3.3 Varieties, dimensions and rational points

Varieties

def: Affine variety

$$\begin{align*} &\sphericalangle \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp} \\ &\exists X' \in \mathbb A^{n-} \cap \mathcal Z^c(\mathbb A^n): X \cong_{Rsp} X' \\ \hline \\ &X \in \mathcal V(\mathbb A^n) \end{align*}$$

note

That is affine variety is a ringed space isomorphic to some irreducible closed set of affine space

Proposition 3.3.1: Affine distinguished subsets are varieties

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp, c, -} \\ &f \in \mathcal O_X(X)=A(X) \\ &X_f:=X \setminus Z(f) \\ \hline \\ &\exists X' \in \mathbb A^{n+1,-} \cap \mathcal Z^c(\mathbb A^{n+1}): (X', T', A(X)_f) \cong_R (X_f, T_{|X_f}, \mathcal O_{|X_f}) \end{align*}$$

note

$\mathcal O_{|X_f}$ is a natural restriction on subsets of $X_f$

Proof

Let $f'$ be a representative of $f$ in $A(X)$. Consider an ideal $J \lhd_R \overline F[x_1, \ldots, x_n, t], J:=\lang I(X), 1-tf'\rang_I$. Next, $X':=Z(J)=\{(P, \lambda), P \in X_f, \lambda = \frac{1}{f'(P)}\}$. The zero set indeed has this form because on $Z(f): 1-tf'$ never goes to $0$ and on $X_f$ it does go to $0$. Also since the last variable $t=1/f'$ it's clear that $A(X')=A(X)_f$. So let's prove:

$$(X', T', A(X)_f) \cong_R (X_f, T_{|X_f}, \mathcal O_{|X_f})$$

Consider a projection map $\pi: X' \to X_f, (P, \lambda) \mapsto P$ and inverse $\pi^{-1}: X_f \to X', P \mapsto (P, \frac{1}{f'(P)})$. Since $A(X)_f$ and $O_{|X_f}(X_f)$ are essentially the same (both have polynomial in numerator and some power of $f$ in denominator), it's easy to see that $\pi$ is indeed a ringed space isomorphism.

$\square$

def: Pre-variety

$$\begin{align*} &\sphericalangle \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp, -} \\ &\exists U_i \in T_o: X = \bigcup_{i=1}^kU_i \\ &\forall i: (U_i, \mathcal O_{X|U_i}) \in \mathcal V(\mathbb A^{n(i)}) \\ \hline \\ &(X, T, \mathcal O_X) \in \mathcal V^{pre} \end{align*}$$

Example: Affine pre-variety

Obvously any affine variety is a pre-variety.

Proposition 3.3.2: Projective variety is a pre-variety

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp, c, -} \\ \hline \\ &(X, T, \mathcal O_X) \in \mathcal V^{pre} \end{align*}$$

Proof

Consider $X_0:= \{[a_0, \ldots, a_n], a_0 \ne 0\}$. If $I(X)= \lang f_1, \ldots, f_r \rang_I$ and define $g_i(x_1, \ldots, x_n):=f_i(1, x_1, \ldots, x_n)$, $Y:=Z(\lang g_1, \ldots, g_r \rang_I)$. It's easy to build a morphism:

$$\phi: X_0 \to Y, [a_0, \ldots, a_n] \to [\frac{a_1}{a_0}, \ldots, \frac{a_n}{a_0}] \\ \phi^{-1}: Y \to X_0, (a_1, \ldots, a_n) \mapsto [1, a_1, \ldots, a_n]$$

In this case:

$$\phi^*(\frac{p(a_1, \ldots, a_n)}{q(a_1, \ldots, a_n)})=\frac{p(\frac{a_1}{a_0}, \ldots, \frac{a_n}{a_0})}{q(\frac{a_1}{a_0}, \ldots, \frac{a_n}{a_0})} \\ \phi^{-1*}(\frac{p(a_0, \ldots, a_n)}{q(a_1, \ldots, a_1)})=\frac{p(1, a_1, \ldots, a_n)}{q(1, a_1, \ldots, a_n)}$$

So these functions are regular and it's easy to see that closed sets goes to closed sets. So

$$X_0 \cong_{Rsp} Y$$

In the same way we can define isomorphism for $X_i$ and we know that $X = \bigcup_i X_i$ so $X \in \mathcal V_q$

