1.3 Morphisms

Definition

Let $V\subseteq k^n$ and $W\subseteq k^m$ be irreducible algebraic sets. A map $\alpha :V\to W$ is called a morphism if there exists $f_1,\cdots ,f_m\in k[x_1,\cdots ,x_n]$ such that $\alpha (x)=(f_1(x),\cdots ,f_m(x))$ for any $x\in V$.

Example. Define prime ideals $P_1\subseteq k[x_1,\cdots ,x_n],P_2\subseteq k[y_1,\cdots ,y_m]$ and $V=V(P_1),W=V(P_2)$. Consider $k$-value functions on $V$

$$\{k\text{mbox}-valuefunctionsonV\}=\{V\to k:\text{mbox}morphism\}=\{f{|}_V:f\in k[x_1,\cdots ,x_n]\}.$$

Note that $f{|}_V=g{|}_V$ iff $(f-g){|}_V=0$ iff $f-g\in P_1$. Therefore, $\{k\text{mbox}-valuefunctionsonV\}\simeq k[x_1,\cdots ,x_n]/P_1$.

$k$-Valued Functions and Coordinate Rings

• A $k$-valued function on an algebraic set $V$ (as defined in the notes) is indeed a morphism from $V$ to $k$.
• The coordinate ring $\Gamma (V)$ can be viewed as the algebraic structure (a ring) formed by all such $k$-valued morphisms from $V$ to $k$, where morphisms that agree on $V$ are considered equivalent.

Equivalence of Categories: Irreducible Algebraic Sets and Coordinate Rings

Proposition

If $V$ and $W$ are two irreducible algebraic sets, then the set of morphisms from $V$ to $W$ and the set of $k$-homomorphisms from $\Gamma (W)$ to $\Gamma (V)$ are canonically isomorphic, that is, $\mathrm{Hom}(V,W)\simeq {\mathrm{Hom}}_k(\Gamma (W),\Gamma (V))$.

\begin{proof} For any morphism $\alpha :V\to W$ and any $k$-value function $g:W\to k$, their composition $g\circ \alpha$ is a $k$-valued function on $V$, as $g\circ \alpha (x)=g(f_1(x),\cdots ,f_m(x))\in k[x_1,\cdots ,x_n]$. Now we have a “pull back”

$$\begin{array}{rl}{\alpha }^{\ast }:\{W\to k\}& \to \{V\to k\},\\ g& \mapsto {\alpha }^{\ast }(g)=g\circ \alpha .\end{array}$$

Since $\{k\text{mbox}-valuefunctionsonV\}\simeq k[x_1,\cdots ,x_n]/P_1$, we have

$$\begin{array}{rl}{\alpha }^{\ast }:\Gamma (W)& \to \Gamma (V),\\ [g]& \mapsto {\alpha }^{\ast }[g]=[g\circ \alpha ],\end{array}$$

where $\Gamma (W)=k[y_1,\cdots ,y_m]/P_2$ and $\Gamma (V)=k[x_1,\cdots ,x_n]/P_1$. Notice that ${\alpha }^{\ast }$ is a $k$-homomorphism of rings, because ${\alpha }^{\ast }([g_1]+[g_2])={\alpha }^{\ast }[g_1]+{\alpha }^{\ast }[g_2]$ and ${\alpha }^{\ast }(c\cdot g)=c{\alpha }^{\ast }(g)$ for any $c\in k$.

Conversely, we claim that for some $k$-homomorphism of rings $k[y_1,\cdots ,y_m]/P_2\stackrel{\beta }{\to }k[x_1,\cdots ,x_n]/P_1$, there exists $\alpha :V(P_1)\to V(P_2)$ such that ${\alpha }^{\ast }=\beta$. If the claim holds, then $\beta (y_1)={\alpha }^{\ast }(y_1)=y_1\circ \alpha =f_1$. Hence, define $[f_1]=\beta (y_1),\cdots ,[f_m]=\beta (y_m)$, and define

$$\alpha :V(P_1)\to V(P_2),x\mapsto (f_1(x),\cdots ,f_m(x)).$$

We can verify that

• $\alpha$ is well-defined, because $[f_1]=[g_1]$ iff $f_1-g_1\in P_1$ iff $f_1(x)=g_1(x)$ for all $x\in V(P_1)$;
• for any $x\in V(P_1)$, there is $\alpha (x)\in V(P_2)$:
  • it suffices to show for any $g\in P_2$, $g(\alpha (x))=0$;
  • $g\circ \alpha =g(f_1,\cdots ,f_m)=g(\beta (y_1),\cdots ,\beta (y_m))=\beta (g)$ as $\beta$ is a $k$-homomorphism of ring;
  • since $g\in P_2$, we have $\beta (g)\in \overline{P_1}$ and so $\beta (g)=0$;
  • thus, $\alpha (x)\in V(P_2)$. \end{proof}

Remark. This proposition tells us the functor $\Gamma$ is fully faithful.

