $I$ a congruence $\sigma$ such that $$I=I_\sigma.$$ Is there a natural congruence relation you can associate to an ideal? (HINT: quotient . . .)