# Thursday, January 12

:::{.remark}
Recall the definition of $\PP^n\slice k$ as a variety: lines in $\AA^{n+1}\slice k$ passing through the origin.
In the classical topology, it is compact since it can be realized as a quotient $S^{2n+1}/S^1$.
One can also cover it by affine charts $U_0 = \spec \CC\adjoin{{x_1\over x_0}, {x_2\over x_0}}$ and $U_2, U_3$ defined similarly, using that $\tv{x_0:\cdots x_k \cdots : x_n} = \tv{{x_0\over x_k}:\cdots 1 : \cdots {x_n\over x_k}}$ on $\ts{x_k\neq 0}$.
Recall that $\OO_{\PP^n}(U) = \ts{f\in \OO_{U_i}(U \intersect U_i) \st \ro{f}{U \intersect U_i} = \ro{f}{U \intersect U_j} }$.
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:::{.example title="?"}
$\OO_{\PP^1}(\PP^1) = \ts{ (f_0, f_1) \in k[s] \times k[t] \st f_0(s) = f_1(s\inv) \text{ on } \AA^1\smz}$ using that $t=s\inv$ on the overlap.
This equals $k[s] \intersect k[s\inv] = k$, so the only global regular functions are constant.
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:::{.proposition title="?"}
Considering $\PP^1\slice \CC$ in the analytic (classical) topology,
\[
\OO_{\PP^1\slice \CC}^\hol(\PP^1\slice \CC) = \ts{f: \PP^1\slice \CC\to \CC \text{ holomorphic}} = \CC
.\]
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:::{.proof title="?"}
Since $\PP^1\slice \CC$ is compact, $f$ achieves a maximum value $m$ at some point $p$, so write $f(p) = m$.
Letting $U\ni p$ be a closed disk containing $p$, then $f$ has a maximum on $U$.
By the maximum modulus principle, $\ro{f}{U}$ is constant, and so $f$ is constant.
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:::{.remark}
Call $\kxnz$ the *projective* coordinate ring of $\PP^n\slice k$.
Recall that $f \in \kxnz$ is not a regular function on $\PP^n$, but if $f$ is homogeneous then $V(f)$ is well-defined since $f(\lambda x_0, \cdots, \lambda x_n) = 0 \iff \lambda^n f(x_0, \cdots, x_n) = 0$.
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:::{.example title="?"}
Consider $V(s^2-st) \subseteq \PP^1\slice k$, then $s(s-t)$ vanishes on $\tv{0: 1}$ and $\tv{1:1}$.
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:::{.example title="?"}
Consider $V(x^3+y^3+z^3) \subseteq \PP^2\slice \CC$ -- topologically this is $S^1\times S^1$ and defines an elliptic curve over $\CC$.
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:::{.example title="of a K3 surface"}
$V(x_0^4 + x_1^4 + x_2^4 + x_3^4) \subseteq \PP^3\slice \CC$ is a K3 surface.
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:::{.remark}
More generally, $V(f_1,\cdots, f_m) \subseteq \PP^n\slice k$ with $f_i$ homogeneous of degrees $d_i$ is a projective scheme, where we use the scheme structure to distinguish e.g. $V(s^2-st)$ and $V(s^3-s^2t)$.
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:::{.remark}
Some recollections:

- The definition of irreducibility: $X$ is reducible if $X = A\union B$ for $A, B$ proper nontrivial closed sets.
  - E.g. $V(xy) = V(x) \union V(y)$ is not irreducible in $\AA^2$.
- Dimension is defined in terms of lengths of chains of closed irreducible subsets.
- $X$ is irreducible iff $k[X] \da \kxn/I(X)$ is a domain iff $I(X)$ is prime 
- Krull's PID theorem, used to show $\dim R/\gens{f} = \dim R - 1$ if $f$ is not a zero divisor 
  - To see why this is, consider $R = \CC[x,y]$, then $\dim R/\gens{xy} = 1 = \dim \CC[x] = \dim R/\gens{x}$.
- $\spec R$ is reduced iff $R$ has no nilpotents
  - E.g. $V(x^2) = \spec k[x]/\gens{x^2}$ is not reduced, since $x$ is nilpotent ($x^2=0$ but $x\neq 0$).
- $X \in \Sch$ is reduced iff $\OO_X(U)$ has no nilpotents, so every regular function $f$ satisfies $f^n\neq 0$ for every $n$.
- $\spec R$ is quasicompact.
  - E.g. $\AA^1$ in the Zariski topology is the cofinite topology, so if $\mcu \covers \AA^1$ the $U_1$ covers all but finitely many points $p_k$ and each $p_k$ is in some $U_k$.
- Open sets are big: they are essentially the whole space, minus lower dimensional things.
- Completeness replaces compactness, where $X$ is (universally) complete iff for all $Y$, the projection $X\times Y\to Y$ is a closed map.
  - $\AA^1$ is not complete: take $Y\da \AA^1$, then $V(xy-1) \mapsto \AA^1\smz$ is a closed set mapping to an open set.
- Producing varieties that aren't manifolds: $V(xy) \subseteq \AA^2\slice \CC$ is singular at the origin and has no local chart to $\CC$ there.
- $V(f) \subseteq \RR^n$ is a manifold when $0$ is a regular value of $f$, so $df: T_p \RR^n\to T_0 \RR$ is surjective at all $p\in V(f)$.
  - Can be formulated as $Jf \da \qty{\dd{f_i}{x_j}}$ has maximal rank everywhere.

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