# Tuesday, September 27 :::{.remark} Types of varieties: - Affine. - Projective: embeds into $\PP^n$. - Quasiprojective: $U \da X\sm Z \embeds X \overset{\text{closed}}\embeds \PP^n$ where $Z \subseteq X$ is closed. - Arbitrary. Being closed in the Zariski topology implies closed in the classical topology, and these are compact in the classical topology. Recall that proper maps are separable and universally closed -- think of proper as essentially projective. ::: :::{.remark} For $k= \bar k$, there is a bijection between $\Aff\Var\slice k$ and $R\in \kalg^\fg$ with no nilpotents, so there is a surjection $\phi: \kxn \surjects R$ with $I \da \ker \phi$ and $I = \sqrt{I}$. The map sends $X$ to $k[X] \da \kxn/I$ the ring of regular functions on $X$. If $k = \kbar$ then $\mspec R$ consists of elements $m = \gens{ x_1-a_1, \cdots, x_n-a_n} \in \mspec \kxn$, corresponding to points $\tv{a_1,\cdots, a_n} \in \AA^{n}\slice k$. Similarly $\Aff\Sch$ bijects with commutative associative unital rings. Projective varieties correspond to $\ZZ_{\geq 0}\dash$graded rings $R$ over $k=\kbar$, so $R = \oplus _{d\geq 0} R_d$ is finitely-generated without nilpotents. The map sends $R$ to its projective spectrum $\mproj R$. For arbitrary $\ZZ_{\geq 0}\dash$graded commutative associative unital rings, one similarly defines $\Proj R$. If $R = k[R_1]$ is generated by degree 1 elements, then there is an embedding $\mproj R \embeds \PP^n$, but an arbitrary projective variety doesn't necessarily come with such an embedding. ::: :::{.remark} For this, take the Veronese subring $R^{(e)} \da \bigoplus _{d\geq 0} R_{d_e}$ for $e > 0$. This corresponds to the Veronese embedding $\PP^n \embeds \PP^N$ for some large $N$, which is defined by \[ V_e: \PP^n &\to \PP^N \\ \tv{x_0: \cdots : x_n} &\mapsto \tv{x_0^e: x_{0}^{e-1}x_1, \cdots} \] where you send a point to all monomials in the coordinates of degree $e$. Here $N+1 = {n+e\choose e}$ is the number of such monomials. Note that e.g. ${x_0^{e-1} x_1 \over x_0^e} = {x_1\over x_0}$, so one can recover the former ratios from the latter. The condition of $R = k[R_1]$ is needed to guarantee $V_e$ is an embedding. ::: :::{.lemma title="?"} \[ \mproj R = \mproj R^{(e)} .\] ::: :::{.lemma title="?"} There exists an $e_0 \gg 1$ such that $R^{(e)}$ is generated in degree 1 for $e > e_0$. ::: :::{.remark} Let $X \subseteq \PP^n$, we want to extract a graded ring. Start with $\PP^n$ corresponding to $\kxn \ni p_d(x_0,\cdots, x_n)$, forms of degree $d$. Then $\ts{p_d = 0} \subseteq \PP^n$ is well-defined, and this defines the Zariski topology on $\PP^n$ Moreover any $p_d/q_d$ is a well-defined regular function on the open subset $\ts{q_d\neq 0} \subseteq \PP^n$. There are several ways to produce the ring $R$: - Consider $CX \subseteq \AA^{n+1}$ with $\kxn \surjects R\da \kxn/I(CX)$. - Consider $R' \da \bigoplus_{d\geq 0} H^0(X; \OO_X(d) )$ whose sections are locally given by ratios of forms $p_{k+d}/q_k$. If $X$ is **projectively normal**, then $R = R'$. - Consider $R'' = R(X, L) = \bigoplus _{d\geq 0} H^0(X, L^d)$, then taking $\Proj$ yields $(X = \Proj R(X, L), \OO(1))$. This is not a bijection; several $R(X, L)$ can yield the same variety (e.g. by leaving out various degrees). ::: :::{.remark} Constructing the correspondence $\Proj R \mapstofrom R = \bigoplus _{d\geq 0} R_d$.