Prime ideals of a localization I have to answer the following question:

Compute the prime ideals of the localization of $\mathbb Z$ given by the multiplicative set of the powers of $n$ (such that $n>1$) and find the nilradical of this ring.

I have thought this:
Let $\mathbb Z _n$ be such localization. I will use the following theorem:

Let $S$ be a multiplicative set. There exists a bijection between the prime ideals $P$ of $A$ such that $P\cap S = \emptyset$ and the prime ideals of $A_S$.

Let $P=(p)$ (with $p$ prime number) be a prime ideal of $\mathbb Z$, then it is bijective to a prime ideal of $\mathbb Z_n$ only if $n\notin (p)$.
Is this reasoning right? How can I approach the part about the nilradical?
 A: Your might translate your answer to the first part, in a more explicit way, as 

The bijection maps the primes non-divisors of $n$ to the prime ideals of $\mathbf Z_n$.

For the nilradical question, remember the localisation of an integral domain is an integral domain.
A: Theorem: if $\varphi$ is a locatization morphism $A\to A_S$, then $spec \varphi$ is a homeomorphism from $spec(A_S)$ onto the subspace $\{p\in SpecA; p\cap S= \phi$ of $spec A \}$. (Q.Liu, Algebraic geometry, p28 )
Theorem: Let $X$ affine scheme. $X$ is a reduced affine scheme $\Leftrightarrow \sqrt{(0)} =(0) $. (Robin Hartshorn, Algebraic geometry, p82)
We know $\mathbb Z$ is an integral domain, then $\mathbb Z_S$ is also an integral domain. Hence $X=spec \mathbb Z_S$ is an integral affine scheme, then  $X=spec \mathbb Z_S$ is reduced and irreducible. i.e. $\sqrt{(0)} =(0)$  
A: Your proof for the first part looks fine, although you might be a little more explicit about enumerating the prime ideals of $\mathbb{Z}_n$. 
For the second part, it might help to remember that the nilradical of a commutative ring with unity is the intersection of all the prime ideals.
