3.1 Topological, affine and projective spaces

Topological spaces

We want to start our discussion of algebraic geometry with defining the most abstract geometrical notion of a topological space.

def: Topological space (closed sets)

$$\begin{align*} &\sphericalangle \\ &X \in \mathcal S \\ &T^c - \text{class of subsets of } X \text{ with properties:}\\ \text{1.} \,\, &\empty, X \in T^c \\ \text{2.} \,\,& Y_i \in T^c \implies \bigcup_{i=1}^nY_i \in T^c \\ \text{3.} \,\,& Y_i \in T^c \implies \bigcap_{i}Y_i \in T^c\\ \hline \\ &(X, T^c) \in \mathcal T - \text{topological space} \\ &Y \in T^c \implies Y - \text{closed set} \\ \end{align*}$$

Each element of $T^c$ is called a closed set. If $Y \in T^c$ then $U = X \setminus Y$ is an open set. We can equivalently define the topology of open sets

def: Topological space (open sets)

$$\begin{align*} &\sphericalangle \\ &X \in \mathcal S \\ &T^o(X) - \text{class of subsets of } X \text{ with properties:}\\ \text{1.} \,\, &\empty, X \in T^o \\ \text{2.} \,\,& U_i \in T^o \implies \bigcap_{i=1}^nU_i \in T^o \\ \text{3.} \,\,& U_i \in T^o \implies \bigcup_{i}U_i \in T^o\\ \hline \\ &(X, T^o ) \in \mathcal T - \text{topological space of open sets} \\ &U \in T^o \implies U - \text{open set} \\ \end{align*}$$

note

The topological space of closed sets and open sets describe essentially the same thing in two different ways. We can always convert topology of open sets to topology of closed sets and vice versa since:

$$U \in T^o \iff X \setminus U \in T^c$$

When we have some topology $T^c$ we will denote the corresponding open topology by $T^o$. If there's ambiguity about what is the underlying topology is we'll write $T^o(T^c)$ where $T^c$ is the underlying closed set topology that is converted to open.

note

Further we'll use the notation $(X, T) \in \mathcal T$ to denote topological space. And then we'll use superscript $T^c$ for closed sets and $T^o$ for open sets.

In a topological space, a topology can be induced on any subset in the following way:

def: Topological subspace

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ &Y \subseteq X \\ & T^c_{|Y}:= \{Y' \cap Y, Y' \in T^c\} \\ & T^o_{|Y}:= \{Y' \cap Y, Y' \in T^o\} \\ \hline \\ & (Y, T_{|Y}) - \text{ topological subspace} \\ & T^c_{|Y} - \text{induced topology of closed sets } \\ & T^o_{|Y} - \text{induced topology of open sets } \\ \end{align*}$$

Next we want to define the basic building blocks of topological spaces akin to primes in numbers and irreducible polynomials in polynomials:

def: Irreducible space

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ &X \ne \empty \\ & \nexists Y_1, Y_2 \in T^c: Y_1 \subset X, Y_2 \subset X, X = Y_1 \cup Y_2 \\ \hline \\ &(X, T) \in \mathcal T^{-} \end{align*}$$

def: Irreducible set

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ & Y \subseteq X \\ &(Y, T_{|Y}) \in \mathcal T^{-} \\ \hline \\ &Y \in S^- \end{align*}$$

def: Closed and open spaces

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ &Y \in T^c \\ &U \in T^o \\ \hline \\ &(Y, T_{|Y}) \in \mathcal T^c - \text{closed space} \\ &(U, T_{|U}) \in \mathcal T^o - \text{open space} \end{align*}$$

Finally, similar to generators of groups and ideals we can define minimal topological sets containing other sets:

def: Set closure

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ &Y \subseteq X \\ \hline \\ &\overline Y := \bigcap_{Y'\in T^c, Y' \supseteq Y} Y' \\ \end{align*}$$

note

By property 3 of topology $\overline Y \in T^c$.

def: Dense set

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ &Y \subseteq X \\ &\overline Y = X \\ \hline \\ &Y - \text{dense} \end{align*}$$

Proposition 3.1.1: Non-empty open set in irreducible closed space is dense and irreducible

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T^{c,-} \\ & U \in T^o, U \ne \empty \\ \hline \\ &\begin{align*} & \overline U = X \hspace{0.5cm} \tag{a}\\ & U \in X^- \hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

Since $X = \overline U \cup (X \setminus U)$ and $X$ is irreducible this means that either $\overline U = X$ or $X \setminus U = X$. The latter is impossible since $U \ne \empty$, so $\overline U = X$.

b.

If we have $U = X_1 \cup X_2=(U \cap X_1) \cup (U \cap X_2) = U \cap (X_1 \cup X_2), X_i \in T^c, X_i \subset X$ then $U \subseteq X_1 \cup X_2, X_1 \cup X_2 \in T_c$ so $X=\overline U = \overline{X_1 \cup X_2}=X_1 \cup X_2$ which is contradiction to irreducibility of $S$.

