# Monday, October 11

:::{.remark}
Chapter 8: actually equivariant \(\K\dash\)theory without saying so!
Also deals with Demazure operators.
Goal: show that $X^Y_w$ is normal.
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:::{.definition title="Normal varieties"}
Let $X\in \Var\slice k$ be irreducible, then $X$ is **normal** at $x\in X$ iff $\mfm_x \da \OO_{X, x}$ is integrally closed in its field of fractions $\ff(\mfm_x)$.
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:::{.remark}
Note that there are implications smooth $\implies$ factorial $\implies$ normal in $\Var\slice k$.
We write $\Sigma(X) \subseteq X$ to be the singular locus, and if $X$ is normal then $\codim_X \Sigma(X) \leq 2$.
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:::{.example title="Whitney's Umbrella"}
Let $f(x,y,z) = x^2-zy^2$ and consider $X \da V(f) \subseteq \AA^3\slice \CC \in \Aff\Var\slice \CC$.
Checking normality for affine varieties just amounts to checking on regular functions, so $X$ is normal iff $\CC[X] \injects \CC(X)$ is integrally closed.
One direction involves checking that localizations are integrally closed, which is an easy exercise in commutative algebra, while the other direction is harder.
Consider $X(\RR)$:

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The claim is that $X$ is not normal.
Setting $\xi = x/y$ is not a regular function on $X$ since $y$ vanishes at some points of $X$, but $\xi^2 = x^2/y^2 = z \in \CC[X]$ is regular.
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:::{.remark}
Motivating question: normality is a local condition, so where can $X$ be non-normal?
There is a process of **normalization** which associates to $X$ a unique normal variety $\tilde X$ with a unique finite birational morphism $\nu: \tilde X\to X$.
Here *finite* means points have finite fibers and the map is proper.

Some properties:
- $\nu$ is unique.

- $\tilde X \mapsvia{\nu} X$ satisfies a universal property:
  For $X\to Y$ for any $Y$ normal, then there exists a unique lift

\begin{tikzcd}
	{\tilde X} \\
	\\
	X && Y
	\arrow["f", from=3-1, to=3-3]
	\arrow["\nu", from=3-1, to=1-1]
	\arrow["{\exists ! \tilde f}", dotted, from=1-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwyLCJYIl0sWzIsMiwiWSJdLFswLDAsIlxcdGlsZGUgWCJdLFswLDEsImYiXSxbMCwyLCJcXG51Il0sWzIsMSwiXFxleGlzdHMgISBcXHRpbGRlIGYiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19XV0=)

  > See also **Stein factorization** for proper morphisms.

- If $f:X\to Y$ is a birational projective morphism between irreducible varieties and $Y$ is normal, then $f_* \OO_X = \OO_Y$.

  > See also **Zariski's main theorem**.

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:::{.example title="?"}
Let $X$ be the umbrella from above.
Consider $\nu(u, v) = \tv{uv, v, v^2}$, so $\AA^2\slice{\CC} \mapsvia{\nu} X \subseteq \AA^3\slice \CC$, and let $f(x,y,z) = x^2 - zy^2$, so $f(uv, v, v^w) = (uv)^2 - u^2v^2 = 0$ is a regular function on $X$.
One can check that $\Im(\nu) \subseteq X$ so this is surjective, and the conclusion is that $X$ is irreducible with 2-dimensional fibers.
Consider the fibers of $\nu$:

1. $\vector x = 0$ yields $\nu\inv(\vector x) = \ts{ \tv{u, v} \in \AA^2 \st \tv{uv, v, u^2} = \vector 0} = \pt$.
2. $\vector x = \tv{0,0, z}$ with $z\neq 0$ yields $\nu\inv(\vector x) = \ts{\tv{uv, v, u^2} = \tv{0,0,z}} = \ts{p_1, p_2}$ which have nonzero 2nd coordinates, by choosing a square root of $u$.
3. $\vector x = \tv{x,y,z}$ with $x\neq 0$ yields $\nu\inv(\vector x) = \ts{\tv{uv, v, u^2} = \tv{x,y,z}}$. 
  This forces $v=y$, and $x = uv= uy$ which is nonzero and can be solved for $u$, so we again get a single point $\nu\inv(\vector x) = \pt$.

Note that just considering the real points misses the entire $-z$ axis.
This can be analyzed by regarding $u,v\in \CC$ as a pair of points in the same plane; then if $u=v=0$ corresponds to (1), $v=0$ with $u$ varying yields (2) (and two-point fibers), and moving $v$ from 0 yields (3).
Here $X$ is normal at the points in (1), but not normal in (2) and (3).

Moral: we can study singularities by looking at fibers.
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:::{.remark}
Next time: Schubert varieties.
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