HW1

Problem(Mumford, page 9). Let $\Sigma \subset {\mathbb{P}}_n(k)$ be a closed algebraic set, and let $H$ be the hyperplane $X_0=0$. Identify ${\mathbb{P}}_n-H$ with $k^n$ in the usual way. Prove that $\Sigma \cap \left({\mathbb{P}}_n-H\right)$ is a closed algebraic subset of $k^n$ and show that the ideal of $\Sigma \cap \left({\mathbb{P}}_n-H\right)$ is derived from the ideal of $\Sigma$ in a very natural way. For details on the relationship between the affine and projective set-up and on everything discussed so far, read Zariski-Samuel, vol. 2, Ch. 7, $\S \S 3,4,5$ and $6$.

\begin{proof} Assume that $I$ is a homogeneous ideal such that $V(I)=\Sigma$. For any homogeneous polynomial $f\in I$ of degree $m$, $f$ can be written as

$$f=x_0^m\cdot (f/x_0^m)=x_0^m{g}_f(y_1,\cdots ,y_n),$$

where $y_i=x_i/x_0$. Hence there is a map $\phi :{}^{h}k[x_0,\cdots ,x_n]\to k[x_1,\cdots ,x_n],f\mapsto g_f$, where ${}^{h}k[x_0,\cdots ,x_n]$ is the set of homogeneous polynomials. Define $J=\phi (I)\subseteq k(x_1,x_2,\cdots ,x_n)$, and we claim that $\Sigma \cap ({\mathbb{P}}^n-H)$ is derived from $J$.

For any $x=[1:x_1:\cdots :x_n]\in \Sigma \cap ({\mathbb{P}}^n-H)$, there is $f(x)=0$ and $g_f(x_1,\cdots ,x_n)=0$ for all $f\in I$. That is, for all $g\in J$, $g(x_1,\cdots ,x_n)=0$ and so $\Sigma \cap ({\mathbb{P}}^n-H)\subseteq V(J)\subseteq k^n$. Conversely, for any $(x_1,\cdots ,x_n)\in V(J)$, there is $g_f(x_1,\cdots ,x_n)=f(1,x_1,\cdots ,x_n)=0$ for any $g_f\in J$. It deduces that $[1:x_1:\cdots :x_n]\in V(I)=\Sigma$. Also it is clear that $[1:x_1:\cdots :x_n]\in {\mathbb{P}}^n-H$. Therefore, if we identify $\Sigma \cap ({\mathbb{P}}^n-H)$ with the subset $S\subseteq k^n$, then $S=V(J)$. That is, $\Sigma \cap ({\mathbb{P}}^n-H)$ is a closed algebraic subset of $k^n$. \end{proof}

1.2.(Ha, page 7) The Twisted Cubic Curve. Let $Y\subseteq {\mathbf{A}}^3$ be the set $Y=\left\{\left(t,t^2,t^3\right)\mid t\in k\right\}$. Show that $Y$ is an affine variety of dimension $1$. Find generators for the ideal $I(Y)$. Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$. We say that $Y$ is given by the parametric representation $x=t,y=t^2,z=t^3$.

\begin{proof} Note that the generator for the ideal $I(Y)$ is $\{x^2-y,x^3-z\}$. Since

$$k[x,y,z]/I(Y)=k[x,y,z]/(x^2-y,x^3-z)\simeq k[x]$$

is a integral domain, $Y$ is a irreducible closed subsets of ${\mathbf{A}}^3$ and so it is an affine variety. If there exists another irreducible closed subsets $Z\subseteq Y$, then $Z=V(I(Z))$ and $I(Z)\supseteq I(Y)$. For any $f\in I(Z)\setminus I(Y)$, $f$ can be identified as a polynomial $f(x)\in k[x]$ and so $Z$ is a set of finitely many points.

For any single point $P\in {\mathbf{A}}^3$, notice that $\{P\}$ is a irreducible closed set as $k[x,y,z]/I(\{P\})\simeq k$ is an integral domain. Hence, if $Z$ is a set of finitely many points and $Z$ is irreducible, then $Z=\{P\}$ for some $P\in {\mathbf{A}}^3$. Therefore, the supremum of $n$ such that there exists a chain $Z_0\subseteq Z_1\subseteq \cdots \subseteq Z_n=Y$ of distinct closed subsets of ${\mathbf{A}}^3$ is $1$, that is, $Y$ is an affine variety of dimension $1$. Since $A(Y)=k[x,y,z]/I(Y)$, it is isomorphic to a polynomial ring in one variable over $k$. \end{proof}

1.3.(Ha, page 7) Let $Y$ be the algebraic set in $A^3$ defined by the two polynomials $x^2-yz$ and $xz-x$. Show that $Y$ is a union of three irreducible components. Describe them and find their prime ideals.