$\square$

Proposition 3.1.2: The closure or irreducible set is irreducible

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ & Z \in S^- \\ \hline \\ &\overline Z \in S^- \end{align*}$$

Proof

Assume $\overline Z = X_1 \cup X_2, X_i \in T^c_{|\overline Z}, X_i \subset \overline Z$. Then $X_1 \cup X_2 = \overline Z \supseteq X \implies Z = Z \cap(X_1 \cup X_2)= (X_1 \cap Z) \cup (X_2 \cap Z) = X_1' \cup X_2', X_i' \in T^c_{|X}$.

Note that if $X_i \supseteq Z$ then $\overline Z \subseteq X_i$ which contradicts $X_i \subset \overline Z$. So $X_i \nsupseteq Z$ and $X_i' \subset Z$.

$\square$

def: Noetherian topological space

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T \\ &Y_i \in T^c \\ &Y_1 \supseteq Y_2 \supseteq \ldots \implies \exists n: Y_n=Y_{n+1}=\ldots \\ \hline \\ &(X, T) \in \mathcal T^{\mathcal N} \\ \end{align*}$$

Proposition 3.1.3: Each closed set is a finite union of irreducible closed sets

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T^{\mathcal N} \\ & Y \in T^c \\ \hline \\ &\exists! Y_1, \ldots Y_k: Y_i \nsubseteq Y_j, Y_i \in S^- \cap T^c, Y = Y_1 \cup \ldots \cup Y_k \end{align*}$$

Proof

Existence

Let $\mathfrak X:= \{V \in T^c: V \ne \empty, \nexists V_i \in S^- \cap T^c: V = V_1 \cup \ldots \cup V_n\}$. Assume $\mathfrak X \ne \empty$. Then since $(X, T) \in \mathcal T^{\mathcal N}$ by Zorn's lemma there's a minimal by inclusion set $Y \in \mathfrak X$. Since by construction $Y \notin S^- \cap T^c$ (otherwise trivially $V_1=Y$) we can write $\exists Y_1, Y_2 \in T^c: Y = Y_1 \cup Y_2$. Since $Y$ is minimal in $\mathfrak X$ it follows that $Y_1, Y_2 \notin \mathfrak X$ so there's a finite set of varities with the union equals to $Y_i$. So $Y$ is a finite union of irreducible closed sets as well which is a contradiction.

Finally in each representation we can throw out unnecessary $Y_i$ to achieve $Y_i \nsubseteq Y_j$.

Uniqueness

Assume $Y = Y_1 \cup \ldots \cup Y_r = Y'_1 \cup \ldots \cup Y'_s, Y_i, Y'_j \in \mathcal V(S)$. Then $Y'_1 \subseteq Y = Y_1 \cup \ldots \cup Y_r \implies Y'_1=(Y'_1 \cap Y_1) \cup \ldots \cup (Y'_1 \cap Y_r)$.

If $\forall i: Y'_1 \nsubseteq Y_i$ then $(Y'_1 \cap Y_i) \subset Y'_1$ and $Y'_1 \notin \mathcal V(S)$ which is a contradiction. So $\exists i: Y'_1 \subseteq Y_i$. In the same way $Y_i \subseteq Y'_j$ so $Y'_1\subseteq Y'_j$ but we know that $Y_i \nsubseteq Y_j$ so $j=1$ and $Y'_1=Y_i$. By excluding $Y'_1$ and $Y_i$ from representation and repeating the process we finish the proof.

$\square$

Proposition 3.1.4: Each open set is a finite union of open sets

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T^{\mathcal N} \\ & U \in T^o \\ \hline \\ &\exists U_1, \ldots, U_k \in T^o: U = U_1 \cup \ldots \cup U_k \end{align*}$$

Proof

Take some cover $\mathfrak Y$ of $U$. If that's infinite then take all distinct $U_i$ from it and consider $V_n = \bigcup_{i=1}^n U_i$. This means that $V_1 \subseteq V_2 \subseteq ...$. and $(X \setminus V_1) \supseteq (X \setminus V_2) \supseteq ...$. Since $(X, T) \in \mathcal T^{\mathcal N}$ and $X \setminus V_i \in T^c$ we know that $\exists n: k\ge n \implies X \setminus V_k=X \setminus V_n \implies V_k=V_n$. So we have a finite cover $U_1, \ldots, U_n$

$\square$

Affine and projective spaces

After defining the abstract notion of topological space we can move to defining more concrete examples of geometrical spaces.

def: Affine space

$$\begin{align*} &\mathbb{A}^n:=\{(x_1, ..., x_n), x_i \in \overline{F}, F \in \mathcal F\} \end{align*}$$

Note that $\mathbb{A}^n \equiv \overline{F}^n$, but we prefer to use the $\mathbb{A}^n$ notation. This is to emphasize that we're talking about a set that doesn't have any additional structure or properties.

