Hi,

the situation is the following: I have a projective scheme $\tilde{P}\rightarrow S=Spec(A)$ with $A$ excellent and $I$-adically complete for some ideal of $A$. A group $Y$ acting on $\tilde{P}$ freely in Zariski topology and $P$ is the quotient by $Y$ and it is proper over $S$. Moreover I know that the fiber $\tilde{P}_0$ of $\tilde{P}$ over $S_0=Spec(A/I)$ is connected. I have to prove that $P$ is irreducible. I read that up to replace $\tilde{P}$ with is normalization it can be assumed that $P$ is normal (this is the first thing I do not understand). Assuming this I read that it is enough to show that $P$ is connected. But why?does normal+connected implies irreducible? I have in mind this example: if we take k-planes, $k>2$, in a $\mathbb{P}^n$, for big n, intersecting only in the origin, this is normal (regular in codimension 1 implies normal right?) and connected but not irreducible. Last problem: I read that since $P$ proper over $S$ and $P_0$ connected then $P$ is connected too.

nottrue: for instance, two planes in $\mathbb{P}^4$ intersecting in a single point are not a normal variety. This follows by taking normalization (two disconnected plains) and then using Zariski's Main Theorem. $\endgroup$ – Francesco Polizzi Mar 21 '11 at 17:07`Moreover I know that the fiber $\tilde P_0$ of $\tilde{P}$ over $S_0 = \text{Spec}(A/I)$ is connected. I have to prove that *this* is irreducible.'' When you say`

I have to prove thatthisis irreducible'', do you mean you need to show that $\tilde P_0$ is irreducible or that $\tilde P$ is irreducible? $\endgroup$ – Karl Schwede Mar 21 '11 at 18:57