What are normal schemes intuitively?

Question

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?

Answer 1

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

Answer 2

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.