Note that we consider affine (and projective) space on top of algebraically closed field $\overline F$.

def: Projective space

$$\begin{align*} &\sphericalangle \\ &(x_0, \ldots, x_n), (y_0, \ldots, y_n) \in \overline{F}^{n+1} \setminus (0, \ldots, 0) \\ &\sim_P: (x_0, \ldots, x_n) \sim_P (y_0, \ldots, y_n) \iff \exists \lambda \in \overline{F}^*: (x_0, \ldots, x_n) = (\lambda y_0, \ldots, \lambda y_n) \\ \hline \\ &\mathbb{P}^n:=\{[x_0, ..., x_n]_{\sim_P}\} - \text{ projective space} \end{align*}$$

Recall that $[x_0, ..., x_n]_{\sim}$ is the notation for equivalence class.

Below is the example of $\mathbb P^2$:

Note that this each point in $\mathbb P^2$ is a line in 3D-space. But we can make a bijection between these lines and the dots of $\mathbb A^2$ via crossings of these lines and the plane $z=1$. This will cover all $\mathbb A^2$ but there will be some points of $\mathbb P^2$ that are not mapped. Namely the lines that lie in $z=0$ plane.

These lines are conceptualized as points of infinity in $\mathbb P^2$ (note they sort of cross $z=1$ in infinity). But they're not just regular $\pm \infty$. Rather they are a set of directional infinities. Note that we can actually project these infinities into half-circle in the plane so each infinity point will correspond to some point of half circle specifying the direction. This means that we don't distinguish between $+\infty$ and $-\infty$ because they will lie in the same line.

Note that in $\mathbb P^1$ the only point of infinity will be $\pm \infty$.

So overall we can consider $\mathbb P^n$ as $\mathbb A^n$ plus some points of infinity. This makes $\mathbb P^n$ closed in some sense like we made algebraic closures and thus having some nice properties and theorems in it.

On the final note - one benefit of projective spaces is we can actually choose any plane to map it to $\mathbb A^2$. In the picture above if we choose $z=1$ for projection then we'll have one infinity point. On the other hand if we choose $y=1$ then all points will be finite in $\mathbb A^2$.

note

When we use illustration for spaces like $\mathbb A^2$ and $\mathbb P^2$ we draw it as plane or a 3d space. However, note that we need 4 dimensions to draw proper $\mathbb A^2$ and 6 dimensions for $\mathbb P^2$. Why? Remeber that we are looking at spaces over algebraically closed fields. Since $\overline \R = \mathbb C$ we actually need to make illustrations over the complex numbers. But we'll stick to traditional 2d and 3d pictures for illustrative purposes.

note

In the following sections we'll use the notation $\mathbb S^n$ for either of $\mathbb A^n$ or $\mathbb P^n$.

Additionally if we want to say that some topological space lives inside affine space (for example) we'll use the following notation:

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

Zero sets

def: Affine zero set

$$\begin{align*} &\sphericalangle \\ &\mathbb{A}^n,\\ &P \subseteq \overline{F}[x_1, ..., x_n] \\ \hline \\ &Z(P) :=\{X \in \mathbb{A}^n: \forall f\in T \implies f(X)=0\} \end{align*}$$

Example: Finite affine vanishing set

Let $F=\mathbb{R}, P=\{x^2+y^2, x-2y-1\}$. Let's find the set $Z(P)$.

$$\begin{cases} x^2+y^2=0 \\ x-2y-1=0 \end{cases} \implies 5y^2+4y+1=0 \implies \begin{cases} x=\frac{1 \pm 2i}{5} \\ y=\frac{-2\pm i}{5} \end{cases}$$ $$Z(P)=\{(\frac{1 + 2i}{5}, \frac{-2 + i}{5}), (\frac{1 - 2i}{5}, \frac{-2 - i}{5})\}$$

Note that $Z(P)$ is composed of complex numbers. This is because, as highlighted at the outset of this chapter, we are always considering $\overline{F}$ rather than $F$. Since $F = \mathbb{R}$, it follows that $\overline{F} = \mathbb{C}$.

Example 2: Infinite affine vanishing set

Let $F=\mathbb{R}, P=\{x^2+y^2\}$. In this case $y=\pm ix$ and $Z(P)$ are two lines in the complex plane.

Next we want to define vanishing sets for projective spaces. However if we do it in the exact same way as for affine spaces we'll have a problem. Namely the polynomials $f([x_0, \ldots, x_n])$ will have different values depending on the representative of the equivalence class $[x_0, \ldots, x_n]$. This can be resolved if we require for polynomials of vanishing set being homogeneous. In other words, if we consider a set of polynomials as a graded ring

$$\overline{F}[x_0, \ldots, x_n]=\bigoplus_{d=0}^{\infty}\overline{F}[x_0, \ldots, x_n]_d, \\ \overline{F}[x_0, \ldots, x_n]_d\overline{F}[x_0, \ldots, x_n]_e \subseteq \overline{F}[x_0, \ldots, x_n]_{d+e}$$

