Ideal $\Leftrightarrow$ ~ is a congruence, for $I \subseteq R$ we define $\sim_{I}$ on $R$: $a,b \in R, a \sim_{I} b$ means $a-b \in I$. Ideal $\Leftrightarrow$ ~ is a congruence.
for $I \subseteq R$ we define $\sim_{I}$ on $R$: $a,b \in R, a \sim_{I} b$ means $a-b \in I$.
"$\Rightarrow$"
Suppose $I$ is an ideal on $R$ which implies $\langle I, + \rangle  \le \langle R, + \rangle$
Reflexive:
$a \in I, r \in R, \,\,a \cdot r \in I \implies a - a \in I \implies a \sim_I a$
Symmetric:
Let $a \sim_I b$ then $a-b \in I$
$a, -b \in I \implies -a,b  \in I$, $-a+b \in I \implies b-a \in I \implies b \sim_I a$
Transitive:
Let $a \sim_I b, b \sim_I c$ then $(a-b),(b-c) \in I$ and $I$ is an additive subgroup thus $(a-b)+(b-c) \in I \implies a+(-b+b)-c \in I \implies a-c \in I \implies a \sim_I c$
"$\Leftarrow$"
suppose $\sim_I$ is a congruence on $R$
Additon compatible:
let $a \sim_I b, c \in R$,
$a-b \in I \implies a - 0 - b \in I \implies a-c+c-b \in I \implies \\ (a-c) -(-c+c) \in I \implies (a-c) -(b-c) \in I \implies (a-c) \sim_I (b-c)$
let $b \sim_I a, c \in R$,
$b-a \in I \implies  b + 0 -a \in I \implies -a + 0 + b \in I \implies -a +c - c +b \in I \implies \\ (-a+c)-(c-b) \in I \implies (c-a) - (c-b) \in I \implies (c-a) \sim_I (c-b)$
Multiplication compatible:
this is where I get a bit messed up. $R$ is a ring so I know left/right distributivity holds.
I need to have $ac\sim_I bc$ and $ca\sim_I cb$
For $a \sim_I b, c \in R$
$a-b \in I$ but how do I make the leap to $(a-b)c = ac-bc \in I$ ?? 
 A: Assuming that both addition and multiplication respect the congruence (mod $I$). 
Then we have $a \sim_I 0$ implies $ra \sim_I r0=0$ (and $ar \sim_I 0r=0$). Thus if $a \in I$ (i.e. $a \sim_I 0$) and $r \in R$ then $ra$ and $ar \sim_I 0$ so $ra,ar \in I$. 
Showing $I$ is closed under subtraction is similar: $a,b \in I$ implies $a \sim_I 0$ and $b \sim_I 0$ so $a-b \sim_I 0-0=0$ thus $a-b \in I$.
Thus $I$ is an ideal.
A: If $I$ is an ideal, it's an additive subgroup of $R$, and subgroups of abelian groups are always normal, so $\sim_I$ is an additive congruence (note that this also means that $\sim_I$ is an equivalence relation, from earlier results on groups).
Now we are given that for any $x \in I$ and any $r \in I$, both $xr,rx \in I$, since $I$ is an ideal. So suppose $a \sim_I b$, and $c \in R$. By our definition (of $\sim_I$) $a - b \in I$, and since $I$ is an ideal, $c(a - b) = ca - cb \in I$. Similarly, $(a-b)c = ac - bc \in I$. This shows that $ca \sim_I cb$ and $ac \sim_I bc$, so that $\sim_I$ is, in fact, a ring congruence.
This is the "other half" of Bill Cook's reply.
