---- Fix $n \in \mathbb{N}$ and let $\mathbb{C}[x_{1},\dots,x_{n}]$ denote the set of [[polynomial|polynomials]] in the variables $x_{1},\dots,x_{n}$ with [[complex numbers|complex coefficients]]. For a subset of polynomials $S \subset \mathbb{C}[x_{1},\dots,x_{n}]$, we denote its **vanishing set** $V(S):=\{ (z_{1},\dots,z_{n}) \in \mathbb{C}^{n} : f(z_{1},\dots,z_{n}) =0 \text{ for all } f \in S \} \subset \mathbb{C}^{n}.$ > [!definition] Definition. ([[Zariski Topology of varieties]]) > $\tau_{Z}:=\{ U \subset \mathbb{C}^{n} : \mathbb{C}^{n} \cut U = V(S) \text{ for some } S \subset \mathbb{C}[x_{1},\dots,x_{n}] \}$ is a topology on $\mathbb{C}^{n}$, called the **Zariski Topology**. Its [[closed set|closed sets]] are exactly the sets $V(S)$ for $S \subset \mathbb{C}[x_{1},\dots,x_{n}]$. > [!justification] > **(i). Let $\{ S_{i} : i \in I \}$ be a collection of subsets of $\mathbb{C}[x_{1},\dots,x_{n}]$. Show that $V(\bigcup_{i\in I}^{}S_{i})=\bigcap_{i \in I} V(S_{i})$.** > Let $z=(z_{1},\dots, z_{n}) \in V(\bigcup_{i=I}^{}S_{i})$. Then $f(z_{1},\dots,z_{n})=0$ for all $f \in S_{i}$, for all $i \in I$. That is to say, $z \in V(S_{i})$ for all $i \in I$ and hence $z \in \bigcap_{i \in I} V(S_{i})$. So $V(\bigcup_{i\in I}^{}S_{i})\subset \bigcap_{i \in I} V(S_{i})$. > Conversely, let $z=(z_{1},\dots,z_{n}) \in \bigcap_{i \in I}^{}V(S_{i})$. So $z \in V(S_{i})$ for all $i \in I$: $f(z_{1},\dots,z_{n})=0$ for all $f \in S_{i}$, for all $i \in I$. It follows that $f(z_{1},\dots,z_{n})=0$ for all $f \in \bigcup_{i \in I}^{}S_{i}$. So $z \in V(\bigcup_{i \in I}^{}S_{i})$ and the reverse inclusion follows. > **(ii). If $S_{1}, S_{2} \subset \mathbb{C}[x_{1},\dots,x_{n}]$, prove that $V(S_{1} \cdot S_{2})=V(S_{1}) \cup V(S_{2})$, where $S_{1} \cdot S_{2}:= \{ f g : f \in S_{1}, g \in S_{2} \}$.** > Let $z=(z_{1},\dots,z_{n}) \in V(S_{1}) \cup V(S_{2})$. Then $f(z_{1},\dots,z_{n})=0$ for all $f \in S_{1}$ or $g(z_{1},\dots,z_{n})=0$ for all $g \in S_{2}$. In either case, it is clear that $(fg)(z_{1},\dots ,z_{n})=f(z_{1},\dots,z_{n})g(z_{1},\dots,z_{n})=0 \text{ for all } f \in S_{1}, g \in S_{2}$ and hence $z \in V(S_{1} \cdot S_{2})$. So $V(S_{1} \cdot S_{2}) \supset V(S_{1}) \cup V(S_{2})$. > Conversely, let $z \in V(S_{1} \cdot S_{2})$: $(fg)(z_{1},\dots ,z_{n})=f(z_{1},\dots,z_{n})g(z_{1},\dots,z_{n})=0 \text{ for all } f \in S_{1}, g \in S_{2}.$ We want to show that $f(z_{1},\dots,z_{n})=0$ for all $f \in S_{1}$ or $g(z_{1},\dots,z_{n})=0$ for all $g \in S_{2}$. > Suppose there exists $f_{*}\in S_{1}$ such that $f_{*}(z_{1},\dots,z_{n}) \neq 0$. Then we may divide both sides of the equation $f_{*}(z_{1},\dots,z_{n})g(z_{1},\dots,z_{n})=0$ by $f_{*}(z_{1},\dots,z_{n})$ to obtain that $g(z_{1},\dots,z_{n})=0$. Since $g$ has remained arbitrary, we can conclude that in this case $g(z_{1},\dots,z_{n})=0$ for all $g \in S_{2}$, i.e., $z \in V(S_{2})$ and thus $z \in V(S_{1}) \cup V(S_{2})$. Likewise, the existence of $g_{*} \in S_{2}$ with $g_{*}(z_{1},\dots,z_{n}) \neq 0$ would imply that $z \in V(S_{1})$ and in turn that $z \in V(S_{1}) \cup V(S_{2})$. Of course, if no such $f_{*}$ exists and no such $g_{*}$ exists then we immediately have $z \in V(S_{1}) \cap V(S_{2})$ and in turn $z \in V(S_{1}) \cup V(S_{2})$. Thus, we have shown the reverse inclusion: $V(S_{1} \cdot S_{2}) \subset V(S_{1}) \cup V(S_{2})$. > **(iii). Prove that the Zariski Topology $\tau_{Z}:=\{ U \subset \mathbb{C}^{n} : \mathbb{C}^{n} \cut U = V(S) \text{ for some } S \subset \mathbb{C}[x_{1},\dots,x_{n}] \}$ is in fact a topology on $\mathbb{C}^{n}$.