# on connectness and normality

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.

• Normal implies regular in codimension 1. The converse is not true: 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. – Francesco Polizzi Mar 21 '11 at 17:07
• @Francesco Polizzi ok sorry! so why normal+connected implies irreducible?I have in mind a solution even if it seems to me there is someting wrong: if $X=Y\cup Z$ with $Y,Z$ irreducible then consider $Y\coprod Z \rightarrow X$. Is this birational?(I think so) then $X$ normal + Zariski's Main thm implies connected fibers but on the intersection ($Z\cap Y\subset X$) fibers are not connected so $X$ must be irreducible.Is it correct? – unknown Mar 21 '11 at 17:56
• @unknown (for normal+connected => irreducible), that sounds right. There is an algebraic approach which doesn't use any real machinary though. All you need to do is notice that a normal local ring must be a domain. See page 64 and Exercise 9.11 in Matsumura's commutative ring theory. – Karl Schwede Mar 21 '11 at 18:11
• @Karl Schwede thanks!!If I have understand a prof coul be the following:the normalization is defined as the disjoint union of the normalization of each irreducible component. So $X$ normal+connected implies $X$ is one of these. If I show that locally this is a $Spec(A)$ with $A$ a domain we have integral implies irreducible. By structure thm for integrally closed domain $A=A/p_1\times\dots\times A/p_r$ for $p_i$ minimal primes, but by connectness of $Spec(A)$ there is only one of these and $A$ is integral. So $X$ is integral and that's all. Is it correct? Any idea for the other questions? – unknown Mar 21 '11 at 18:36
• For your other questions, can you clarify something for me? You say 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 that this is irreducible'', do you mean you need to show that $\tilde P_0$ is irreducible or that $\tilde P$ is irreducible? – Karl Schwede Mar 21 '11 at 18:57

Here is one justification for this: suppose that a reduced ring $A$ is integrally closed in its total quotient ring (i.e. in the localization of $A$ with respect to its non-zero divisors). If $K(A)$ denotes this total quotient ring, then $K(A) = \prod K_i$ is a product of field $K_i$. Let $e_i$ denote the element of $K(A)$ which is $1$ in the $i$th place and $0$ everywhere else. Then $e_i^2 = e_i = 0$, thus $e_i$ is integral over $A$, thus $e_i \in A$ by the assumption, and so $A$ has a decomposition $A = \prod A_i$ into a product of domains corresponding to the decomposition of $K(A)$ into a product of fields.
In particular, if Spec $A$ is connected, then $A$ is itself a domain, and so Spec $A$ is in fact irreducible.