One-to-one correspondence of ideals in the quotient also extends to prime ideals? I'm beginning to learn some Grothendieck's algebraic geometry and I have a doubt about a property of commutative algebra.

For a comm. ring $A$ and an ideal $I$ of $A$, does the one-to-one correspondence between ideals of the quotient $A/I$ and ideals of $A$ containing $I$ extends to a correspondence of prime ideals ?

My guess would be no, because I never learned it and couldn't find it on internet. However, where is the mistake in :
Let $I \leq J \leq A$ be an ideal. We have $J$ prime iff $A/J \cong (A/I) / (J/I)$ integral domain iff $\bar{J} \leq A/I$ prime.
Also can someone provide a counter example if this is not true ?
Thanks !
 A: There is no mistake; the one-to-one, inclusion preserving correspondence between ideals of a ring $R$ that contain the ideal $I$ and ideals of the quotient ring $R/I$ also gives a correspondence between prime ideals.
For the commutative-with-unity case, your proof works.
If you want a proof for the more general case, and to prove a bit more:
Definition. Let $R$ be a ring; an ideal $I$ of $R$ is completely prime if and only if whenever $ab\in I$, either $a\in I$ or $b\in I$. 
Definition. Let $R$ be a ring; an ideal $I$ of $R$ is prime if and only if whenever $JK\subseteq I$, where $J$ and $K$ are ideals, either $J\subseteq I$ or $K\subseteq I$.
If $R$ is commutative, then an ideal is prime if and only if it is completely prime. The two notions are not equivalent for arbitrary rings, though. For example, in the ring of $2\times 2$ matrices over a field, the only ideals are the trivial ideal and the whole ring; in particular, the zero ideal is a prime ideal, but it is not completely prime because you can find two nonzero matrices whose product is the zero matrix.
Theorem. Let $R$ be a ring and let $I$ be an ideal of $R$. Then the natural correspondence between ideals of $R$ that contain $I$ and ideals of $R/I$ identifies prime ideals with prime ideals and identifies completely prime ideals with completely prime ideals.
Proof. Let $J$ be an ideal that contains $I$. If $J$ is completely prime, suppose that $a+I, b+I\in R/I$ are such that $(a+I)(b+I)\in J/I$. Then $ab+I\in J/I$, hence $ab\in J$ (since $I\subseteq J$); since $J$ is completely prime, either $a\in J$ or $b\in J$, so either $a+I\in J/I$ or $b+I\in J/I$. Thus, $J/I$ is completely prime. Conversely, suppose that $J/I$ is completely prime, and let $a,b\in R$  be such that $ab\in J$. Then $ab+I\in J/I$, hence either $a+I\in J/I$ or $b+I\in J/I$. If $a+I\in J/I$, then there exists $i\in I$ such that $a+i\in J$, hence $a\in J$ (since $I\subseteq J$); likewise, if $b+I\in J/I$, then $b\in J$. Thus, $J$ is completely prime.
Now suppose that $J$ is a prime ideal. Let $K/I$ and $L/I$ be ideals of $R/I$ such that $(K/I)(L/I)\subseteq J/I$, with $K/I$ corresponding to the ideal $K$ of $R$ that contains $I$, and the ideal $L/I$ corresponding to $L$. Then $(K/I)(L/I) = (KL)/I\subseteq J/I$, hence, by the inclusion-preserving correspondence, $KL\subseteq J$, so either $K\subseteq J$ or $L\subseteq J$, hence $K/I\subseteq J/I$ or $L/I\subseteq J/I$. Thus, $J/I$ is a prime ideal. Conversely, if $J/I$ is a prime ideal, let $K$ and $L$ be ideals such that $KL\subseteq L$. Then $K+I$ and $L+I$ are ideals that contain $I$, and $(K+I)(L+I)=KL+KI+IL+I^2\subseteq KL+I\subseteq KL+J=J$; therefore, $(K+I)/I$ and $(L+I)/I$ are ideals of $R/I$ whose product is contained in $J/I$, so either $K+I\subseteq J$ or $L+I\subseteq J$. But $K\subseteq K+I$ and $L\subseteq L+I$, so either $K\subseteq J$ or $L\subseteq J$. Thus, $J$ is prime. QED
A: Yes, this is a standard result. The proof you give is the same as that in Zariski and Samuel, Commutative Algebra I, S.3.8, Theorem 11. See also the slightly more general presentation on extension and contraction, p.9 in Atiyah and MacDonald, Introduction to Commutative Algebra. $\ $ I had no problem locating it by web searches, e.g. the Lemma at the bottom of p.8 Boocher's notes. 