$\square$

def: Variety

$$\begin{align*} &\sphericalangle \\ &(X, T, \mathcal O_X) \in \mathcal V^{pre} \\ &\forall (Y, T', \mathcal O_Y) \in \mathcal V^{pre}, f_1, f_2: Y \rightsquigarrow_{Rsp}X \implies \\ &\{P \in Y: f_1(P)=f_2(P)\} \in T'^c \\ \hline \\ &(X, T, \mathcal O_X) \in \mathcal V \end{align*}$$

note

We define the general notion of variety for the sake of completness. Affine and projective varieties (meaning irreducible closet sets) are varieties.

def: Non-singular point

$$\begin{align*} &\sphericalangle \\ &\mathbb S^n \\ &X \in \mathcal V(\mathbb S^n) \\ &I(X) = \lang f_1, \ldots, f_k \rang_I \\ &P \in X \\ &\text{rank } || \partial f_i/\partial x_j(P)||=n-\dim X \\ \hline \\ &P \in \cancel S(X) \\ &P - \text{ non-singular point of }X \end{align*}$$

def: Non-singular variety

$$\begin{align*} &\sphericalangle \\ &\mathbb S^n \\ &X \in \mathcal V(\mathbb S^n) \\ &X=\cancel S(X) \\ \hline \\ &X - \text{non-singular variety} \end{align*}$$

Dimensions

def: Dimension of a topological space

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T\\ \hline \\ &\dim X:=\sup_n\{X_0 \subset X_1 \subset \ldots \subset X_n, X_i \in X^- \cap T^c \} \end{align*}$$

Proposition 3.3.3: Dimension of closed set is a dimension of it's coordinate ring

$$\begin{align*} &\sphericalangle \\ &\mathbb S^n \\ &(X, T) \in \mathcal T^{c} \\ \hline \\ &\dim X = \dim A(X) \end{align*}$$

Proof

By $(3.1.13)$ each chain of irreducible closed sets $X_0 \subset \ldots \subset X_n \subseteq X$ correspond to prime ideals $P_0 \supset \ldots \supset P_n \supseteq I(X)$ and by $(2.3.12)$ these correspond to prime ideals in $A(X)$. Since these two correspondences are bijections, we finish the proof.

We'll state two propositions from commutative algebra without a proof

Proposition 3.3.4: Krull dimension and transcendence degree

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &B \in \mathcal R^{\mathcal {ID}}, B - \text{finitely generated F-algebra} \\ \hline \\ &\begin{align*} & \dim B = \text{trdeg }\mathfrak F(B)/F\hspace{0.5cm} \tag{a}\\ & \mathfrak p \in \mathfrak P_I(B) \implies \text{ht } \mathfrak p + \dim B/\mathfrak p = \text{dim } B\hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proposition 3.3.5: UFD and prime ideals criterion

$$\begin{align*} &\sphericalangle \\ &N \in \mathcal R^{\mathcal {N}} \\ \hline \\ &N \in \mathcal R^{\mathcal {UFD}} \iff \forall \mathfrak p \in \mathfrak P_I(N), \text{ht } \mathfrak p = 1: \exists f \in N: \mathfrak p = \lang f \rang_I \end{align*}$$

Proposition 3.3.6: Krull's hauptidealsatz

$$\begin{align*} &\sphericalangle \\ &N \in \mathcal R^{\mathcal N} \\ &f \in N \\ &\mathfrak p \in \mathfrak P_I(N): f \in \mathfrak p, \nexists \mathfrak p' \in \mathfrak P_I(N): f \in \mathfrak p' \subseteq \mathfrak p \\ \hline \\ &\text{ht }\mathfrak p = 1 \end{align*}$$

Proposition 3.3.7: Dimension of an open set equals the dimension of its closure

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T^{c,-} \\ &U \in T^o \\ \hline \\ &\dim U = \dim \overline U \end{align*}$$

Proof

Note that by $(3.1.1)$ $(U, T_{|U}) \in \mathcal T^-$ and by $(3.1.2)$ we have $(\overline U, T_{|\overline U}) \in \mathcal T^{c,-}$. Consider a chain $Z_0 \subset \ldots \subset Z_n, Z_i \in T^{c,-}_{|U}$. Now take the closure of these sets in $X$ to get the chain $\overline Z_0 \subseteq \ldots \subseteq \overline Z_n, \overline Z_i \in T^{c}$. First, let's prove that $Z_i=\overline Z_i \cap U$:

$$U \cap \overline Z_i = U \cap \bigcap_{X_i \supseteq Z_i, X_i \in T^c} X_i = \\ \bigcap_{X_i \supseteq Z_i, X_i \in T^c} (U \cap X_i) = \bigcap_{U \cap X_i \supseteq U \cap Z_i, X_i \in T^c} (U \cap X_i)= \\ \bigcap_{X_i' \supseteq Z_i, X_i' \in T^c_{|U}} X_i' = Z_i \text{ since } (Z_i \in T^c_{|U})$$

