I am going through the following proof and I need some explanation

Thm: Let A be a commutative ring, an ideal $\displaystyle P\subset A$ is prime iff $\displaystyle A/P$ is an integral domain.

Proof:

$\displaystyle (\Rightarrow)$ Assume that P is prime and pick $\displaystyle a+P, b+P\in A/P$ suppose that $\displaystyle (a+P)(b+P)=0+P$ is a zero in $\displaystyle A/P$

we need to show that either $\displaystyle a+P=0+P \wedge b+P=0+P$

we have now $\displaystyle (a+P)(b+P)=ab+P=0+P$ so $\displaystyle ab\in P$(how do we know that???) $\displaystyle a,b\in A$ right? then by def of prime ideal $\displaystyle a\inP \wedge b\in P$ thus $\displaystyle a+P=0+P\wedge b+P=0+P$ no zero divisors and $\displaystyle A/P$ is int domain

$\displaystyle (\Leftarrow)$ suppose $\displaystyle ab\in P$ then $\displaystyle (a+P)(b+P)=ab+P=0+P$ as we have no zero divisors in $\displaystyle A/P$ we get $\displaystyle a+P=0+P \wedge b+P=0+P$ so $\displaystyle a\in P\wedge b\in P$

why does $\displaystyle ab+P=0+P$ implies that $\displaystyle ab\in P$? and

why does $\displaystyle ab\in P$ implies that $\displaystyle ab+P=0+P$?

thanks for the explanaition