Proposition

The assignment

$$V\to \Gamma (V)$$

extends to a contravariant functor $\Gamma$ :

$$\left\{\begin{array}{l}\text{ Category of irreducible }\\ \text{ algebraic sets }+\text{ morphisms }\end{array}\right\}\to \left\{\begin{array}{l}\text{ Category of finitely generated integral }\\ \text{ domains over }k+k\text{-homomorphisms }\end{array}\right\}$$

which is an equivalence of categories.

\begin{proof} ^23e635 shows that $\Gamma$ is a fully faithful factor, so it remains to check every finite generated integral domain $R$ over $k$ occurs as $\Gamma (V)$ for some $V$. Since $R$ is finitely generated, it can be written as $R\simeq k[x_1,\cdots ,x_n]/I$ and then $\Gamma (V(I))=R$. \end{proof}

Corollary

For a morphism between affine varieties $f:X\to Y$ and its induced map $f^{\ast }:S\to R$, if $f^{\ast }$ is injective, then $f(X)$ is dense in $Y$.

\begin{proof} \end{proof}

Topology on $k^n$

If $f\in k[x_1,\cdots ,x_n]$ is continuous under this topology, then $f^{-1}(0)=V(f)$ is closed. Define Zariski topology of $k^n$ consists of closed sets $\{V(I):\text{mbox}idealI\subseteq k[k_1,\cdots ,k_n]\}$. For algebraic set $V(I)\subseteq k^n$, Zariski topology on $V(I)$ is the induced topology from $k^n$, that is, its closed sets are defined as

$$\{V(I)\cap V(J)\}=\{V(I\cup J)\}=\{V(J):J\supseteq I\}=\{V(J):\overline{J}\subseteq \Gamma (V(I))\}.$$

It coincides with

$$\Gamma (k^n)\to \Gamma (V(I)),J\mapsto \overline{J}.$$

For any $I=(g_1,\cdots ,g_m)$, consider $f^{-1}(V(I))=\{x:f(x)\in V(I)\}=\{x:g_i(f_1,\cdots ,f_n)=0\}$. Hence $f^{-1}(V(I))=V(J)$ with $J=(g_1(f_1,\cdots ,f_n),\cdots ,g_m(f_1,\cdots ,f_n))$.

A basis of Zariski topology is

$$V_f=\{x\in k^n:f(x)\ne 0\}=k^n-V((f))$$

because any open set $U=k^n-V(I)=\{x:\exists i,f_i(x\ne 0)\}={\cup }_{i=1}^m(k^n-V(f_i))={\cup }_{i=1}^m{V}_{f_i}$.

Definition

A topological space is Noetherian if its closed sets satisfy the DCC, that is, there does not exist infinite sequence of closed sets $V_1⊋V_2⊋\cdots$.

Notice that $V(I_1)⊋V(I_2)⊋\cdots$ corresponds to $I_1⊊I_2⊊\cdots$, then Zariski topology is Noetherian as $k[x_1,\cdots ,x_n]$ is Noetherian and ideals is ACC.

Definition

In ${\mathbb{P}}^n$,

• Zariski closed sets = $\{V(I)\subseteq {\mathbb{P}}^n:I\text{mbox}isahomogeneousideal\}$;
• For projective algebraic set $V(J)$, we use the induced topology;
• $\{V_f\}$ is a topological basis;
• For a homogeneous coordinate $x_i$ on ${\mathbb{P}}^n$, $V_{x_i}\simeq {\mathbb{C}}^n=U_i,[x_0,\cdots ,x_i,\cdots ,x_n]\mapsto (x_1/x_i,\cdots ,\widehat{x_i/x_i},\cdots ,x_n/x_i)$ is a homeomorphism.

Recall in affine algebraic set, we have

$$\{f:V(I)\to k\}=\{p\in k[x_1,\cdots ,x_n]:p_{V(I)}=f\}.$$

We cannot use it to define functions on projective algebraic set. For example, $f([x,y])=x^2/y^2$ is homogeneous of degree $0$, and it has no definition on $[1,0]$.