Next, $\overline Z_i \in T^{c,-}_{|\overline U}$. It's closed simply because $\overline U \subseteq X$. If $\overline Z_i$ was reducible, that is $\overline Z_i = X_1 \cup X_2, X_{1,2} \subset \overline Z_i \implies Z_i=U \cap \overline Z_i = (U \cap X_1) \cup (U \cap X_2) \implies (X_1 \supseteq U) \vee (X_2 \supseteq U)\implies (X_1=\overline X_1 = \overline U) \vee (X_2=\overline X_2 = \overline U)$. But this is a contradiction since $X_{1,2} \subset \overline Z_i \subseteq \overline U$. So we establised:

$$Z_i \in T^{c, -}_{|\overline U}$$

So we know the chain $Z_0 \subset \ldots \subset Z_n, Z_i \in T^{c,-}_{|U}$ implies the existence of chain $\overline Z_0 \subset \ldots \subset \overline Z_n, \overline Z_i \in T^{c,-}_{|\overline U}$. There could not be equivalence $\subseteq$ in this chain as we know that $Z_i = U \cap \overline Z_i$ so $\overline Z_i = \overline Z_{i+1} \implies U \cap \overline Z_i = U \cap \overline Z_{i+1} \implies Z_i = Z_{i+1}$.

Now we can go the other way round and project some chain $X_0 \subset \ldots \subset X_n, X_i \in T^{c,-}_{|\overline U}$ to $U$ by taking $X_i'=U \cap X_i$ which will be closed and irreducible (the latter can be proving by assuming otherwise and taking closure similarly to the above). So we have a bijection of irreducible closed chains. The last thing we need to deal with is what if $U \cap X_i = U \cap X_{i+1}$? Then we'll have a shorter chain of $X_i'$. If $U \cap X_i = U \cap X_{i+1}$ then $X_{i+1} \setminus X_i \in \overline U \setminus U$. On the other hand $X_{i+1}=X_i \cup \overline {X_{i+1} \setminus X_i} \overset{X_{i+1} \in T^{c,-}_{|\overline U}}\implies \overline {X_{i+1} \setminus X_i} = X_{i+1}$. At the same time $\overline U \setminus U \in T^c_{|\overline U}$ since $U \in T^o$. So we have $X_{i+1}=\overline {X_{i+1} \setminus X_i} \subseteq \overline{\overline U \setminus U}=\overline U \setminus U$. But it cannot be since $X_{i+1} \cap U \ne \empty$.

So we have fully matched chains of irreducible in $U$ and $\overline U$ and $\dim U = \dim \overline U$

$\square$

Proposition 3.3.8: Dimension of a variety with a principal vanishing ideal

$$\begin{align*} &\sphericalangle \\ &\mathbb S^n \\ &(X, T) \in \mathcal T^{c,-}\\ \hline \\ &\dim X = n-1 \iff \exists f \in \overline F[x_1, \ldots, x_n]^-: X = Z(f) \end{align*}$$

Proof

$\implies$

$$\dim X = n-1 \implies n-1 \overset{(3.3.3)}= \dim A(X) = \\ \dim \overline F[x_1, \ldots, x_n] / I(X) \overset{(3.3.4.b)}= \\ \dim \overline F[x_1, \ldots, x_n] - \text{ht }I(X) \implies \text{ht }I(X) = 1 \overset{(3.3.5)}\implies \\ \exists f \in \overline F[x_1, \ldots, x_n]: I(X) = \lang f \rang_I \implies X = Z(f)\\$$

$\impliedby$

$f \in \overline F[x_1, \ldots, x_n]^- \implies I(X) = \lang f \rang_I \in \mathfrak P_I(\overline F[x_1, \ldots, x_n]) \overset{(3.3.6)}\implies \text{ht } I(X) = 1 \overset{(3.3.4.b)}\implies \dim A(X)=\dim \overline F[x_1, \ldots, x_n]/I(X) = n-1 \overset{(3.3.3)}\implies \dim X = n-1$

