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\title{Algebraic Geometry, PS 5}
\author{Bryden Cais\thanks{in collaboration with Eric, Chris, Cornelia, and Jesse}}
\begin{document}
\maketitle

\section{Lecture 10}

\begin{enumerate}
\item
\begin{enumerate}
\item Let $Y=V(Z_1^2-Z_2^3)$ and $X=\A^1$.  Define $f:X\rightarrow Y$ via $f(x)=(x^3,x^2)$.  Then clearly,
$\O(X)=k[Z_2]$ and $\O(Y)=k[Z_1,Z_2]/(Z_1^2-Z_2^3)$ and $f^*:\O(Y)\rightarrow \O(X)$ is given by
$f^*(z_1)=Z_2^3,\ f^*(z_2)=Z_2^2$, where $z_i$ is the image of $Z_i$ under the natural homomorphism
$k[Z_1,Z_2]\rightarrow \O(Y)$.  Obviously, $f^*$ is injective.  Moreover, observe that
$$\O(X)=f^*(\O(Y))[Z_1]$$ and that $Z_1$ is a root of the monic polynomial
$$Z^2-f^*(z_1)=0.$$  It follows that $\O(X)$ is integral over $f^*(\O(Y))$ and hence that $f$
is finite.

\item Let $X=Y=\A^2,$ and suppose $f:X\rightarrow Y$ is given by $f(x,y)=(xy,y)$.  By Proposition 3
of Lecture 10, we know that a finite map is surjective.  But observe that $f$ is not surjective:
the fibre over any point $(a,0)$ with $a\in\A^1\setminus\{0\}$ is clearly empty.  It follows that
$f$ is not finite.  Indeed, we have $\O(X)=\O(Y)=k[Z_1,Z_2]$ and $f^*:\O(Y)\rightarrow \O(X)$
is given by $f^*(Z_1)=Z_1Z_2,\ f^*(Z_2)=Z_2$.  Observe that
$$\O(X)=f^*(\O(Y))[Z_1]$$ and that $Z_1$ is a root of the degree one equation
$$f^*(Z_2)Z-f^*(Z_1)=0,$$
which is not monic.
\end{enumerate}

\item Let $f:X\rightarrow Y$ be a finite map.  We show that the image of any closed subset $Z\subset X$ under $f$ is closed.
First, it is enough to show this for irreducible closed subsets of $X$ since every closed subset is a finite union of irreducible
closed subsets.  So assume that $Z$ is irreducible.  If we show that $f:Z\rightarrow \overline{f(Z)}$ is a finite map then we are done,
since by Proposition 3 any finite map is surjective (i.e. we will have shown that $f(Z)=\overline{f(Z)}$ so that $f(Z)$ is closed).
Let $I$ be the prime ideal such that $$\O(Z)=\O(X)/I$$ and let $I^c=(f^*)^{-1}(I)$.  We claim that $\O(Y)/I^c=\O(\overline{f(Z)})$.
Observe that $p\in \O(Y)$ vanishes on $\overline{f(Z)}$ if and only if $p$ vanishes on
$\O(f(Z))$, if and only if $p\circ f=f^*(p)\in \O(X)$ vanishes on $Z$, that is, if and only if $f^*(p)\in I$.  It follows that we have the
following commutative diagram:
$$
\begin{CD}
    \O(Y)@>f^*>>\O(X)\\
    @VVV        @VVV\\
    \O(\overline{f(Z)})@>>f^*>\O(Z)
\end{CD},
$$
where the vertical maps are the obvious quotient maps.  Since $\O(X)$ is integral over $f^*(\O(Y))$, it follows that
$\O(Z)$ is integral over $f^*(\O(\overline{f(Z)}))$ and hence that $f^*:Z\rightarrow \overline{f(Z)}$ is finite.

