Riemann surface
matlab

Problem 4. [Hartshorne II.2.16 and II.2.17]
Let $X$ be any variety and $f \in k[X]$ a regular function.
(a) If $h$ is a regular function on $D(f) \subset X$ then $f^n h$ can be extended to a regular function on all of $X$ for some $n>0$. [Hint: Let $X=U_1 \cup \cdots \cup U_m$ be an open affine cover. Start by showing that some $f^n h$ can be extended to $U_i$ for each $i$.]
(b) $k[D(f)]=k[X]f$ (c) Let $R$ be a $k$-algebra and let $f_1, \ldots, f_r \in R$ be elements that generate the unit ideal, $\left(f_1, \ldots, f_r\right)=R$. If $R{f_i}$ is a finitely generated $k$-algebra for each $i$, then $R$ is a finitely generated $k$-algebra.
(d) Suppose $f_1, \ldots, f_r \in k[X]$ satisfy $\left(f_1, \ldots, f_r\right)=k[X]$ and $D\left(f_i\right)$ is affine for each $i$. Then $X$ is affine.

Solution: (a) If $f$ is a regular function on a variety $X$ we will write $X_f=D(f)$. If $h$ is regular on $X_f$ and $X=U_1 \cup \cdots \cup U_m$ where each $U_i$ is open affine, then for each $i$ we have $h \in k\left[\left(U_i\right)f\right]=k\left[U_i\right]_f$, so $h=g / f^n$ for some $g \in k\left[U_i\right]$, i.e. $f^n h=g \in k\left[U_i\right]$. If we take $n$ large enough to work for all $1 \leq i \leq m$, we obtain $f^n h \in k[X]$ (b) Using (a) it is easy to check that the obvious $k$-algebra homomorphism $k[X]_f \rightarrow k\left[X_f\right]$ is an isomorphism. (c) We first prove that $k[X]$ is a finitely generated $k$-algebra. By the assumptions we can write $g_1 f_1+\cdots+g_r f_r=1$ with $g_i \in k[X]$. Since $X{f_i}$ is affine, we know that $k[X]{f_i}=k\left[X{f_i}\right]$ is a finitely generated $k$-algebra for each $i$. Choose $h_{i, 1}, \ldots, h_{i, N} \in$ $k[X]$ so that $k[X]{f_i}$ is generated by $h{i, 1}, \ldots, h_{i, N}, f_i^{-1}$. We claim that $k[X]$ is generated by the elements $f_i, g_i, h_{i, j}$ for $1 \leq i \leq r$ and $1 \leq j \leq N$. In fact, if

Problem 3. Let $X$ be a pre-variety such that for each pair of points $x, y \in X$ there is an open affine subvariety $U \subset X$ containing both $x$ and $y$.
(a) Show that $X$ is separated.
(b) Show that $\mathbb{P}^n$ has this property.

Solution: (a) Given two morphisms $f, g: Y \rightarrow X$ we show that $D={y \in Y \mid$ $f(y) \neq g(y)}$ is open in $Y$. If $f(y) \neq g(y)$ then take an open affine $U \subset X$ such that $f(y), g(y) \in U$. Then $V=f^{-1}(U) \cap g^{-1}(U) \subset X$ is open and $f$ and $g$ restrict to morphisms $f, g: V \rightarrow U$. Since $U$ is separated, it follows that ${v \in V \mid f(v) \neq g(v)}$ is open in $Y$. Finally, $D$ is the union of such sets.

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