We want the polynomials to be in $\overline{F}[x_0, \ldots, x_n]_d$ for some $d$ ($d$ can have different values in the same set of polynomials). In this case $[x_0, \ldots, x_n] \sim [y_0, \ldots, y_n] \implies (x_0, \ldots, x_n)=(\lambda y_0, \ldots, \lambda y_n), \lambda \ne 0$. And if $f \in \overline{F}[x_0, \ldots, x_n]_d$ then $f(x_0, \ldots, x_n)=0 \iff \lambda^d f(x_0, \ldots, x_n)= f(\lambda x_0, \ldots, \lambda x_n) = f(y_0, \ldots, y_n)$. So we have the following definition:

def: Projective zero set

$$\begin{align*} &\sphericalangle \\ &\mathbb{P}^n,\\ &\overline{F}[x_0, \ldots, x_n]=\bigoplus_{d=0}^{\infty}\overline{F}[x_0, \ldots, x_n]_d \\ &P \subseteq \bigcup_{d=0}^{\infty} \overline{F}[x_0, \ldots, x_n]_d \\ \hline \\ &Z(P) :=\{Y \in \mathbb{P}^n: \forall f\in P \implies f(Y)=0\} \end{align*}$$

Next, we want to expand the definition of zero set to ideals. For affine zero set this is trivial as ideal is just some set of polynomials over the affine space. So the definition is exactly the same.

For ideals in the projective space it is more complicated. Our requirement for the polynomials of zero set was that $P \subseteq \bigcup_{d=0}^{\infty} \overline{F}[x_0, \ldots, x_n]_d$. This doesn't hold if $P \lhd_R \overline{F}[x_0, \ldots, x_n]$ because $P$ should be closed under multiplication on any polynomial including non-homogeneus so naturally it will contain non-homogeneous polynomials. To overcome this difficulty let's recall that by $(2.5.13) \implies \overline{F}[x_0, \ldots, x_n] \in \mathcal R^{\mathcal N} \overset{(2.4.9)}\implies \exists p_1, \ldots, p_k: \lang p_1, \ldots, p_k \rang_I = P$. So if $\{p_1, \ldots, p_k\} \subseteq \bigcup_{d=0}^{\infty} \overline{F}[x_0, \ldots, x_n]_d$ then we can say $Z(P)=Z(\lang p_1, \ldots, p_k \rang_I):=Z(\{ p_1, \ldots, p_k \})$. Consider some polynomial $g \in P, g(x_0, \ldots, x_n) = f(x_0, \ldots, x_n)p_i(x_0, \ldots, x_n)$. Note that if $Y \in Z(P)$ then $p_i(Y)=0$. It's true that $f(x)$ might be not homogeneous and thus undefined at $Y$. But we "sort of" don't care about it because $g(Y)=f(Y)p_i(Y)$ and $p_i(Y)=0$ and no matter what value $f(Y)$ takes we always have $g(Y)=0$. So we introduce the following definitions:

def: Homogeneous ideal

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &I \lhd_R \overline{F}[x_0, \ldots, x_n] \\ &\exists P \subseteq \bigcup_{d=0}^{\infty} \overline{F}[x_0, \ldots, x_n]_d, |P| < \infty: I = \lang P \rang_I \\ \hline \\ &I \lhd_{gr} \overline{F}[x_0, \ldots, x_n] \end{align*}$$

note

Homogeneous ideal is still an ideal. That is $I \lhd_{gr} \overline{F}[x_0, \ldots, x_n] \implies I \lhd_{R} \overline{F}[x_0, \ldots, x_n]$

Proposition 3.1.5: Homogeneous ideals properties

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &I_1, I_2 \lhd_{gr} \overline F[x_0, \ldots, x_n] \\ \hline \\ &\begin{align*} &I_1I_2 \lhd_{gr} F[x_0, \ldots, x_n] \hspace{0.5cm} \tag{a}\\ &I_1+I_2 \lhd_{gr} F[x_0, \ldots, x_n] \hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

Note that by $(2.5.13)$ $\overline F[x_0, \ldots, x_n] \in \mathcal R^{\mathcal N}$ so each ideal is finitely generated. So in $(a)$ we'll have an ideal that is generated by cross products, each cross product will be homogeneous.

b.

$(2.3.4.b) \implies I_1+I_2 = \lang I_1 \cup I_2 \rang_I$ so the generating set for $I_1+I_2$ is homogeneous.

$\square$

def: Projective zero set of an ideal

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &I \lhd_{gr} \overline{F}[x_0, \ldots, x_n] \\ &P \subseteq \bigcup_{d=0}^{\infty} \overline{F}[x_0, \ldots, x_n]_d, |P| < \infty: I = \lang P \rang_I \\ \hline \\ &Z(I):=Z(P) \end{align*}$$

We'll make a final note that any set $P$ can be turned into ideal $\lang P \rang_I$ and as we discussed above $Z(P)=Z(\lang P \rang_I)$. So instead of discussing zero sets of arbitrary polynomials subsets we can limits disucssions only to the zero sets of ideals for $\mathbb A^n$ and homogeneous ideals in $\mathbb P^n$.

