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Let $A$ be a complete regular local noetherian ring of dimension $d>1$ and $B$ an $A$-algebra, finite and free as $A$-module. Assume moreover that there exists an open subset $U$ of $\textrm{Spec}\ A$ of primes of height $d-1$ such that there is exactly one prime in $\textrm{Spec}\ B$ over a prime of $U$. Is it then true that $\textrm{Spec}\ B$ is an irreducible topological space?

Under various choices of ancillary hypotheses, results like this one follow from Zariski's connectedness principle; see e.g A useful form of principle of connectedness.

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up vote 3 down vote accepted

Yes. Assume not. Since $B$ is finite flat over $A$, each irreducible component of $\mathrm{Spec}(B)$ is pure $d$-dimensional, finite and surjective over $\mathrm{Spec}(A)$. If $Y\neq Z$ are two of them, then the image of $Y\cap Z$ in $\mathrm{Spec}(A)$ is a proper closed subset, hence does not contain $U$. Every point of $U$ has a preimage in $X$ and one in $Y$.

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Merci beaucoup. – Olivier Oct 5 '10 at 13:17

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