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I was studying Algebraic Geometry and I found the following result:

If $X$ is a normal variety, the set of singular points $Sing(X)$ has codimension $\geq 2$.

I understand this result and its proof, however my teacher said that this is not only a inequality but an equality, that is $\text{codim}(Sing(X))=2$. I do not undestand why this is true and I have not found this result in any book, which is weird because every book on algebraic geometry proves the first inequality.

I would be grateful if you can explain me why this is true, or give me a counterexample in case it is not true.

Marcos
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    Well, if $X$ is smooth, $\operatorname{Sing}(X) = \emptyset$ so I don't know if $\operatorname{codim} \operatorname{Sing}(X) = 2$ makes any sense. – red_trumpet Oct 19 '21 at 07:19
  • @red_trumpet indeed it doesn't - the codimension of the empty set is often taken to be $\infty$ in order to make our intuition about "codimension adds under intersections" work. Good easy counterexample! – KReiser Oct 19 '21 at 07:25

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The first statement, i.e. that the codimension of the singular locus is at least two on a normal variety, is correct. It might be the case that someone said "the singular locus is codimension two" with an implied "or more" at the end - I know sometimes when I'm talking fast in person I can sometimes drop statements like that without meaning to.

For an example to see that the second statement cannot hold as written, consider the subvariety of $\Bbb A^n$ cut out by $x_1^2+\cdots+x_m^2$. When $n\geq m\geq 3$, this is normal, so taking $n=m=10$ for instance we find a 9-dimensional normal variety with singular locus a single (closed) point. That's definitely not of codimension two.

KReiser
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  • Thanks, you are absolutely right. Somehow I was expecting the result to be true and me to be stupid not being able to understand it... But I guess that teachers can have mistakes too. – Marcos Oct 19 '21 at 07:30