Proposition 3.1.6: Zero sets properties

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n (\text{or } \mathbb P^n) \\ &I_1, I_2, \ldots \lhd_R \overline F[x_1, \ldots, x_n] (\text{or } I_1, I_2, \ldots \lhd_{gr} \overline F[x_0, \ldots, x_n]) \\ \hline \\ &\begin{align*} &I_1 \subseteq I_2 \implies Z(I_1) \supseteq Z(I_2) \hspace{0.5cm} \tag{a} \\ &Z(I_1 I_2 \ldots I_n) = Z(I_1) \cup Z(I_2) \cup \ldots \cup Z(I_n) \hspace{0.5cm} \tag{b} \\ &Z(I_1 + I_2 + \ldots) = Z(I_1) \cap Z(I_2) \cap \ldots \hspace{0.5cm} \tag{c} \\ \end{align*} \end{align*}$$

Proof

a.

Follows from definition

b.

By $(2.5.13)$ we know that $\overline F[x_1, \ldots, x_n] \in \mathcal R^{\mathcal N}$ so each ideal is finitely generated. So $I_1\ldots I_n$ is generated by the cross products of the generating sets. It's easy to verify that in this case $(b)$ holds true.

For projective case we know from $(3.1.5)$ that $I_1I_2 \ldots I_n \lhd_{gr} \overline F[x_0, \ldots, x_n]$, so the proof above is the same.

c.

$I_1+I_2+\ldots$ is a finite linear combinations of terms from generating sets of $I_1, I_2, \ldots$ (see $(2.4.2), (2.3.4)$). So we can always take polynomials from $I_n$ to prove $Z(I_1 + I_2 + \ldots) \subseteq Z(I_n) \implies Z(I_1 + I_2 + \ldots) \subseteq Z(I_1) \cap Z(I_2) \cap \ldots$. On the other hand if $Q \in Z(I_1) \cap Z(I_2) \cap \ldots$ it will be a zero for each $Z(I_n)$ and so of $Z(I_1 + I_2 + \ldots)$.

For projective case we know from $(3.1.5)$ that $I_1+I_2 +\ldots \lhd_{gr} \overline F[x_0, \ldots, x_n]$, so the proof above is the same.

$\square$

Lemma 3.1.7: Field extension that is finitely generated F-algebra is algebraic

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &E \in \mathcal A_{\text{id}, F} \\ &E - \text{finitely generated } F\text{-algebra} \\ \hline \\ &E/_AF \end{align*}$$

Proof intuition

$E/F$ means that $E=F(x_1, \ldots, x_n)$ so $x_i$ should be in denominators. On the other hand since it's a finitely-generated algebra it should be polynomials so $1/x_i$ should be represented as a polynomial. It cannot be if $x_i$ is transcendental but can be if it's algebraic (like in $\mathbb Q(\sqrt 2)/\mathbb Q$ we have $1/\sqrt 2 = \sqrt 2 / 2$)

Proof

We will assume that $E/_TF$ and prove that it cannot be a finitely generated $F$-algebra.

Assume $\text{trdeg}(E/F)=1$, that is $E/_AF(x)$, $x$ is the element of transcendence basis. Note that each element in $E$ is some polynomial over finite set of elements in $E$ with coefficents in $F$.

$$e \in E \implies e = p(\alpha_1, \ldots, \alpha_n, x)$$

If we consider $E/F(x)$ then each element in $E$ is a polynomial of algebraic elements with coefficients in $F(x)$. Since each algebraic element have finite degree it follows that $[E:F(x)]< \infty$. Let $\{e_1, \ldots, e_l\} \in \mathfrak B_{F(x)}(E)$ and write multiplication table:

$$e_ie_j=\sum_k\frac{a_{ijk}(x)}{b_{ijk}(x)}e_k, a_{ijk}, b_{ijk} \in F[x]$$

Now consider some arbitrary $f_1, \ldots f_m \in E$ and add $f_0 = 1$ to it. Let $A$ be the $F$-algebra they generate. We want to prove that it's smaller than $E$, so $E$ cannot be finitely-generated $F$-algebra. In terms of our basis:

$$f_i = \sum_j \frac{c_{ij}(x)}{d_{ij}(x)}e_j, c_{ij}, d_{ij} \in F[x]$$

Consider some element $a \in A$ - it's a linear combination of $f_0=1$ and some products of $f_1, \ldots, f_m$. If we apply the multiplication table of $e_ie_j$ we'll see that $a$ is some $F(x)$-linear combination of $e_i$ where denominators are only $b_{ijk}$ and $d_{ij}$. In other words for any $a \in A$ as a linear combination of $e_i$ all irreducible factors in denominators are from $b$ and $d$.

Next note that the set of $f_i$ is finite so we'll have a finite number of irreducibles in denominator. So $z:=1/\text{some other irreducible}$ will have it in denominator and therefore cannot be expressed as a polynomial in $f_i$. Note that by $(2.4.12)$ irreducible = prime in $F[x]$ which is a UFD. So we can use similar argument showing there are infinitely many primes (or irreducibles) in $F[x]$ (if there's a finite number of primes $p_1, \ldots, p_n$ then $p_{n+1}=p_1 \cdot \ldots \cdot p_n+1$ is a prime).

