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 $X$?


2 Answers 2


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Jun 30, 2011 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).

  • $\begingroup$ I hadn't read Pete's answer while writing mine: the similarity is remarkable! $\endgroup$ Jun 30, 2011 at 10:19
  • $\begingroup$ it is pretty darned close, isn't it? +1. $\endgroup$ Jun 30, 2011 at 10:22
  • $\begingroup$ Yes, Pete. And as far as I'm concerned, I'm quite flattered that my answer is so close to yours:+1 too! $\endgroup$ Jun 30, 2011 at 10:26
  • $\begingroup$ And I would sure love to accept both your answers, but I can't ... hence, I flipped a coin. $\endgroup$ Jun 30, 2011 at 11:26
  • $\begingroup$ Thanks, rattle, that is nicely put. And you fairly lifted the indeterminacy... $\endgroup$ Jun 30, 2011 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.