\item Let $f:X\rightarrow Y$ and $g:X^{\prime}\rightarrow Y^{\prime}$ be two finite regular maps.  Then $f,g$ are surjective
and $\O(X),\ \O(X^{\prime})$ are finitely generated modules over $f^*(\O(Y)),\ g^*(\O(Y^{\prime})),$ respectively.
We showed on a previous problem set that $f\times g$ is regular.
Since $f,g$ are surjective, so is $f\times g:X\times X^{\prime}\rightarrow Y\times Y^{\prime}$.
Moreover, we have seen on a previous homework that
$$\O(X\times X^{\prime})=\O(X)\otimes\O(X^{\prime}),\ \O(Y\times Y^{\prime})=\O(Y)\otimes\O(Y^{\prime}).$$
Suppose that $\O(X)$ is generated by $a_1,\ldots,a_m$ over $f^*(\O(Y))$ and that
$\O(X^{\prime})$ is generated by $b_1,\ldots,b_n$ over $g^*(\O(Y^{\prime})).$  Then we see that
$\O(X)\otimes\O(X^{\prime})$ is generated by $\{a_i\otimes b_j\}$ for $1\leq i\leq m$ and $1\leq j\leq n$
over $f^*(\O(Y))\otimes g^*(\O(Y^{\prime}))=f^*\times g^*(\O(Y)\otimes \O(Y^{\prime}))$, and hence that
$\O(X)\otimes \O(X^{\prime})$ is integral over $f^*\times g^* (\O(Y)\otimes \O(Y^{\prime}))$.  It follows that
$f\times g$ is a finite regular map.

\item Let $X\subset \A^2$ be given by the equation $T_2^2T_1-T_2=0$ and let $Y=\A^1$.
It is not hard to see that $X$ consists of two irreducible components: a copy of $\A^1$
given by $(T_1,0)$ and the hyperbola $T_1T_2=1$.
Define
$f:X\rightarrow Y$ by
$$f(a,b)=a.$$
It is clear that $f$ is a surjective regular map (since, in particular, the point $a$
of $K$ has preimage containing $(a,0)\in X(K)$; regularity follows from the fact that $f$ is given by
polynomials).  Moreover, the fibres of $f$ are finite: the preimage of a point $a$ consists
of the points $(a,0),\ (a,1/a)$ if $a\neq 0$ and of the single point $(0,0)$ if $a=0$.
However, $f$ is not a finite map.  If it were, it would take closed sets to closed sets by problem 2.  However,
the algebraic subset of $X$ given by $T_1T_2=1$ is obviously a closed set in $X$, but its image in
$\A^1$ is not closed since its complement is a single point, which is clearly closed.

\item \begin{enumerate}
    \item Let $A$ be an integral domain and let $Q(A)$ denote the field fractions of $A$.  Let $\bar{A}$ be the integral
    closure of $A$ in $Q(A)$.  We claim that $\bar{A}$ is normal.  First, it is obvious that $Q(A)=Q(\bar{A})$.  Now suppose that $x\in Q(A)$
    is integral over $\bar{A}$.  Then $\bar{A}(x)$ is an integral extension of $\bar{A}$.  Since $\bar{A}$ is integral over $A$, it
    follows that $\bar{A}(x)$ is integral over $A$ and hence that $x$ is integral over $A$.  Thus $x\in \bar{A}$ by definition.
    \item Let $A=k[Z_1,Z_2]/(Z_1^2-Z_2^2(Z_2+1))$.  We claim that $Q(A)=k(z_1/z_2)$, where $z_1,z_2$ are the images of $Z_1,Z_2$ in the quotient.
    Certainly we have $Q(A)=k(z_1,z_2).$  But $(z_1/z_2)^2-1=z_2$ and $(z_1/z_2)^3-(z_1/z_2)=z_1$ so that $k(z_1,z_2)\subset k(z_1/z_2)$.
    On the other hand, we certainly have $k(z_1/z_2)\subset Q(A)=k(z_1,z_2)$.  Our claim follows.  Observe, however, that $z_1/z_2$
    is integral over $A$: it is obviously a root of $X^2-1-z_2$.  Thus, $k[z_1/z_2]\subset \bar{A}$.  But since
    $k[z_1/z_2]\simeq k[T]$ by the obvious isomorphism, (and $k[T]$ is a PID) it follows that $k[z_1/z_2]$ is integrally closed in $k(z_1/z_2)=Q(A)$
    and by our above remarks, $A\subset k[z_1/z_2]$.  It follows that $\bar{A}=k[z_1/z_2]\simeq k[T]$, as required.