Finally for $\text{trdeg}(E/F)>1$ we can choose subextension of $\text{trdeg}(E/F)=1$ that will be algebraic by the above, hence contradiction.

$\square$

Proposition 3.1.8: Weak Zero point theorem (Nullstellensatz)

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ & \mathfrak m \in \mathfrak M_I(\overline F[x_1, \ldots, x_n]) \\ &P \subseteq \overline F[x_1, \ldots, x_n]\\ \hline \\ &\begin{align*} &\exists a_1, \ldots a_n \in \overline F: \lang x_1-a_1, \ldots, x_n-a_n\rang_I=\mathfrak m \hspace{0.5cm} \tag{a}\\ &Z(P) = \empty \implies \lang P \rang_I = \overline F[x_1, \ldots, x_n] \hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

By $(2.4.21)$ we have $E:=\overline F[x_1, \ldots, x_n] / \mathfrak m$ is a field. By $(2.4.6)$ and definition of maximal ideal $\mathfrak m \cap \overline F = \empty$ so $E/\overline F$. On the other hand $\overline F[x_1, \ldots, x_n]$ is a finitely-generated $\overline F$-algebra and so is $E$. So by $(3.1.7)$ we have $E/_A \overline F$. But since $\overline F$ is algebraically closed that means $E=\overline F$. So under the natural homomorphism map $\overline F[x_1, \ldots, x_n] \to \overline F[x_1, \ldots, x_n]/\mathfrak m$ we have $x_i \mapsto a_i \in \overline F$ so $x_i-a_i \in \mathfrak m$ and

$$\lang x_1-a_1, \ldots, x_n-a_n\rang_I\subseteq \mathfrak m$$

Denote $I:=\lang x_1-a_1, \ldots, x_n-a_n\rang_I$ and consider evaluation homomorphism $\varepsilon_{(a_1, \ldots, a_n)}: \overline F[x_1, \ldots, x_n] \to \overline F$. Since $\ker \varepsilon_{(a_1, \ldots, a_n)} = I$ by $(2.3.9)$ we have $\overline F[x_1, \ldots, x_n]/I \cong \varepsilon_{(a_1, \ldots, a_n)}(\overline F[x_1, \ldots, x_n])=\overline F$. So $\overline F[x_1, \ldots, x_n]/I \in \mathcal F \overset{(2.4.21)}\implies I \in \mathfrak M_I(\overline F)$. And thus we have

$$\lang x_1-a_1, \ldots, x_n-a_n\rang_I = \mathfrak m$$

b.

$$\lang P \rang_I \ne \overline F[x_1, \ldots, x_n] \implies \\ \lang P \rang_I \subseteq \mathfrak m = \lang x_1-a_1, \ldots, x_n-a_n\rang_I \implies \\ (a_1, \ldots, a_n) \in Z(P) \implies Z(P) \ne \empty$$

$\square$

Proposition 3.1.9: Zero point theorem (Nullstellensatz)

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &g, p_1, \ldots, p_m \in \overline F[x_1, \ldots, x_n] \\ &\forall z \in Z(p_1, \ldots, p_m): g(z) = 0 \\ \hline \\ &g \in \sqrt{\lang p_1, \ldots, p_m \rang_I} \end{align*}$$

Proof

Consider polynomials $P:=\{p_1, \ldots, p_m, x_{n+1}g-1\} \subseteq \overline F[x_1, \ldots, x_{n+1}]$. It's obvious that $Z(P)=\empty$ so by $(3.1.8)$ we have $\lang P \rang_I = \overline F[x_1, \ldots, x_{n+1}]$ and thus:

$$1=f_1p_1+\ldots+f_mp_m+f_{m+1}(x_{n+1}g-1)$$

Now consider a homomorphism $\phi: \overline F[x_1, \ldots, x_{n+1}] \to \overline F(x_1, \ldots, x_{n}), x_{n+1} \to 1/g$. The image of the equation above under this homomorphism becomes:

$$1=f_1(x_1, \ldots, x_n, 1/g)p_1+\ldots+f_m(x_1, \ldots, x_n, 1/g)p_m$$

If we clear the $g$ denominators we'll have a $g^k$ on the left-hand side and a linear combination of polynomials $p_i$ and with coefficents in $\overline F[x_1, \ldots, x_n]$ on the right hand side which belongs to $\lang p_1, \ldots, p_m \rang_I$.

