What is the meaning of normalization of varieties in complex geometry?

Question

There is a question already asked here about this. But I know almost nothing of algebraic geometry, nothing fancy to understand the answer. So I would highly appreciate an elementary explanation to my question.

I encountered the term normalization while I was trying to understand that a particular algebraic curve is smooth. My questions are:

1) What is the meaning of normalization?

2) Why do we perform it?

3) How is it related to smoothness of algebraic curves? To singularities of curves?

4) Is normalization cannonical? If so, how?

Answer 1

0) Recall that a domain $A$ is said to be normal if it is integrally closed in its fraction field $K=Frac(A)$.
This means that any element $q\in K$ killed by a monic polynomial in $A[T]$, i.e. such that for some $n\gt 0, a_i\in A$ one has $$q^n+a_1q^{n-1}+\cdots+a_n=0$$ already satisfies $ q\in A$ .
A variety $V$ is said to be normal if it can be covered by open affines $V_i\subset V$ whose associated rings of functions $A_i=\mathcal O(V_i)$ are normal.

1) The normalization of an irreducible variety $X$ is a morphism $n:\tilde X\to X$ such that $\tilde X$ is a normal variety and there exists a closed subvariety $Y\subsetneq X$ such that $n|(\tilde X\setminus n^{-1}(Y))\stackrel {\cong}{\to}X\setminus Y$ is an isomorphism.

2) We perform normalization because normal varieties have better properties than arbitrary ones.
For example in normal varieties regular functions defined outside a closed subvariety of codimension $\geq 2$ can be extended to regular functions defined everywhere ("Hartogs phenomenon") .

3) A curve is non-singular (=smooth if the base field is algebraically closed ) if and only if it is normal, so that normalization=desingularization for curves.
In higher dimensions normal varieties, alas, may have singularities.
Getting rid of these is tremendously difficult in characteristic zero (Hironaka) and is an unsolved challenge in positive characteristics.

4) Yes, normalization of $X$ is canonical in the sense that if $n': X'\to X$ is another normalization we have an isomorphism $j:\tilde X \stackrel {\cong}{\to} X'$ commuting with the normalization morphisms, namely $n'\circ j=n$ .
At the basis of this canonicity is the fact that there is a (trivial) canonical procedure for enlarging a domain to its integral closure in its fraction field.

Answer 2

I like Georges Elencwajg's answer, but I think it's useful to see some topological intuition for what normalization does over $\mathbf{C}$.

Note we say a variety is normal if its local rings are integrally closed in their fraction field.

Riemann Extension Theorem

This fleshes out 2) in Georges Elencwajg's answer. Most of what follows is from Kollár's article "The structure of algebraic threefolds". I also enjoy the discussion around p. 391 in Brieskorn and Knörrer's book Plane algebraic curves.

In complex analysis, you learn about the Riemann extension theorem, which says a bounded meromorphic function on any open set $U \subset \mathbf{C}$ that is holomorphic on $U \setminus \{p\}$ is in fact holomorphic on $U$. In (complex) algebraic geometry, we want something similar to hold (let's say, for curves): that a bounded rational function that is regular on $U \setminus \{p\}$ is in fact regular on $U$.

This fails in general, however:

Example (cuspidal cubic). Let $V = \{x^2 - y^3 = 0\} \subset \mathbf{C}^2$, and let $f = (x/y)\rvert_V$. $f$ is a rational function on $V$, regular away from $0$. You can of course demand $f(0,0) = 0$ to make $f$ continuous at $(0,0)$, but this does not make $f$ regular. For, suppose $x/y = a(x,y)/b(x,y)$ for some polynomials $a,b$ such that $b(0,0) \ne 0$. Then, $xb(x,y) - ya(x,y) = 0$ on $V$, so $x^2 - y^3$ divides it. But there is a nonzero constant term in $b(x,y)$ which contributes a nonzero coefficient for $x$ in $xb(x,y) - ya(x,y)$, so it can't be zero. Note, though, that $(x/y)^2 = y$ is regular on $V$, which shows $V$ is not normal.

The question then becomes: can we modify the curve $V$ so that the Riemann extension theorem does hold? The answer is that yes, the normalization in fact does this for us: it gives another variety $\tilde{V}$ such that the rational functions on $V$ and $\tilde{V}$ agree, but an extension property like the one above holds. This extension property is the content of

Hartog's Theorem. Let $V$ be a normal variety and let $W \subset V$ be a subvariety such that $\dim W \le \dim V - 2$. Let $f$ be a regular function on $V - W$. Then $f$ extends to a regular function on $V$.

But returning to our example: the map $\mathbf{C} \to V$ sending $z \mapsto (z^3,z^2)$ is in fact a normalization. The function $x/y$ then pulls back to $z$ on $\mathbf{C}$, which is obviously regular!

Remark. It is possible to define normality as saying every rational function that is bounded in a neighborhood $U$ of a point $p$ is in fact regular on $U$, in direct analogy to the Riemann extension theorem. But the equivalence of these definitions is hard: see Kollár, Lectures on resolution of singularities, §1.4, especially Rem. 1.28.

Separating Branches

What follows is from Mumford's The red book of varieties and schemes, III.9.

Normality can be understood as a way to separate the "branches" of an algebraic variety at a singular point. Consider the following

Example (nodal cubic). Let $V = \{x^2(x+1) - y^2\} \subset \mathbf{C}^2$. It is not normal at $(0,0)$ since it's singular there. Consider a small analytic neighborhood $U = \{(x,y) \mid \lvert x \rvert < \epsilon,\ \lvert y \rvert < \epsilon\}$. Points in $U \cap V$ satisfy $\lvert x - y \rvert \lvert x + y \rvert = \lvert x \rvert^3 < \epsilon \lvert x \rvert^2$ hence $\lvert x - y \rvert < \sqrt{\epsilon} \lvert x \rvert$ or $\lvert x + y \rvert < \sqrt{\epsilon} \lvert x \rvert$, but both can't occur simultaneously for small enough $\epsilon$. Thus, near the origin $V$ splits into two "branches" containing points satisfying $\lvert x - y \rvert \ll \lvert x \rvert$ and $\lvert x + y \rvert \ll \lvert x \rvert$. Each piece is connected, but there is no algebraic way to separate each branch.

The normalization $\pi\colon \tilde{V} \to V$ ends up fixing this, in the following way: for each point $p \in V$, the inverse image $\pi^{-1}(p)$ is in 1-1 correspondence with the set of branches at $p$. In our particular example, it is given by $\mathbf{C} \to V$ where $z \mapsto (z^2-1,z(z^2-1))$; the two branches correspond to $z=\pm1$.

So perhaps a variety $V$ is normal if and only if at every point $p \in V$, there is only one branch. The forward direction is essentially the content of Zariski's main theorem; see pp. 288–289 in Mumford. But the converse is false: the cuspidal cubic only has one branch but is not normal.