(Assuming the rings are commutative with identity, else the statement is false.)
As Norbert says, your two useful facts immediately combine to say that the set of nilpotent elements is equal to the intersection of prime ideals containing the ideal $\{0\}$, and that means all prime ideals.
First, I need to show that the nilpotent elements are in fact an ideal
Well sure, this is possible, although it is not necessary. You can find how to do this in a few other questions on the site, starting here
The set of all nilpotent elements is an ideal
and then I need to show it's prime.
That... is not in the question at all. Showing something is equal to an intersection of prime ideals does not mean it itself is prime. Most of the time, the ideal of nilpotent elements is not prime.
I just don't know how to make this proof flow and exactly what leads to what.
Ok, fair enough. The outline of the proof is: show the two sets are equal by showing each one is a subset of the other.
On one hand, each nilpotent element is contained in each prime ideal. Thus the nilpotent elements are in the intersection of primes.
The other direction is a little harder, and usually we use a lemma: every ideal maximal with respect to being disjoint from a multiplicative subset of R is a prime ideal of R. Using this, we can show that for each nonnilpotent element, there is a prime ideal disjoint from the powers of that element, and hence the prime does not contain the element. This shows how nonnilpotents are excluded from the intersection, so that everything there is nilpotent.
As a corollary, you get that the nilpotents form an ideal, and it is not really necessary to prove this ahead of time.
Yes, it is.
If $I$ is prime, then the conclusion follows easily.
For the converse, write $I=\bigcap_{j\in J} P_j$ with $P_j$ prime ideals. (This can be done since $I$ is radical.) Let's prove that $I$ is prime. Let $ab\in I$ such that $b\notin I$. If $b\notin P_j$ for all $j\in J$, then $a\in I$ and we are done. Otherwise, set $J_1=\{j\in J:b\in P_j\}$. We have $I=I_1\cap I_2$ where $I_1=\bigcap_{j\in J_1} P_j$ and $I_2=\bigcap_{j\in J-J_1} P_j$. Note that $I_1$ and $I_2$ are radical ideals, and moreover $I\subsetneq I_1$ (why?). Since $I$ can not be an intersection of two radical ideals properly containing it, it follows $I=I_2$, so $a\in I$ (why?).
Best Answer
First, we need to show that your homomorphism is surjective. Since these two ideals are relatively prime, you can find $e_{1} \in I$ and $e_{2} \in J$ such that $e_{1}+e_{2}=1$ $f(e_{1})=(e_{1}+I,e_{2}+J)=(e_{1}+I,1-e_{2}+J)=(\overline{0},\overline{1})$. You can do the similar thing to show $f(e_{2})=(\overline{1},\overline{0})$. Can you deduce surjectivity then? The kernel is $I \cap J$. We need to show $I \cap J =IJ$. Notice that $IJ \subset I \cap J$ is clear. Now let $x \in I \cap J$. Write $x=x \times 1=x(e_{1}+e_{2})=xe_{1}+xe_{2}$. Could you see what to do next ?