$\square$

Zariski topology

In the previous section we figure out that ideals in $\overline F[x_1, \ldots, x_n]$ can be mapped to zero sets in $\mathbb A^n$ and $\mathbb P^n$. It turns out that these sets form a topology on $\mathbb A^n$ and $\mathbb P^n$.

def: Zariski topology

$$\begin{align*} &\sphericalangle \\ &T:=\mathcal Z(\mathbb A^n) \\ &T':=\mathcal Z(\mathbb P^n) \\ &X \in T^c \iff \exists I \lhd_R \overline F[x_1, \ldots, x_n]:Z(I)=X \\ &X \in T'^c \iff \exists I \lhd_{gr} \overline F[x_0, \ldots, x_n]:Z(I)=Z \\ \end{align*}$$

Proposition 3.1.10: Zariski topology is a topology of closed sets

$$\begin{align*} &\sphericalangle \\ & \\ \hline \\ &\begin{align*} & (\mathbb A^n, \mathcal Z(\mathbb A^n)) \in \mathcal T\hspace{0.5cm} \tag{a}\\ & (\mathbb P^n, \mathcal Z(\mathbb P^n)) \in \mathcal T\hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

First, $\empty =Z(\overline F[x_1, \ldots, x_n]), \mathbb A^n = Z(\{0\})$. Then if we have $X_i \in T^c$ then we can take $I_i \lhd \overline F[x_1, \ldots, x_n]: X_i=Z(I_i)$. Then by $(3.1.5)$ $X_1 \cup \ldots \cup X_n=Z(I_1\ldots I_n) \in \mathcal Z^c(\mathbb A^n)$ and $X_1 \cap X_2 \cap \ldots = Z(I_1 + I_2 + \ldots) \in \mathcal Z^c(\mathbb A^n)$.

b.

Proved exactly the same

$\square$

def: Algebraic set

Algebraic set is any closed set in Zariski topology.

note

We will refer to algebraic sets and closed sets interchangeably. However note that there could be other types of closed sets under other topologies.

In the previous sections on zero sets we defined the mapping from ideals to sets. Now when we have a topology on $\mathbb S^n$ we want to define the reverse mapping:

def: Vanishing ideal

$$\begin{align*} &\sphericalangle \\ &X \subseteq \mathbb A^n\\ &Y \subseteq \mathbb P^n\\ \hline \\ &I(X) :=\{f \in \overline{F}[x_1, ..., x_n]:\forall Q \in X \implies f(Q)=0\} \\ &I(Y) :=\lang f \in \bigcup_{d=0}^{\infty} \overline{F}[x_0, \ldots, x_n]_d:\forall Q \in X \implies f(Q)=0\rang_I \end{align*}$$

note

In affine case the set of polynomials that vanishes on $X$ naturally forms an ideal. In projective case it's not true so note we're taking an ideal closure.

Example: Affine vanishing ideal

Let $k=\mathbb{R}, X=\{(\frac{1 + 2i}{5}, \frac{-2 + i}{5}), (\frac{1 - 2i}{5}, \frac{-2 - i}{5})\}$. Let's find the ideal $I(X)$. As can be seen from example above:

$$I(X) = \lang x^2+y^2, x-2y-1 \rang_I$$

Example: Projective vanishing ideal

Consider $Y:=\{[1,0,0]\}\subseteq \mathbb P^2$. For this set $I(Y):=\lang y, z \rang_I$.

Corrolary 3.1.11: Affine and projective spaces are irreducible

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

Proof

Since $I(S)=\{0\} \in \mathfrak P_I(\overline F[x_1, \ldots, x_n])$ then by $(3.1.13.i)$ we have $S \in S^-$ so $(X, T) \in \mathcal T^{-}(S)$.

$\square$

Proposition 3.1.12: Affine and projective spaces are noetherian

$$\begin{align*} &(\mathbb S^n, T := \mathcal Z(\mathbb S^n)) \in \mathcal T^{\mathcal N} \end{align*}$$

Proof

Consider $X_i \in T^c, X_1 \supseteq X_2 \supseteq \ldots$. Then $I(X_1) \subseteq I(X_2) \subseteq \ldots$. Since $\overline F[x_1, \ldots, x_n] \in \mathcal R^{\mathcal N}$ we know that $\exists k: I(X_k)=I(X_{k+1})=\ldots$. So $Z(I(X_k))=Z(I(X_{k+1})) = \ldots$. Since $Z(I(X))=X$ for closed sets, we finish the proof.

$\square$

Proposition 3.1.13: Algebraic sets and vanishing ideals relations

$$\begin{align*} &\sphericalangle \\ &(\mathbb S^n, T := \mathcal Z(\mathbb S^n)) \in \mathcal T \\ &X, X_1, X_2, \ldots \subseteq \mathbb S^n \\ &J, J_1, J_2, \ldots \lhd_R \overline{F}[x_1, ..., x_n] (\lhd_{gr} \overline{F}[x_0, ..., x_n] \text{ for } \mathbb P^n) \\ \hline \\ &\begin{align*} &J_1 \subseteq J_2 \implies Z(J_1) \supseteq Z(J_2) \tag{a}\\ &X_1 \subseteq X_2 \implies I(X_1) \supseteq I(X_2) \tag{b}\\ &I(Z(J)) = \sqrt{J} \,\,(\text{additional requirement } J \ne \lang x_0, \ldots, x_n \rang_I \text{ for }\mathbb P^n) \tag{c} \\ &Z(I(X)) = \overline X \tag {d} \\ &Z(J_1\ldots J_k) = Z(J_1) \cup \ldots \cup Z(J_k) \hspace{0.5cm} \tag{e} \\ &Z(J_1 + J_2 + \ldots) = Z(J_1) \cap Z(J_2) \cap \ldots \hspace{0.5cm} \tag{f}\\ &I(X_1 \cup \ldots \cup X_k) = I(X_1) \cap \ldots \cap I(X_k) \hspace{0.5cm} \tag{g} \\ &X_i \in T^c \implies I(X_1 \cap X_2 \cap \ldots) = \sqrt{I(X_1) + I(X_2) + \ldots }\hspace{0.5cm}\tag{h} \\ &X \in T^c, X \in \mathbb S^{n-} \iff I(X) \in \mathfrak P_I(\overline{F}[x_1, ..., x_n]) (\mathfrak P_I(\overline{F}[x_0, ..., x_n] \text{ for } \mathbb P^n) \hspace{0.5cm}\tag{i} \\ \end{align*} \end{align*}$$