** > >1. $\emptyset \in \tau_{Z}$ because $\mathbb{C}^{n} \cut \emptyset=\mathbb{C}^{n}$ is the vanishing set of the singleton $\{ f \}$ containing the trivial polynomial $f=f(z_{1},\dots,z_{n}) \equiv 0$. $\mathbb{C}^{n} \in \tau_{Z}$ because its complement, $\emptyset$, is the vanishing set of the singleton $\{ c \}$ containing the constant polynomial $c=c(z_{1},\dots ,z_{n}) \equiv i$. > >2. Let $\{ U_{\alpha} \}_{\alpha \in J}$ be a collection of sets in $\tau_{Z}$ indexed by $\alpha \in J$ for $J$ some index set. For each $\alpha \in J$, denote by $S_{\alpha}$ a subset of $\mathbb{C}[z_{1},\dots,z_{n}]$ for which $\mathbb{C}^{n}-U_{\alpha}$ is the vanishing set. Now, that $\tau_{Z}$ is stable under arbitrary unions follows from the following computation (we use **(i)**): $\begin{align} \mathbb{C}^{n} - \bigcup_{\alpha \in J}^{}U_{\alpha} = & \bigcap_{\alpha \in J}^{} (\mathbb{C}^{n} - U_{\alpha}) \\ = & \bigcap_{\alpha \in J}^{} V(S_{\alpha}) \\ = & V(\bigcup_{i \in I}^{} S_{\alpha}). \end{align}$ >3. Let $U_{1},\dots,U_{n}$ be a finite collection of sets in $\tau_{Z}$. For each $k \in [n]$, denote by $S_{k}$ the subset of $\mathbb{C}[z_{1},\dots,z_{n}]$ for which $\mathbb{C}^{n}-U_{k}$ is the vanishing set. Now, that $\tau_{Z}$ is stable under finite intersection follows from the following computation (where the last equality follows from **(ii)** and induction): >$\begin{align} \mathbb{C}^{n}- \bigcap_{k=1}^{n} U_{k} = & \bigcup_{k=1}^{n} (\mathbb{C}^{n} - U_{k}) \\ = & \bigcup_{k=1}^{n} V(S_{k}) \\ = & V\left( \prod_{k=1}^{n} s_{k} \right). \end{align}$ > **(iv). Prove that the complement of any point in $\mathbb{C}^{n}$ is open in the Zariski Topology.** > Let $z'=(z_{1}',\dots,z_{n}') \in \mathbb{C}^{n}$. We want to show that $\mathbb{C}^{n} - \{ z' \}$ is an element of $\tau_{Z}$, ie., that $\mathbb{C}^{n}- (\mathbb{C}^{n} - \{ z' \}) = \{ z'\} = V(S)$ for some $S \subset \mathbb{C}[z_{1},\dots,z_{n}]$. For $i \in [n]$, define $S:=\{ f_{1},\dots,f_{n} \}$, where $\begin{align} f_{1}(z_{1},\dots,z_{n}):= & z_{1} - z_{1}' \\ f_{2}(z_{1},\dots,z_{n}) := & z_{2} - z_{2}' \\ \vdots & \\ f_{n}(z_{1},\dots, z_{n}):= & z_{n} - z_{n}'. \end{align}$ Notice that $\begin{align} f_{1}(z_{1}',\dots, z_{n}') = & z_{1}' - z_{1}'=0 \\ f_{2}(z_{1}',\dots, z_{n}') = & z_{2}' - z_{2}' = 0 \\ \vdots & \\ f_{n}(z_{1}',\dots, z_{n}') = & z_{n}' - z_{n}' = 0, \end{align}$ from which it is clear that $\{ z' \} \in V(S)$. Also notice that given $z^{*} \neq z'$, we have $z^{*}_{j} \neq z'_{j}$ for some $j \in [n]$ and therefore $f_{j}(z_{1}^{*}, \dots, z_{n}^{*}) = z_{j}^{*} - z_{j}' \neq 0,$ meaning that $z^{} \notin V(S)$ for all $z \neq z'$. Hence, $\{ z \}=V(S)$. ^46ebf1 ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```