Solution. Notice that

$$Y=\{(x,y,z):x=0,y=0\}\cup \{(x,y,z):x=0,z=0\}\cup \{(x,y,z):x^2=y,z=1\}:=Y_1\cup Y_2\cup Y_3$$

where $I(Y_1)=(x,y)$, $I(Y_2)=(x,z)$ and $I(Y_3)=(x^2-y,z-1)$. Since $k[x,y,z]/I(Y_1)\simeq k[z]$ and $k[x,y,z]/I(Y_2)\simeq k[y]$ are integral domains, $Y_1$ and $Y_2$ are irreducible closed subsets. Furthermore, as $k[x,y,z]/I(Y_3)\simeq k[x]$, $Y_3$ is also irreducible. Therefore,

$$Y=V(I(Y_1))\cup V(I(Y_2))\cup V(I(Y_3))$$

where $I(Y_1)=(x,y)$, $I(Y_2)=(x,z)$ and $I(Y_3)=(x^2-y,z-1)$. \end{proof}

1.9.(Ha, page 7) Let $a\subseteq A=k\left[x_1,\dots ,x_n\right]$ be an ideal which can be generated by $r$ elements. Then every irreducible component of $Z(a)$ has dimension $⩾n-r$.

\begin{proof} Decompose $Z(\mathfrak{a})$ into finitely many irreducible components $Z(\mathfrak{a})=Y_1\cup \dots \cup Y_n$. Each $Y_i$ corresponds to a prime ideal ${\mathfrak{p}}_i$ which is minimal over $\mathfrak{a}$. By Krull’s Principal Ideal Theorem, the height of ${\mathfrak{p}}_i⩽r$. We also know

$$\text{ ht. }{\mathfrak{p}}_i+\dim A_n/{\mathfrak{p}}_i=\dim A_n$$

thus $\dim Y_i=\dim A_n/{\mathfrak{p}}_i⩾n-r$. \end{proof}

2.16.(Ha, page 14)

• (a) The intersection of two varieties need not be a variety. For example, let $Q_1$ and $Q_2$ be the quadric surfaces in ${\mathbf{P}}^3$ given by the equations $x^2-yw=0$ and $xy-zw=0$, respectively. Show that $Q_1\cap Q_2$ is the union of a twisted cubic curve and a line.
• (b) Even if the intersection of two varieties is a variety, the ideal of the intersection may not be the sum of the ideals. For example, let $C$ be the conic in ${\mathbf{P}}^2$ given by the equation $x^2-yz=0$. Let $L$ be the line given by $y=0$. Show that $C\cap L$ consists of one point $P$, but that $I(C)+I(L)\ne I(P)$.

\begin{proof} i) Note that $Q_1\cap Q_2=Z(x^2-yw,xy-zw)$. Since

$$y(xy-zw)=x{y}^2-x^{2}z=x(y^2-xz)=0,$$

either $x=0$ or $y^2=xz$. If $x\ne 0$, then $x^2-yw=xy-zw=y^2-xz=0$. Otherwise, if $x=0$, then $yw=zw=0$. If $w\ne 0$, then $y=z=0$. If $w=0$, then $y,z$ are in arbitrary. Hence

$$Q_1\cap Q_2=\{[0:0:0:1]\}\cup \{[0:y:z:0]\mid y,z\in k\}\cup \{[x:y:z:w]\mid x^2-yw=xy-zw=y^2-xz=0\}$$

Remark that $\{[0:0:0:1]\}\subseteq \{[x:y:z:w]\mid x^2-yw=xy-zw=y^2-xz=0\}$ and $\{[0:y:z:0]\mid y,z\in k\}$ is a line. So it remains to show $\{[x:y:z:w]\mid x^2-yw=xy-zw=y^2-xz=0\}$ is a twisted cubic curve. Define $t_1=y/x$, $t_2=z/x$ and $t_3=w/z$, then we have $t_1{t}_3=1$, $t_1^2=t_2$ and $t_1=t_2{t}_3$. It deduces that

$$\{[x:y:z:w]\mid x^2-yw=xy-zw=y^2-xz=0\}=\{[1:t:t^2:1/t]:t\in k\}=\{[t:t^2:t^3:1]:t\in k\}$$

and so it is indeed a twisted cubic curve.

ii) Define $C=Z(x^2-yz)$ and $L=Z(y)$, then $I(C)+I(L)=(x^2-yz,y)$. Since $P:=[0:0:1]\in C\cap L$, we have $I(P)=(x,y)\ne I(C)+C(L)$. In fact, $C\cap L=\{P\}$, so there is $I(C)+I(L)\ne I(C\cap L)$. \end{proof}

2.17.(Ha, page 14) Complete intersections. A variety $Y$ of dimension $r$ in ${\mathbf{P}}^n$ is a (strict) complete intersection if $I(Y)$ can be generated by $n-r$ elements. $Y$ is a set-theoretic complete intersection if $Y$ can be written as the intersection of $n-r$ hypersurfaces.

• (a) Let $Y$ be a variety in ${\mathbf{P}}^n$, let $Y=Z(a)$; and suppose that $a$ can be generated by $q$ elements. Then show that $\dim Y⩾n-q$.
• (b) Show that a strict complete intersection is a set-theoretic complete intersection.
• *(c) The converse of (b) is false. For example let $Y$ be the twisted cubic curve in ${\mathbf{P}}^3$ (Ex. 2.9). Show that $I(Y)$ cannot be generated by two elements. On the other hand, find hypersurfaces $H_1,H_2$ of degrees 2,3 respectively, such that $Y=H_1\cap H_2$.
• **(d) It is an unsolved problem whether every closed irreducible curve in ${\mathbf{P}}^3$ is a set-theoretic intersection of two surfaces. See Hartshorne [1] and Hartshorne $[5$, III, §5] for commentary.

\begin{proof} a) By Exercise 1.9, each irreducible component of $Y$ has dimension $⩾n-q$ and so for $Y$.

b) If $Y$ is a strict complete intersection, then $\dim Y=r$ and $I(Y)$ can be generated by $n-r$ elements. By a) we have $r=\dim Y⩾n-(n-r)=r$.

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