Proof

a., b.

Obvious from definitions

c.

Since $I(Z(J))$ vanishes on all points where $J$ vanishes, by $(3.1.9)$ we have $I(Z(J)) \subseteq \sqrt J$. On the other hand if $p \in \sqrt J \implies p^k \in J \implies \forall x \in Z(J): p^k(x)=0 \implies \forall x \in Z(J): p(x)=0 \implies p(x) \in I(Z(J)) \implies \sqrt J \subseteq I(Z(J))$.

For projective case there's one exception, namely the ideal $J:=\lang x_0, \ldots, x_n \rang_I$ is radical, i.e. $\sqrt J = J$. The zero set $Z(J)$ should have been $\{(0, \ldots, 0)\}$ but the problem is this point is not in $\mathbb P^n$. So $Z(J)=\empty$ and $I(\empty)=\overline F[x_0, \ldots, x_n]$. If zero point was in $\mathbb P^n$ then this problem wouldn't exist. So we make additional requirement that $J \ne \lang x_0, \ldots, x_n \rang_I$ which is also called irrelevant ideal.

d.

Note that $x \in X \implies \forall p \in I(X): p(x)=0 \implies x \in Z(I(X)) \implies X \subseteq Z(I(X))$. By definition $Z(I(X)) \in \mathcal Z^c(\mathbb S^n)$ so $\overline X \subseteq Z(I(X))$.

On the other hand consider some $Y\in \mathcal Z^c(\mathbb S^n), Y \supseteq X$. Then $\exists I \lhd_R \overline{F}[x_1, ..., x_n] (\lhd_{gr} \overline{F}[x_0, ..., x_n] \text{ for } \mathbb P^n ), : Y = Z(I) \supseteq X \implies I(Z(I)) \subseteq I(X)$. But since $I \subseteq I(Z(I))$ so $Y=Z(I) \supseteq Z(I(Z(I))) \supseteq Z(I(X))$. So $\overline X \supseteq Z(I(X))$

e., f.

Follows from $(3.1.6)$

g.

On the left hand side and right hand side we state that some $p$ vanishes on all $X_i$.

h.

$$I(Z(I(X_1)+\ldots+I(X_k))) = \sqrt{I(X_1)+\ldots+I(X_k)}$$

i.

$\implies$

Assume $I(X) \notin \mathfrak P_I(\overline F[x_1, \ldots, x_n]))$ and $\exists pq \in I(X), p,q \notin I(X)$. Then $Z(pq) \supseteq X$ and $X=X\cap Z(pq)= X \cap (Z(p) \cup Z(q)) = (X\cap Z(p))\cup(X\cap Z(q))$. Since $p, q \notin I(X)$ we have $X \cap Z(p) \subset X, X \cap Z(q) \subset X$ and so $X \notin \mathbb S^{n,-} \cap T^c$

$\impliedby$

If $X \notin \mathbb S^{n,-} \cap T^c$ then $X=X_1 \cup X_2, X_i \subset X$ then $I(X_i) \supset I(X)$. Consider $p \in I(X_1)\setminus I(X)$, $q \in I(X_2)\setminus I(X)$. Then $pq$ is zero on $X_1$ and $X_2$ so $pq \in I(X_1) \cap I(X_2)=I(X_1 \cup X_2)=I(X)$ so $pq \in I(X)$.

$\square$

From the statement above we see that there's a natural bijection $J \to Z(J)$ and $X \to I(X)$ from algebraic sets to radical ideals (the ideals such that $J = \sqrt J$). By $(3.1.8)$ points corrsepond to maximal ideals. And by $(3.1.13.i)$ Prime ideal cossespond to irreducible closet sets. We summarize this in the following table:

$$\def\arraystretch{1.5} \begin{array}{c:c} \textbf{Geometry} & \textbf{Algebra} \\ \hline \mathbb S^n & \empty \\ \hline \mathbb \empty & \overline F[x_1, \ldots, x_n] \\ \hline \text{Closure} & \text{Radical} \\ \hline \text{Closed set} & \text{Radical ideal} \\ \hline \text{Irreducible closed set} & \text{Prime ideal} \\ \hline \text{Point} &\text{Maximal ideal} \end{array}$$