Characterizing prime ideals in $S^{-1}R$ Let $R$ be a Noetherian ring, let $S$ be the set of non-zero divisors.


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*Is there a way to characterize the prime ideals in $S^{-1}R$?

*How does the answer to the above question change when we add the assumption that $R$ is reduced?

 A: Assuming you have commutative rings in mind:
There is a bijection between the set of prime ideals of the localization $S^{-1}R$ and the set of prime ideals of $R$ which do not intersect the set $S$ at which you localize. The bijection is in fact induced by map $R\to S^{-1}R$, by extending and contracting ideals.
This is true for all admissible $S$, by the way, and easy to prove.
Now: let's drop commutativity. (And keep in mind that in the non.comm. context, prime means something weaker than in the commutative context; the usual defining property of prime ideals in commutative rings defines completey prime ideals in the general context.
The statement above is still true whenever $S$ is a (right) denominator set in a right noetherian $R$. You can find the (slightly more involved proof) in the book by J.C. McConnell and J.C. Robson on Noncommutative noetherian rings, where it is Proposition 2.1.16(vii)
A: For a general multiplicative set $S$ there is nothing to add to (nor to subtract from!)  Mariano's answer.
Here are  a few remarks in the special case you're interested in: the case where $S$ is the set of non-zerodivisors of $R$. All rings here are commutative.
I) The natural morphism $R\to S^{-1}R$ is injective. Actually, $S$ is the largest multiplicative set with this property. Indeed  if a multiplicative set $T\subset R$ contains a zerodivisor $t\in T$, then  there exists  $t'\neq0 \in R$ with $tt'=0$
and $t'$ maps to zero under $R\to  T^{-1}R$, which is thus not injective.
II) The set $Zdiv(R)$  of zerodivisors of a reduced (not necessarily noetherian) ring $R$ is exactly the union of the set $Specmin(R)$ of minimal ideals of $R$:
$$Zdiv(R)=\bigcup\{\mathfrak p| \mathfrak p  \in Specmin(R) \}     $$
For non reduced rings (noetherian or not)  we still have
$$\bigcup\{\mathfrak p| \mathfrak p  \in Specmin(R) \} \subset Zdiv(R)                  $$
III) The set $Specmin(R)$ of minimal ideals of the noetherian (not necessarily reduced) ring $R$ is finite. 
IV) If $R$ is reduced and noetherian, then the ring of fractions $S^{-1}R$ (S=non zero divisors of R) is artinian and in particular $Spec(S^{-1}R)$ is finite. 
Edit At Jean's request, here is a proof of IV).
 A prime ideal of $S^{-1}R$ corresponds (cf. II) to a prime ideal   $\mathfrak p\subset \bigcup \mathfrak p_i$ of $R$ included in the union of the finitely many (cf. III) minimal primes of $R$. By prime avoidance $\mathfrak p=\mathfrak p_i$ for some $i$. This already proves finiteness of $Spec(S^{-1}R)$.
But since the $\mathfrak p_i$'s are minimal and thus incomparable  the ring $S^{-1}R$ has dimension zero. Recalling that it is noetherian (because $R$ is), it must be artinian.
Addendum (added later) The conclusion of IV) is not necessarily true for a non reduced noetherian ring. For example, take $R=k[X,Y]/(XY,X^2)=k[x,y] $ ($k$ a field).Its zero divisors coincide with the maximal ideal ${\frak m} =(x,y)$. Hence  the total ring of fractions $S^{-1}R$ of $R$ is (a little surprisingly) the localized ring $R_{\frak m}$,  which  is of dimension one and thus not artinian. 
