A basic example in commutative algebra: Let $A$ $B$ be noetherian rings, with $B$ simple noetherian. Suppose that for every element $b$ in $B$, there exists a power $b^{n}$ belonging to the image of the map. Is the nilradical in $A$ trivial ? Can one say something about prime ideals in $A$?

$\begingroup$ What do you mean by a simple Noetherian commutative ring? A field? $\endgroup$– Simon WadsleyDec 7, 2011 at 17:00

$\begingroup$ If the kernel of the map is additionally contained in the radical of A, then the map is called a "Fisomorphism". It induces an bijection from Spec(B) on Spec(A). $\endgroup$– RalphDec 7, 2011 at 18:36

$\begingroup$ By a simple commutative ring i mean with trivial nilradical. Simon, some reference on this? $\endgroup$– user19792Dec 7, 2011 at 19:17

$\begingroup$ What if $B$ is $A_\text{red} = A/nil(A)$? $\endgroup$– Graham LeuschkeDec 7, 2011 at 19:36

$\begingroup$ Graham, I actually would like conditions for this to be the case. $\endgroup$– user19792Dec 7, 2011 at 19:58
1 Answer
In general, the nilradical of $A$ isn't trivial. As an example let $B =k$ be a field an $A=k[X]/(X^2)$ be the exterior algebra over $k$. Then $f: A \to k, X \to 0$ is surjective and the radical of $A$ is $(X)$.
Concerning the primes: Let $f:A \to B$ be a homomorphism of rings such that for each $b \in B$ a power of $b$ is in the image of $f$. Then there is an injection $$\text{Spec}(B) \to \text{Spec}(A), P \mapsto f^{1}(P)$$ Proof: $f$ always induces a mapping between the spectra. So it's enough to show injectivity: Let $P,Q$ be primes of $B$ with $f^{1}(P) = f^{1}(Q)$. Let $b \in Q$. There is $n > 0$ and $a \in A$ such that $f(a)=b^n \in Q$. In particular, $a \in f^{1}(Q) = f^{1}(P)$. Thus $b^n=f(a) \in P$ and since $P$ is prime, $b \in P$. Hence $Q \subseteq P$. By symmetry $P \subseteq Q$ holds as well.