show that if ~ is a congruence on a ring $R$, then the equivalence class of $0$ is an ideal of $R$ show that if ~ is a congruence on a ring $R$, then the equivalence class of $0$ is an ideal of $R$
Im not sure where to start here. when we did groups the congruence ~ was defined as if $H \le G$ then $a \sim b$ iff $ab^{-1} \in H$
 A: Let $\sim$ be a congruence on $R$ and let $[0]=\{r\in R: 0\sim r\}$.


*

*$0\in [0]$ because $\sim$ is reflexive.

*Suppose $0\sim r$ and $0\sim s$; then $(0+0)\sim(r+s)$, so $r+s\in[0]$.

*Suppose $0\sim r$; then, since $-r\sim -r$, we have $(0-r)\sim(r-r)$ and so $-r\sim 0$, so $-r\in[0]$.

*Suppose $0\sim r$ and $s\in R$. Since also $s\sim s$, we have $0s\sim rs$ and $s0\sim sr$, so $rs\in[0]$ and $sr\in[0]$.


We have of course used that $\sim$ is an equivalence relation and that $a\sim b$ and $c\sim d$ imply $a+c\sim b+d$ and $ac\sim bd$.
A: Maybe I've somewhat figured it out.
Let $\sim$ be a congruence relation on $\langle R, + , \cdot \rangle$ and $S = [0]_{\sim}$ 
$0 \in S, S \neq \emptyset$
$0 \sim 0$
Let $x,y \in S$ then,
if $0 \sim x, 0 \sim y$ then $0+0 \sim x+y$ and $0 \cdot 0 \sim x \cdot y$ $S$ is closed
and $0 \sim x \implies 0^{-1} \sim x^{-1}$ but $0=0^{-1}$ thus $0 \sim x^{-1}$ inverses exist in $S$
further,
$0 \sim y \implies  0 - x  \sim y - x$, and $ \implies x- 0 \sim x-y \in S$  thus $\langle S, + \rangle \le \langle R, + \rangle$
$0 \sim y \implies 0 \cdot x \sim y \cdot x \implies 0 \sim y \cdot x$
That's probably a pretty rough start but maybe on the right track?
