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Consider a separated, reduced scheme $X$ of finite type over some algebraically closed field $\Bbbk$. Let $X=X_1\cup\cdots X_r$  be its irreducible components, each of which is then a variety. Assume that $P\in X_i \cap X_j$  for $i\ne j$. I then wonder when $X$ is singular in $P$ - is this always the case or do I need some additional assumption? Also, can I weaken the assumptions on $X$?

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up vote 7 down vote accepted

For any Noetherian scheme $X$, if $P$ is a nonsingular point of $X$ then the local ring at $P$ is UFD. In particular it's a domain (which is not entirely trivial to prove!). But where two irreducible components cross the local ring will not be a domain, as you will have more than one minimal prime ideal.

Maybe you don't even need Noetherian: I guess if the local ring at $P$ is not Noetherian then by definition the point cannot be nonsingular. But for some reason I am uncomfortable calling it a "singular point": it just seems out of bounds of the singular / nonsingular dichotomy.

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I like your description "out of bounds of the singular/ nonsingular dichotomy" very much. It is definitely subtler than my comment that regularity "doesn't make much sense" in the non-noetherian case. – Georges Elencwajg Jun 30 '11 at 10:35

Your problem is local at $P$, so it suffices to study the local ring $R=\mathcal O_{X,P}$ of your scheme at $P$. The irreducible components passing through $P$ correspond to the minimal primes of $R$. If $R$ is regular it is a domain: this is not trivial and is proved for example in Matsumura's Commutative Ring theory, theorem 14.3, page 106 . So $R$ has zero as unique minimal prime ideal and your scheme is locally irreducible at $P$.

1) If $R$ is not reduced it will definitely not be regular, so you can assume your scheme is reduced.
2) That $X$ is a $k$-scheme for some field $k$ is irrelevant.You should just assume that $X$ is locally noetherian, so that $R$ is a noetherian local ring (else the notion of regular doesn't make much sense).

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I hadn't read Pete's answer while writing mine: the similarity is remarkable! – Georges Elencwajg Jun 30 '11 at 10:19
it is pretty darned close, isn't it? +1. – Pete L. Clark Jun 30 '11 at 10:22
Yes, Pete. And as far as I'm concerned, I'm quite flattered that my answer is so close to yours:+1 too! – Georges Elencwajg Jun 30 '11 at 10:26
And I would sure love to accept both your answers, but I can't ... hence, I flipped a coin. – Jesko Hüttenhain Jun 30 '11 at 11:26
Thanks, rattle, that is nicely put. And you fairly lifted the indeterminacy... – Georges Elencwajg Jun 30 '11 at 12:15

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