    \item Let us show that $A=k[Z_1,Z_2,Z_3]/(Z_1Z_2-Z_3^2)$ is normal.  We have an isomorphism of $A$ with $k[T_1^2,T_2^2,T_1T_2]$
    given by
    \begin{align*}
        Z_1&\mapsto T_1^2\\
        Z_2&\mapsto T_2^2\\
        Z_3&\mapsto T_1T_2.
    \end{align*}
    (Surjectivity is obvious; injectivity just as trivial).  But this shows that $A$ is isomorphic to a direct summand of $k[T_1,T_2]$
    (recalling that we can write $k[T_1,T_2]=\bigoplus_{i=0}^{\infty}S_i$, where $S_i$ consists of all homogenous polynomials of degree $i$,
    we see that $k[T_1,T_2]=k[T_1^2,T_2^2,T_1T_2]\oplus \bigoplus_{r=1}^{\infty}S_{2i-1}$).  Since $k[T_1,T_2]$ is normal, it follows from
    commutative algebra that any direct summand of a normal ring is normal, and hence that $A$ is normal.

\end{enumerate}

\item Let $B=k[Z_1,Z_2]/(Z_1Z_2^2+Z_2+1),$ and let $A=k[z_2+z_1z_2]$, where $z_i$ is the image of $Z_i$
under the natural homomorphism $k[Z_1,Z_2]\rightarrow B$.  Then we claim that $B$ is finite over $A$.
Indeed, first notice that $A[z_2]$ contains $z_2$ (trivially) and $z_1$ since we have
$$(z_1z_2)^2+z_1z_2=-z_1$$ and $z_1z_2\in A[z_2]$, clearly.  It follows that $A[z_2]=B$.
On the other hand, $z_2$ is a root of the monic polynomial
$$X^2-(z_2+z_1z_2+1)X-1=0$$ whose coefficients are clearly in $A$, so that $B$ is integral (and hence finite) over $A$.
Finally, we have an isomorphism
$$A\rightarrow k[T]$$ given by
$$z_2+z_1z_2\mapsto T.$$

\end{enumerate}

\section{Lecture 11}

\begin{enumerate}
\item
\begin{enumerate}
    \item Let $X$ be the two point space $\{a,b\}$ with $a$ open and $b$ closed.  This is obviously a topological space.  Moreover,
    $a$ is dense in $X$ since it is open.  Further, it is clear that $X,b$ are closed and irreducible.  However, $\dim a=0$
    while $\dim X=1$ since we have the chain $b\subset X$.

    \item Let $X=\Z^+$ with the discrete topology and let $Y=\Z^+$ with the following topology: the closed sets are the sets
    $$C_N=\{n\in\Z^+:n\leq N\}.$$  By part c), $Y$ is infinite dimensional while $X$ is clearly one dimensional since the only closed
    irreducible subsets are single points.  Moreover, the identity map $\id:X\rightarrow Y$ is obviously surjective and continuous,
    since every finite subset of $X$ is closed.

    \item Let $Y$ be as in part b).  We claim that $C_n$ is irreducible for all $n$.  For we have the trivial property
    $$C_i\cup C_j=C_{\max(i,j)},$$ from which it follows that any union of closed sets that is equal to $C_n$ contains
    $C_n$ as one of the elements of the union.
    Certainly $Y$ is noetherian: if $Z_0\supset Z_1\supset\ldots$ is any descending chain of
    closed irreducible subsets then either every $Z_i=Y,$ in which case there is nothing to prove, or there is some $i$ such that
    $Z_i=C_m$ for some $m$.  Since $C_m$ has dimension $m$ (by considering the chain $C_m\supset C_{m-1}\supset\ldots\supset C_1$
    it follows that the chain $\{Z_i\}$ must stabilize.  However, $Y$ is infinite dimensional since it contains
    closed subsets of dimension $m$ for any positive integer $m$ (namely $C_m$).
\end{enumerate}

