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Let $A \to B$ be a surjective homomorphism between (unital) noetherian commutative rings with the same Krull dimension. Is the kernel of this map nilpotent ?

Thanks to Makoto Kato and Martin Brandenburg, it seems that the answer to the question is trivially false.

Now assume $A$ is a quotient formal power serie rings over a DVR, say $\mathbb{Z}_p$ the ring of $p$-adic integers. Can we say something ?

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Do you want to assume the Krull dimension to be finite? Probably it's more interesting then. Also one should look at the case that $A,B$ are integral domains, otherwise there are trivial counterexamples (see the answer by Makoto Kato). –  Martin Brandenburg Dec 12 '13 at 10:13
What I had in mind was the case of $A$ being a quotient of a power serie rings with coefficients in a DVR. –  user65490 Dec 12 '13 at 10:20
Power series in one variable? –  Martin Brandenburg Dec 12 '13 at 11:46
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2 Answers

Geometrically, you are asking whether a closed immersion $\operatorname {Spec} B\hookrightarrow \operatorname {Spec} A$ of affine schemes of the same dimension is surjective.
This can be seen to be false (algebraic geometry is a visual art!): just inject the $x$-axis of the plane $\mathbb A^2$ into the union of the coordinate axes of that plane $\mathbb A^2$.
Translating back to algebra, the above counterexample corresponds to the the surjective algebra morphism $A=k[X,Y]/(X\cdot Y)\to B=k[X]$ sending $\bar X\mapsto X$ and $\bar Y\mapsto 0$ .

The translation between algebra and geometry is based on the following results:
(i) A surjective ring morphism $\phi:A\to B$ gives rise to a closed immersion $f:\operatorname {Spec} B\hookrightarrow \operatorname {Spec} A$ of affine schemes
(ii) The image $\operatorname {Image}(f)$ of $f$ is the closed subset $V(\operatorname {ker}(\phi))$ .
(iii) The equality of $\operatorname {Image}(f) =V(\operatorname {ker}(\phi))$ with $\operatorname {Spec} A=V(0)$ holds if and only if $\operatorname {ker}(\phi)\subset \sqrt {(0)}=\operatorname {Nil}(A)$.

New Edit
If $A$ is a domain with finite Krull dimension, the closed immersion $\operatorname {Spec} B\hookrightarrow \operatorname {Spec} A$ is necessarily surjective since a strict closed subset $F\subsetneq \operatorname {Spec} A$ must have dimension smaller than that of $ \operatorname {Spec} A$.
Since $Nil(A)=0$ for a domain, the translation above implies that $\operatorname {ker}(\phi)\subset \operatorname {Nil}(A)=0$ , i.e. that $\phi$ is injective (as well as surjective).
In other words, a surjective morphism $A\to B$ of rings of the same finite Krull dimension is an isomorphism as soon as $A$ is a domain.

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Or just the inclusion from one point into two points :-) (see also my comment to Makoto's answer). Nice answer by the way. –  Martin Brandenburg Dec 12 '13 at 11:47
Thanks, @Martin. Yes, your comment certainly gives the simplest possible example, but I wanted to give an example with connected schemes . –  Georges Elencwajg Dec 12 '13 at 11:57
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Let $A$ be an Artinian ring which is not a local ring. Let $M$ be one of its maximal ideals. dim $A$ = dim $A/M = 0$. But $M$ is not nilpotent.

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So an explicit example is $K \times L \twoheadrightarrow K$ for two fields $K,L$. –  Martin Brandenburg Dec 12 '13 at 10:15
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