This is a very standard result of Ring Theory, but I'm having trouble
with it and I would like some clarificaiton.  Basically the result is
that the Third (Second, if you prefer) Isomorphism Theorem tells us
that we have a correspondence of Ideals between two rings R and S
using a surjective ring homomorphism.  Also however, there is a
correspondence between prime ideals.  This coorespondence of prime
ideals is what I need help with.  A clear explanation or a detailed
(read: don't wave your hands) proof is all I need to understand the
relationship.  The two statements below are the precise statements of
the problem.

Note: f^-1(S) is the inverse image of f

Let f:R->S be a surjective ring homomorphism of commutative rings.

Prove or Explain: If I is a prime ideal of R such that I is contained
in the Ker(f) then f(I) is a prime ideal.

Prove or Explain: If J is a prime ideal of S then f^-1 is a prime
ideal of R which contains the Ker(f).

Further Note: The Second Isomorphism tells us that for a. f(I) is an
ideal all you need to do is prove primeness.  For b. the Second
Isomorphism Theorem gives us the containment, all is needed is to
prove primeness.

I hope this is clear.  If not, feel free to ask questions.

The first question is poorly formulated. Any ideal or set contained in
the kernel maps to {0}, which of course is a prime ideal.

Change "contained in" to "contains". Then it becomes a non-trivial
true statement. The image of an ideal I containing the kernel is an
ideal. To show f[I] is prime is not hard: Take f(a) in f[I] (that is,
a is in I) and f(b) in S. Then f(a)f(b)=f(ab) is in f(I) since ab is
in I.

Regarding the second question, the inverse image of any ideal J is an
ideal I containing the kernel. This proof is straightforward. To show
I is prime if J is, take a in I and b in R. Then f(a) is in J and so
f(ab)=f(a)f(b) is in J. So ab is in I.

That answered my question berkeleychocolate.  If you will post that as
an answer then I will give you credit for the question.

Thanks, but unless google's policy has changed recently since I am not
on the official answer list I can only make comments. By the way, my
comment can be improved. It wasn't quite right. Here's the correction:

That f[I] is an ideal follows since f(ab)=f(a)f(b), if a is in I then
ab is in I as well. So since f(a) is in f(I),  f(a)f(b) is in f(I) as
well. Show f(I) is prime as follows: Given f(ab)=f(a)f(b) in f(I). We
must show f(a) or f(b) is in f(I). So f(ab)=f(c) for some c in I. So
abc^-1 is in kerf and therefore in I. Since I is prime, a or b or c^-1
is in I. If c^-1 is in I then so is ab since abc^-1 is in I. So again
since I is prime, a or b is in I. So f(a) or f(b) is in f[I].

The correction of the second statement is similar but even easier: If
ab is in f^-1[J], then f(ab)=f(a)f(b) is in J. Since J is prime,
either f(a) or f(b) is in J. So a or b is in f^-1[J}. so f^-1[J] is
prime.

Hope this is clear.