How to prove "ideal $I$ is prime iff $A/I$ is a integral domain "? $A$ is a commutative ring with identity.
$I$ is a ideal of $A$.
then ideal $I$ is prime iff $A/I$ is a integral domain.
here is what I thought
$(\Rightarrow)$ We want to prove $A/I$ is a integral domain. It's equivalent to prove there are no nonzero element $a+I$ can divide $0$ such that $a\in A$. Let $a,b\in A$. Then $(a+I)(b+I)=ab+I$. Let $ab+I=I$, with $a+I\neq I$. So we need prove $b=0$ if $a\neq 0$ and $ab=0$. It's equivalent to prove $A$ is a integral domain.
$A$ is a integral domain iff the zero ideal is prime.
I stuck at here and doubt something is wrong.
($\Leftarrow$) 
$A/I$ is a integral domain $\iff$ A is a integral domain
then I have no ideal how to prove.
 A: You started off well and then got a little bit lost towards the end. By definition, $A/I$ is an integral domain if and only if the following statement holds.

If $a,b\in A$ are such that $(a+I)(b+I) = I$, then either $a + I = I$ or $b+I = I$. Also, we require $A/I\neq 0$, or in other words, $I\subsetneq A$.

Let's first make sure we know what the statement $(a+I)(b+I) = I$ means. You correctly point out that $(a+I)(b+I) = ab + I$. Thus it means $ab +I = I$, or equivalently, that $ab\in I$. Similarly, the statement $a+I = I$ is equivalent to $a\in I$, and $b+I = I$ is equivalent to $b\in I$. So we can rewrite the boxed statement above as 

If $a,b\in A$ are such that $ab\in I$, then either $a\in I$ or $b\in I$. Also, we require $A/I\neq 0$, or in other words, $I\subsetneq A$.

Notice, this last boxed statement is exactly the definition of $I$ being a prime ideal of $A$. In this way we see that $I$ being prime is equivalent to $A/I$ being an integral domain.

EDIT: Just one additional comment. I should emphasize that the statement $A$ is an integral domain $\iff A/I$ is an integral domain is false. For instance, if $A = \mathbb{Z}$ and $I = 4\mathbb{Z}$, then $A$ is an integral domain, but $A/I = \mathbb{Z}/4\mathbb{Z}$ is not an integral domain!
A: Let $\mathcal{R}$ be our ring in question.
Suppose that $a,b\in\mathcal{R}$. Then $(a+I)(b+I)=0+I$ if and only if $ab\in\mathcal{R}$. Thus, if $\mathcal{R}/I$ is an integral domain, then $(a+I)(b+I)=0+I$ means that either $a+I=0+I$ or $b+I=0+I$. That is, either $a$ or $b$ lies in $I$, which shows that $I$ is prime.
The converse can then be proved by a similar argument.