[^Groth] One can safely assume the rings are finitely-generated, the general construction goes exactly the same way. Define $\Proj R$ to be the set of prime homogeneous ideals $p \subseteq R$ which are not contained in a certain ideal: considering $\kxn$, note that $\gens{x_0,\cdots, x_n}$ does not define a point of $\PP^n$, so define $R_+ \da \bigoplus _{d\geq 1} R_d$ to be the **irrelevant ideal**. One can define fundamental closed subsets: for $\PP^n$ these are of the form $\ts{f_d = 0}$, so generalize to $f_d\in R_d$ and define $Z(f_d) \da \ts{ [p] \st f([p]) = 0\in R/p}$. Note that $f \equiv 0 \mod p$ in $R$ iff $f\in p$. Define fundamental closed sets as intersections $\Intersect_\alpha Z(f_\alpha)$ and fundamental open sets as $D(f_d) = \ts{f_d\neq 0}$. [^Groth]: This construction is in EGA II. ::: :::{.example title="?"} If $R\in\kalg^\fg$ is generated in degree 1, $I \injects \kxn \surjects R$ and $X \subseteq \PP^n$ corresponds to $Z(I) \da \Intersect_{f\in I} Z(f)$. ::: :::{.remark} Sections of $\OO$ are locally of the form $f_k/g_k$, and for $\OO(d)$ of the form $f_{d+k}/g_k$. It remains to define local sections of the following on $\Proj R$: - $\OO$: $f_k/g_k$ - $\OO(1)$: $f_{k+1}/ g_k$ - $\OO(d)$: $f_{d+k}/f_k \in R\localize{g}_{\deg = d}$, where ${f\over g^n} \sim {f'\over g^m} \iff g^N(fg^m - f'g^n) = 0$ for some $N$. ::: :::{.example title="?"} Consider $\mu_2 \actson k[x,y]$ by $(x,y)\mapsto (x, -y)$. Then $k[x, y]^{\mu_2} = k[x, y^2]$, which is a graded subring of $k[x,y]$ which is not generated in degree 1. Note that $\Proj k[x,y]^{\mu_2} = \PP^1(1, 2)$ is a weighted projective space, which turns out to be isomorphic to $\PP^1$. Moreover $0 = \tv{0: 1}$ and $\infty = \tv{1: 0}$ have nontrivial stabilizers, while the action is free elsewhere, and remembering the stabilizers yields a quotient stack. ::: :::{.example title="?"} Consider the moduli of elliptic curves $(E, 0)$ over $\CC$. Realize $(E, 0)\embeds \PP^2$ as a cubic curve by its Weierstrass equation: - In $\AA^2$: $y^2 = x^3 + Ax + B$ - In $\PP^2$: $zy^2 = x^3 + Az^2 x + Bz^3$. Regrade the first equation to total homogeneous degree $6$ by setting $\deg y = 3, \deg x = 2, \deg A = 4, \deg B = 6$ and rescale \[ (x,y,A,B) \mapsto (\lambda^2 x, \lambda^3 y, \lambda^4 A, \lambda^6 B) .\] This makes it unique up to rescaling, so the moduli of such equations is $\PP(4, 6)$ corresponding to $A$ and $B$, which is the $j\dash$line $\PP^1_j$. Every point has a stabilizer of size at least 2 since 2 divides both 4 and 6, which comes from the involution $z\mapsto -z$. This corresponds to two lattices with automorphism groups $C_4$ and $C_6$: ![](figures/2022-09-27_14-04-24.png) The former is $\CC/\gens{1, i}$ which has the extra automorphism $z\mapsto iz$, which has CM. The latter is $\CC/\gens{1, \zeta_3}$ which has $z\mapsto \zeta_3 z$. :::