HW4

When $X,Y$ are affine varieties, we have defined morphisms $X\to Y$:

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 $x\in V$.

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We also define morphism between prevarieties:

g \in \Gamma\left(V, \underline{o}_Y\right) \Longrightarrow g \cdot f \in \Gamma\left(f^{-1} V, \underline{o}_X\right) .

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For $f:X\to Y$ with $X,Y$ affine varieties, show that the two definitions are compatible.

\begin{proof} i)→ii). Assume that $X\subseteq k^n$ and $W\subseteq k^m$. If $f:X\to Y$ satisfies $f(x)=(f_1(x),\cdots ,f_m(x))$ for any $x\in X$ with $f_i\in k[x_1,\cdots ,x_n]$, then $f$ is continuous. It suffices to show for any open $V\subseteq Y$, there exists $f^{\ast }:\Gamma (V,Y)\to \Gamma (f^{-1},X),s\mapsto s\circ f$.

For any $s\in \Gamma (V,{\underline{o}}_Y)$, $s$ can be written as $s=s_1/s_2$, where $s_1,s_2\in \Gamma (Y)$, $s_2\notin I(Y)$ and $s_2(V)\ne 0$. (Here we take $V$ as distinguished open sets.) Then $s\circ f\in \Gamma (f^{-1}(V),{\underline{o}}_X)$ as $s_1\circ f,s_2\circ f\in \Gamma (X)$, $s_2\notin I(X)$ and $s_2\circ f(f^{-1}(V))\ne 0$. Therefore, $f$ is a morphism in the second definition.

ii)→i). Assume that $f:X\to Y$ is continuous and for all open sets $V$ in $Y$,

$$g\in \Gamma \left(V,{\underline{o}}_Y\right)⟹g\cdot f\in \Gamma \left(f^{-1}V,{\underline{o}}_X\right).$$

Take $V=Y$, which is an open set of $Y$. For $g_i=y_i$, we have $g_i\circ f:=f_i\in \Gamma (X,{\underline{o}}_X)=\Gamma (X)$ and $f(x)=(f_1(x),\cdots ,f_n(x))$. Therefore, $f$ is a morphism in the first definition. \end{proof}

Let $X$ be a prevariety, and let $U,V$ be open sets with $V\subseteq U$. Prove that ${\underline{o}}_X(U)\to {\underline{o}}_X(V)$ is injective.

\begin{proof} Note that ${\underline{o}}_X(U)={\cap }_{x\in U}{\underline{o}}_x$ and ${\underline{o}}_X(V)={\cap }_{x\in V}{\underline{o}}_x$, then ${\underline{o}}_X(U)\subseteq {\underline{o}}_X(V)$ and inclusion map $\iota :{\underline{o}}_X(U)\to {\underline{o}}_X(V)$ is injective. \end{proof}

Let $X$ be a prevariety, and let $Y\subseteq X$ be an irreducible closed subset. For an open set $\Omega \subseteq X$, $Y\cap \Omega$ is still irreducible.

\begin{proof} Since $Y$ is an irreducible closed subset, $(Y,{\underline{o}}_Y)$ is also a prevariety by Proposition 1.5.5 and then $Y$ is irreducible. Since $\Omega \cap Y\subseteq Y$ is an open set of $Y$, $(\Omega \cap Y,{\underline{o}}_Y{|}_{\Omega \cap Y})$ is also a prevariety by Proposition 1.5.4 and then $\Omega \cap Y$ is irreducible. \end{proof}

Problem on page 32. Generalize the result by which we covered ${\mathbb{P}}_n$ by open affine sets as follows: for all homogeneous polynomials $H\in k\left[X_0,\dots ,X_n\right]$ of positive degree, show that ${\mathbb{P}}_n(k{)}_H$ is an affine variety.

\begin{proof} Assume that $\mathrm{deg}H=d$. Let $\mathcal{S}=\{f_1,\cdots ,f_m\}$ be the set of all monomials of degree $d$ in $k[X_0,\cdots ,X_n]$, where $|\mathcal{S}|=m$. Define

$$\phi :{\mathbb{P}}_n(k)\to {\mathbb{P}}_m(k),x\mapsto (f_1(x),\cdots ,f_m(x)),$$

then $\phi$ is a morphism by definition. For each nonzero $x\in {\mathbb{P}}_n(k)$, WLOG assume $x_0\ne 0$ and $f_{i_1}(x)/f_{i_2}(x)=x_i/x_0$, then ${\phi }^{-1}(f_1(x),\cdots ,f_m(x))=(1,f_{1_1}(x)/f_{1_2}(x),\cdots ,f_{n_1}(x)/f_{n_2}(x))$ and so $\phi$ is an isomorphism.

Since ${\mathbb{P}}_n(k{)}_H=\{x\in {\mathbb{P}}_n(k):H(x)\ne 0\}$, $\phi ({\mathbb{P}}_n(k{)}_H)=\{y\in {\mathbb{P}}_m(k):{\sum }_{i=1}^m{a}_i{y}_i\ne 0\}$ with $y_i=f_i$. It deduces that $\phi ({\mathbb{P}}_n(k{)}_H)$ is an affine variety and so for ${\mathbb{P}}_n(k{)}_H$. \end{proof}

Check that $(Z_1\times Z_2,{\underline{o}}_{X_1\times X_2}{|}_{Z_1\times Z_2})$ is isomorphic to $(Z_1\times Z_2,{\underline{o}}_{Z_1\times Z_2})$, where $Z_i\subseteq X_i$ are irreducible closed set.

\begin{proof} Assume that $\Gamma (Z_1)=\Gamma (X_1)/p_1$ and $\Gamma (Z_2)=\Gamma (X_2)/p_2$, then it suffices to show

$$\Gamma (Z_1)\otimes \Gamma (Z_2)\simeq \Gamma (X_1)\otimes \Gamma (X_2)/(p_1\otimes B\oplus A\otimes p_2).$$

It is easy to check and we finish the proof. \end{proof}