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A ring is called integrally closed if it is an integral domain and is equal to its integral closure in its field of fractions. A scheme is called normal if every stalk is integrally closed.

Some theorems on normality:

  1. A local ring of dimension 1 is normal if and only if it is regular.

  2. (Serre's criterion) A scheme is normal if and only if it is nonsingular in codimension 0 and codimension 1 and every stalk at a generic point of an irreducible closed subset with dimension $\ge 2$ has depth at least 2.

  3. Every rational function on a normal scheme with no poles of codimension 1 is regular.

  4. (Zariski connectedness): If $f:X\rightarrow Y$ is a proper birational map of noetherian integral schemes and $Y$ is normal, then every fiber is connected.

  5. Normal schemes over $C$ are topologically unibranched.

But the proofs I've seen are fairly ad-hoc, and I was wondering if there's some geometric perspective that would clarify these results. The only result here thats an "iff" is Serre's criterion, but I don't understand depth geometrically so I'm not sure how to interpret it.

Is there some nice geometric perspective on normality?

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  • $\begingroup$ In (4), suppose $f$ is proper and birational. $\endgroup$ – user18119 Oct 2 '12 at 18:56
  • $\begingroup$ Sorry, I have edited the post. $\endgroup$ – only Oct 2 '12 at 20:05
  • $\begingroup$ I don't have the "Red book on varieties and schemes" of Mumford under hands, but I remember there is a page on normal varieties. $\endgroup$ – user18119 Oct 2 '12 at 23:32
  • $\begingroup$ Shouldn't Serre's criterion just be R1 and S2? You have "nonsingular in codimension 0" in there which means the whole thing is nonsingular? R1=regular in codimension 1 and S2=depth condition you wrote. $\endgroup$ – Matt Oct 3 '12 at 4:32
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    $\begingroup$ @Matt: points of codimension 0 are the generic points of $X$. So regular in codimension $0$ meance $X$ (when noetherian) contains a reduced dense open subscheme. Condition $(R_1)$ means regular in codimension $\le 1$. $\endgroup$ – user18119 Oct 3 '12 at 6:31
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In the world of locally Noetherian schemes, Serre's criterion can be made quite geometric.

Let $X$ be a reduced, locally Noetherian scheme. Then...

  • $X$ is $R_1$ iff the singular locus has codimension at least 2.
  • $X$ is $S_2$ iff, for each $Y\subset X$ of codimension at least $2$, the regular functions on the complement $X-Y$ extend to regular functions on $X$.

This second fact can be found in Ravi Vakil's notes (Theorem 12.3.10), or in this MathOverflow post.

Roughly speaking, normalizing a variety improves singularities as follows.

  • In codimension $1$, it completely resolves them.
  • In codimension $\geq 2$, it improves them enough so that rational functions defined on their complement can be extended to the singularity.
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As a number theorist, I would first think about normality in terms of orders in algebraic number fields.

Consider the number field $K$ defined by adjoining $\sqrt{-3}$ to the rationals. What is the ring of integers in this field? At first glance, the "obvious" answer is $\mathbb{Z}[\sqrt{-3}]$, but the element

$$\alpha = \frac{1 + \sqrt{-3}}{2}$$

is integral over $\mathbb{Z}$, with minimal polynomial $x^2 - x + 1$. Thus, $\mathbb{Z}[\sqrt{-3}]$ is not integrally closed in its quotient field.

What, then, is the difference between $\mathbb{Z}[\sqrt{-3}]$ and $\mathbb{Z}[\alpha]$? As they're isomorphic as schemes over $\text{Spec } \mathbb{Z}[\frac{1}{2}]$, the problem, if any, is with 2.

Because $x^2 - x + 1$ is irreducible mod 2, the prime ideal (2) of $\mathbb{Z}$ remains prime in $\mathbb{Z}[\alpha]$. However, $x^2 + 3 \equiv (x-1)^2 \ (\text{mod }2)$, and it follows that (2) is not prime in $\mathbb{Z}[\sqrt{-3}]$. So, the normal scheme $\text{Spec }\mathbb{Z}[\alpha]$ gives the correct description of the arithmetic of this number field $K$.

One other way to think about these objects using your theorem 3 above: What is the divisor of $\alpha$ considered as an element of the fraction field of $\mathbb{Z}[\sqrt{-3}]$ (i.e. the function field of $\text{Spec }\mathbb{Z}[\sqrt{-3}]$)?

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  • $\begingroup$ Dear Thom, this is an interesting perspective to me! Just a nit-picky point: it seems to me that when you say $\Bbb Z[\sqrt{-3}]$ and $\Bbb Z[\alpha]$ are isomorphic over $\Bbb Z[\frac{1}{2}]$, that what you really mean is $\Bbb Z[\sqrt{-3}]$ (properly) contains an open neighbourhood $D(2)$ that is isomorphic to $\Bbb Z[\alpha].$ Is this right? Also, I'm wondering, what is the "order" you refer to in this case? Regards, $\endgroup$ – Andrew Oct 2 '12 at 23:49
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    $\begingroup$ Alas, (orders in) algebraic number fields are one-dimensional, so the notions of "normal" and "regular" coincide, and we cannot use this example to understand how the geometric concept of "normal" differs from the geometric concept of "nonsingular". $\endgroup$ – user14972 Oct 3 '12 at 0:39
  • $\begingroup$ @Andrew The two schemes are isomorphic along their respective D(2)'s. If we invert 2 in $\mathbb{Z}[\sqrt{-3}]$ we can generate $\alpha$, but in the process any ideal that contained 2 is made trivial (the ring/ideal structure changes). $\text{Spec } \mathbb{Z}[\alpha]$ still has a prime ideal that contains 2, so the two schemes could not be isomorphic as schemes over $\mathbb{Z}$. To consider the schemes as "over $\text{Spec } \mathbb{Z}[\frac{1}{2}]$" we base change with a fiber product. In doing so we invert 2, just as in the set D(2). $\endgroup$ – Thom Tyrrell Oct 3 '12 at 2:41
  • $\begingroup$ @ThomTyrrell, I see now, my mistake was thinking that $2$ is invertible in $\Bbb Z[\alpha]$! Makes sense now... thanks, $\endgroup$ – Andrew Oct 3 '12 at 2:58
  • $\begingroup$ @ThomTyrrell: Perhaps I am a little slow. May I ask why the polynomial factorizes implies that $2$ is no longer irreducible in $\mathbb{Z}[\sqrt{-3}]$? I could not get it. $\endgroup$ – Bombyx mori Jan 12 '15 at 9:03

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