$\square$

Proposition 3.3.9: Dimension of the local ring

$$\begin{align*} &\sphericalangle \\ &\mathbb S^n \\ &X \in \mathcal V(\mathbb S^n) \\ &P \in X \\ &\mathfrak m_{P} \in \mathfrak M_I(\mathcal O_{X,P})\\ \hline \\ &\dim \mathcal O^r_{X,P} = \dim X \end{align*}$$

Proof

First let's note that chain of prime ideals in $\mathfrak m_{P}$ corresponds to a chain of prime ideals in $A(X)_{\mathfrak m_{P}} \cong \mathcal O^r_{X,P}$ so $\text{ht }\mathfrak m_{P}=\dim \mathcal O^r_{X,P}$. But from $(3.3.4.b)$ we know that

$$\text{ht }\mathfrak m_{P} + \dim A(X)/\mathfrak m_{P} = \dim A(X) = \dim X$$

Next $A(X)/\mathfrak m_{P} \in \mathcal F$ implies $\dim A(X)/\mathfrak m_{P} = 0$ (note we're talking about Krull dimension here) so

$$\dim \mathcal O^r_{X,P}=\text{ht }\mathfrak m_{P}=\text{ht }\mathfrak m_{P} + \dim A(X)/\mathfrak m_{P}=\dim X$$

$\square$

Proposition 3.3.10: Non-singular point criterion

$$\begin{align*} &\sphericalangle \\ &\mathbb S^n \\ &X \in \mathcal V(\mathbb S^n) \\ &P \in X \\ &\mathfrak m_{P} \in \mathfrak M_I(\mathcal O_{X,P})\\ \hline \\ &P \in \cancel S(X) \iff \dim X=\dim_{\overline F} \mathfrak m_{P}/\mathfrak m^2_{P} \end{align*}$$

note

Since $\mathcal O_{X,P}/\mathfrak m_{P} \cong_F \overline F$ we know that $\dim_{\mathcal O_{X,P}/\mathfrak m_{P}} \mathfrak m_{P}/\mathfrak m^2_{P}=\dim_{\overline F} \mathfrak m_{P}/\mathfrak m^2_{P}$

Proof

We'll make a proof for $\mathbb A^n$. Assume $P=(a_1, \ldots, a_n) \in \mathbb A^n$ then by $(3.1.8)$ we have a maximal ideal $\mathfrak a_{P}=\lang x-a_1, \ldots, x-a_n \rang_I$. Consider a homomorphism:

$$\theta_P: \overline F[x_1, \ldots, x_n] \rightsquigarrow_M \overline F^n, f \to (\partial f/\partial x_1(P), \ldots, \partial f/\partial x_n(P))$$

If we consider a polynomal from $\mathfrak a_p: q(x)=(x-a_i)p(x)$ then $q'(x)=p(x)+(x-a_i)p'(P) \implies q'(P)=p(P)$. So $\theta_P(\mathfrak a_{P})=\overline F^n$ and $\ker \theta_P=\mathfrak a_{P}^2$ we have

$$\mathfrak a_{P} / \mathfrak a^2_{P} \overset{\phi_P}\cong_M \overline F^n$$

Next, consider $I(X)= \lang f_1, \ldots, f_k \rang_I$ and $J:=\lang \partial f_i/\partial x_j(P) \rang_I$. Consider some function $f_ig \in I(X)$ then $\theta_P(f_ig)=(\ldots, f_i'(P)g(P) + f_i(P)g'(P), \ldots)=(\ldots, f_i'(P)g(P), \ldots) = g(P)J_i$. So $\theta_P(I(X)) = \lang J_1, \ldots, J_k \rang_M$ and thus $\text{rank }J=\dim_{\overline F} \theta_P(I(X))$.

The image $\theta_P(I(X))$ is isomorphic to it's image under natural homomorphism in $\mathfrak a_P^2/\mathfrak a_P^2$ which is $(I(X)+\mathfrak a_P^2)/\mathfrak a_P^2$. So $\text{rank }J=\dim_{\overline F} \theta_P(I(X)) = \dim_{\overline F} (I(X) + \mathfrak a_{P}^2) / \mathfrak a_{P}^2$.