\item Let $X$ be a closed irreducible subset of $\P^n(K)$ or $\A^n(K)$ of codimension $1$.  Since $X$ is not all of $\P^n(K)$
(resp. $\A^n(K)$) since $\dim X=n-1$, there is some nonzero polynomial $f$ vanishing on $X$.  Moreover, since $X$ is irreducible,
there is some irreducible factor $g$ of $f$ vanishing on $X$.  Let $Y=V(g)$.  Then $\dim Y=n-1$ by Theorem 4 of Lecture 11.
Moreover, by construction, we have $X\subset Y$.  But $X,Y$ are irreducible and have the same dimension.  It follows that
the inclusion map $X\rightarrow Y$ is finite and hence surjective; that is, $X=Y$.  Thus, $X$ is a hypersurface.
\stepcounter{enumi}
\item Let $X$ be an irreducible algebraic set of dimension $n-1$.  Then we have seen that $I(X)$ is a prime ideal
of Krull dimension $n-1$, that is, there exists a chain of $n$ prime ideals
$$I(X)=\wp_0\subsetneq \wp_1\subsetneq\ldots \subsetneq\wp_n.$$  To any closed irreducible subset $Z$ of $X$ we can
associate the prime ideal $\q=I(Z)\supsetneq I(X)$.  Then as a consequence of the well-known Jordan H\"{o}lder
Theorem, there exists a chain of prime ideals of length $n+1$ containing $\q$, say
$$I(X)=\q_0\subsetneq \q_1\subsetneq\ldots\subsetneq \q_n$$
with $\q_i=\q$ for some $i$.  Then $$V(\q_n)\subsetneq V(\q_{n-1})\subsetneq\ldots\subsetneq X$$
is a chain of strictly decreasing closed irreducible subsets of $X$ with $Z$ as a member of the chain.
Define
$$\codim(Y,X)=\max\{k:\exists \text{a chain of closed irreducible subsets } Y=Z_0\subsetneq\ldots\subsetneq Z_k=X \}.$$
Suppose that $\dim Y=d-1$.  Then we have a chain of closed irreducible subsets
$$Y_0\subsetneq Y_1\subsetneq\ldots\subsetneq Y_d=Y.$$
This gives a chain of {\em maximal} length $=k+d+1$ (by the definition of codimension and dimension) given by
$$Y_0\subsetneq Y_1\subsetneq\ldots\subsetneq Y_d\subsetneq Z_1\subsetneq\ldots\subsetneq Z_k=X.$$
It follows from the above that $k+d=\dim X$, that is,
$$\dim Y+\codim(Y,X)=\dim X$$
\stepcounter{enumi}
\stepcounter{enumi}
\item Recall that a point of the veronese curve $v_r(\P^1)\subset \P^n(K)$ is given by
$$(X_0^r,X_0^{r-1}X_1,X_0^{r-2}X_1^2,\ldots,X_1^r)$$ for $(X_0,X_1)\in\P^1$.  Let
$$a_0T_0+a_1T_1+\ldots+a_nT_n$$ be a hyperplane in $\P^n$.  Then the number of points of intersection
of $v_r(\P^1)$ with this hyperplane is clearly the number of solutions to
$$a_0X_0^r+a_1X_0^{r-1}X_1+\ldots+a_nX_1^r=0.$$
Observe that if $X_1=0$ we are forced to take $X_0=0$ (since we may assume that $a_0\neq 0$, as we shall see
momentarily), and this is not a point of projective space.  We may therefore dehomogenize to obtain a polynomial $p$
of degree $r$ in one variable.  Now if the coefficients $a_i$ belong to the dense, zariski open subset of the space of coefficients
given by the nonvanishing of the discriminant of the $a_i$, then the polynomial $p$ has $r$ distinct roots.
The degree is therefore $r$, since this dense open set nontrivially intersects the dense open set as in problem 6.

\end{enumerate}






\end{document}