Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
In (4), suppose $f$ is proper and birational. – user18119 Oct 2 '12 at 18:56
Sorry, I have edited the post. – only Oct 2 '12 at 20:05
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. – user18119 Oct 2 '12 at 23:32
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. – Matt Oct 3 '12 at 4:32
@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 $\le 1$. – user18119 Oct 3 '12 at 6:31

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 $R1$ iff the singular locus has codimension at least 2.
  • $X$ is $S2$ 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.
share|cite|improve this answer

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 $\text{Spec }\mathbb{Z}[\sqrt{-3}]$)?

share|cite|improve this answer
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, – Andrew Oct 2 '12 at 23:49
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". – Hurkyl Oct 3 '12 at 0:39
@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). – Thom Tyrrell Oct 3 '12 at 2:41
@ThomTyrrell, I see now, my mistake was thinking that $2$ is invertible in $\Bbb Z[\alpha]$! Makes sense now... thanks, – Andrew Oct 3 '12 at 2:58
@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. – Bombyx mori Jan 12 '15 at 9:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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