Consider $M_P=\mathfrak a_P/I(X)$ - the image of $\mathfrak a_P$ in $A(X)$ under natural homomorphism. Then $\forall m \in M_P^2: \exists x \in \mathfrak a_P: m = (x+I(X))^2= x^2 + I(X)$ so $M_P^2=(\mathfrak a_p^2+I(X))/I(X)$. Thus by third isomopshism theorem $M_P/M_P^2 \cong \mathfrak a_P/(\mathfrak a_p^2+I(X))$. Next note that $\mathfrak m \in \mathfrak M_I(\mathcal O_{X, P})$ then $\mathfrak m = (A(X) \setminus M_P)^{-1}M_P, \mathfrak m^2 = (A(X) \setminus M_P)^{-1}M_P^2$ (we don't double $-1$ power because $M_P$ is prime so any two functions not in $M_P$ will still be not in $M_P$ and obviously we can take one function to be $1$ to get the whole $(A(X) \setminus M_P)^{-1}$) so

$$\mathfrak m / \mathfrak m^2 \cong \mathfrak a_P / (I(X) + \mathfrak a_P^2)$$

Finally

$$\dim_{\overline F} \mathfrak m / \mathfrak m^2 + \text{rank }J= \\ \dim_{\overline F} \mathfrak a_P / (I(X) + \mathfrak a_P^2) + \dim_{\overline F} (I(X) + \mathfrak a_{P}^2) / \mathfrak a_{P}^2 = \dim_{\overline F} \mathfrak a_{P} / \mathfrak a_{P}^2 = n$$

So

$$P \in \cancel S(X) \iff \text{rank }J = n - \dim X \iff \\ \dim_{\overline F} \mathfrak m / \mathfrak m^2 = \dim X$$

$\square$

Rational points

Starting from this point we'll always assume that $F \in \mathcal F^{\mathcal P}$.

So far we always dealt with $\overline F$ but it's natural when we want to consider varieties over $F$ as well. This is where Galois theory comes very handy.

Recall that if we have a Galois extension $E/_{Gal}F$ then we can define $F$ as $F=\text{Fix}_{\text{Gal}(E/F)}E$. We'll soon see that many notions in algebraic geometry is Galois invariant, meaning we can quickly see how Galois automorphisms acts on these notions. So this is a good tool for analyzing the structure of varieties over $F$.

Notice that we always consider Galois extension $\overline F / F$ since by defintion $\overline F /_{\lhd} F$ and since $F \in \mathcal F^p$ we have $\overline F /_{\boxminus} F$ so by $(2.9.15)$ $\overline F /_{\text{Gal}} F$.

We'll start by stating very simple object - a point $P \in \mathbb A^n$. Then Galois automorphism acts trivially on $P$:

$$P^\sigma = (x_1^\sigma, \ldots, x_n^\sigma), x_i^{\sigma}=\sigma(x_i)$$

Then we can say that:

$$\mathbb A^n_F=\{P \in \mathbb A^n:\sigma \in \text{Gal}(\overline F / F) \implies P^{\sigma}=P\}= \\ \{(x_1, \ldots, x_n): x_i \in F\}$$

Next if we consider a variety $X$ then we say that $X$ is defined over $F$ if vanishing ideal $I(X)=\lang f_1, \ldots, f_n \rang_I, f_i \in F[x]$.

note

We will use notation $X/F$ or $X \in \mathcal V/F$ for variety defined over F.

In this case indeed we can consider $X_F=Z_F(I(X))$ in the field $F$ and each solution of polynomials in the field $F$ will be a solution in $\overline F$ and vice versa. We say that $X_F$ is a set of $F$-rational points.

Next we can define a vanishing ideal:

$$I(X)_F=I(X)\cap F[x_1, \ldots, x_n]$$

Then we'll have:

$$X/F \iff I(X)=I(X)_F \overline F[x_1, \ldots, x_n]$$

We can consider how Galois automorphisms act on functions. If $f \in \overline F[x_1, \ldots, x_n]$ then

$$f(P)^\sigma=f^{\sigma}(P)^\sigma$$

So if $f \in F[x_1, \ldots, x_n]$

$$f(P)^\sigma=f(P^\sigma)$$

We can summarize that if $X/F$ then

$$X_F=X \cap \mathbb A^n_F=\\\{P \in X: \sigma \in \text{Gal}(\overline F / F) \implies P^{\sigma}=P \}$$

Finally we can define coordinate ring:

$$A(X)_F:=F[x_1, \ldots, x_n]/I(X)_F$$

And the function field $F(X)_F:=\mathfrak F(A(X)_F)$