Apologies if this is an obvious question. I've really gotten my head tangled up in knots trying to approach it from the right angle, and I'm not getting anywhere - so I thought I'd ask.

A scheme is said to be regular in codimension 1 if every local ring $\mathcal{O}_x$ of $X$ of dimension one is regular.

Let $X$ be noetherian, integral, separated and regular in codimension 1.

Then every algebraic geometry reference ever says that if a subscheme of $X$ has codimension 1, the local ring of its generic point $\eta$ is of Krull dimension 1 $(*)$.

Nowhere seems to give this any thought, and I can't for the life of me see why it's true. Moreover, I cannot see how the dimension of the local ring of a point, that is, the dimension of a bunch of functions on the point, is in any way at all related to the dimension of the actual space it sits in.

If anyone could answer

  1. Why $(*)$ is true.

  2. Why these two things are actually related.

I'd be very grateful. Thanks!

Edit: My commutative algebra has been quite wobbly in recent times, so the more anything of this nature is spelt out, very much the better!


2 Answers 2


If Y is a closed irreducible subspace of a scheme X, $y \in Y$ the generic point, then one has always

$$ \mathrm{codim}(Y,X) = \dim(\mathscr{O}_{X,y}). $$

The closed irreducible subspaces of $X$ containing $y$ are in bijective correspondence with the closed irreducible subspaces of the local scheme $\mathrm{Spec}(\mathscr{O}_{X,y})$, hence with the prime ideals of $\mathscr{O}_{X,y}$. EGA is a reference that does have details for this (and most other things), see (EGA, IV_2, 5.1.2).

P.S. Happy belated birthday to Grothendieck!

  • 3
    $\begingroup$ Thanks for this. For future lost souls: a good reference for this is also Liu Qing's Algebraic geometry, which I discovered after I'd gotten most of the way with EGA. $\endgroup$
    – Hock
    Mar 30, 2013 at 12:39

That was also my question which Martin answered it here

Hint: Precisely, the dimension of $\mathcal{O}_{X,x}$ is the codimension of $\overline{\{x\}}$ in $X.$

  • 1
    $\begingroup$ You have just stated the fact Hock asked for a proof of. $\endgroup$
    – user314
    Mar 29, 2013 at 12:49
  • $\begingroup$ Yes, indeed, since it's more instructive to prove this fact on his own. Refering to EGA, is excessive, of course. $\endgroup$ Mar 29, 2013 at 14:42
  • $\begingroup$ You could give him a hint then, instead of just restating his question. Why would referring to EGA be "excessive"? He specifically expressed his frustration that every reference he looked at lacked a proof. $\endgroup$
    – user314
    Mar 29, 2013 at 15:21
  • $\begingroup$ Dear Adeel, true, I added "hint." I was also frustrated, since couldn't find the connection between these things. For me when I read the fact Martin gave me I started to think on my own and it was quite instructive. I believe at the moment of frustration the real development happens :) $\endgroup$ Mar 29, 2013 at 15:39
  • $\begingroup$ Hello both. I ended up doing a combination of the two, and half-proving it myself, referring to EGA and trying to decode it with my terrible broken French. As Adeel says, I was extremely frustrated as to the general approach that everyone leaves it to everyone else, and I couldn't quite get a handle on the thing by myself. I now have a proof that I think by and large works. If it's the done thing, I could also post this as response, to aid any future lost souls. Would this be right thing to do? $\endgroup$
    – Hock
    Mar 30, 2013 at 12:34

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