Critique of my proof of the structure of ideals in a quotient ring I am wondering if there are slicker ways to prove this:
If let $A$ be a commutative ring and $I \subseteq A$ be an ideal, and let $\nu : A \rightarrow \frac{A}{I}$ be the quotient homomorphism. Then $\nu$ induces a bijection between prime ideals in $\frac{A}{I}$ and prime ideals in $A$ containing $I$.
I'll just sketch my proof, since these ideas are well-known.
First, we show prime ideals in $\frac{A}{I}$ pullback under $\nu^{-1}$ to prime ideals in $A$ containing $I$. This is just straightforward application of the definitions, taking advantage of the fact that $\nu$ is a ring-hom. 
Next, I show that $\nu$ also sends prime ideals in $A$ containing $I$ to prime ideals in $\frac{A}{I}$ by pushforward. Again, this is just definitions and abusing the fact that $\nu$ is a ring-hom, and additionally, that all elements of $\frac{A}{I}$ have the form $\nu(a)$ for some $a \in A$.
All that is left, then, is to show that these two operations are inverse. 
Take $J' \subseteq \frac{A}{I}$ to be an ideal. Then $\nu \nu^{-1} J'$ is equal to $J' \cap \text{Im }\nu$. Since $\nu$ is surjective, the image is all of $\frac{A}{I}$ and so we get $\nu \nu^{-1} J' = J'$.
Finally, the last step caused me trouble for a bit before I got it.
Take $J \subset A$ to be an ideal containing $I$. If $J = I$, the matter is trivial, since $\nu I = 0 \in \frac{A}{I}$, and $\nu^{-1} 0 = I$. 
But suppose instead that $J \neq I$. Then $\tilde{J} := \nu^{-1} \nu J$ must be a superset of $J$. So we have $I \subsetneq J \subseteq \tilde{J}$. We want to show this last inclusion is actually equality.
Take $\tilde{j} \in \tilde{J}$. Since $I \neq \tilde{J}$, we know that $\nu(\tilde{j}) \neq 0$. And since $0 \in \nu(J)$, we must have an element $j \in J$ such that $\nu(j) = \nu(\tilde{j})$. This means that $\nu(\tilde{j} - j) = 0$, and thus, $\tilde{j} - j \in \text{ker } \nu = I \subseteq J$. And since $\tilde{j} - j$ and $j$ are both in $J$, so is $\tilde{j}$. And equality holds.
 A: I can advise to read the introduction of "Eléments de la géométrie algébrique" (Springer version of the IHES article EGA1) by Grothendieck and Dieudonné if you happen to read French. There it is explained how the set of prime ideals $\mathrm{Spec}(A)$ of a ring $A$ can be canonically identified with the set of equivalences classes of all "geometric points" of $A$. 
A geometric point of $A$ is simply a morphism $t$ from $A$ into a field $k$. Later, by "geometric point" one understands such a morphism with $k$ algebraically closed, but here $k$ does not need to be so. Two geometric points $t\colon A\rightarrow  k$ and $t'\colon A\rightarrow k'$ are equivalent if there are morphisms of fields $f\colon k\rightarrow K$ and $f'\colon k'\rightarrow K$ such that $f\circ t=f'\circ t'$. The correspondence between prime ideals in $A$ and equivalence classes of geometric points of $A$ is the following. If $P\subseteq A$ is a prime ideal, then the corresponding geometric point of $A$ is the morphism $t\colon A\rightarrow k(P)$, where $k(P)$ is the fraction field of the integral domain $A/P$. Conversely, if $t\colon A\rightarrow k$ is a geometric point of $A$, then the corresponding prime ideal of $A$ is $P=\mathrm{ker}(t)$. 
In fact, the motivation for considering, in algebraic geometry, the set $\mathrm{Spec}(A)$ of prime ideals $P$ of $A$ comes exactly from this. What one is interested in is the set of geometric points of $A$, since it generally is the set of solutions of a polynomial system. The set $\mathrm{Spec}(A)$ is the underlying set of the geometric object defined by $A$, as Grothendieck and Dieudonné put it.
As for your question, it now has become a tautology by considering the analogous question on sets of equivalence classes of geometric points: the map $\nu\colon A\rightarrow A/I$ induces a map from the set of equivalence classes of geometric points of $A/I$ into the set of equivalence classes of geometric points of $A$ whose kernel contain $I$. Indeed, if $t\colon A/I\rightarrow k$ is a geometric point of $A/I$, then $t\circ\nu\colon A\rightarrow k$ is a geometric point of $A$ whose kernel contains $I$. The induced map is indeed a bijection, which is a trivial matter to check.
A: The bijective correspondence between ideals (not necessarily prime) is the exact statement of the Lattice Isomorphism theorem. The contraction of prime ideals is always prime so it just remains to show that if $p \subset R$ is prime, then so is $p + I \subset R/I$ the image of $p$ under the quotient map. 
Take $ab \in p + I$ then $ab + i \in p$ for some $i \in I$. But $I \subseteq p$ so $i \in I \Rightarrow i \in p$ and thus $ab \in p$ so $a \in p$ or $b \in p$ which implies that $a$ or $b$ is in $p + I$. 
Your proof seems to be trying to combine all the steps into one. I think it'd be easier to break it up into steps as I have done. Even if you cannot use the Lattice Isomorphism theorem for rings, you can easily deduce it from groups by checking that the group isomorphism is a multiplicative map, hence gives a ring isomorphism.  
