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\begin{document}
\pagestyle{plain}
\title{Intro to Algebraic Geometry, Problem Set 2}
\date{September 21, 1999.}
\maketitle

{\parskip=12pt % closing bracket is just before the bibliography 

Due Tuesday September 28 in class.  No lates will be accepted, so Ana-Maria will be able to return it in time for
Thursday's class.  Throughout this set, we will only deal with morphisms between affine algebraic sets, so 
use the ``naive'' definition which works in this case.


\begin{enumerate}

\item Prove that any non-empty open subset of an irreducible space is irreducible and dense.

\item Suppose $X$ is an algebraic subset of $\A^n$.  Show that
$I(X)$ is a prime ideal if and only if $X$ is irreducible.

\item Suppose $Y$ is a Noetherian topological space.  Prove that it can be expressed as a finite union
$Y = Y_1 \cup \dots \cup Y_r$ of irreducible closed subsets $Y_i$ in a unique way (up to relabelling) assuming
no redundancies (i.e. $Y_i \subseteq Y_j$ implies $i=j$).

\item (a)  Let $Y$ be the twisted cubic that is the subset $(t,t^2,t^3)$ in  $\A^3(\kb)$.  Show that $A(Y)$ (the ring of
regular functions on $Y$, which you essentially calculated in the last
set) is isomorphic to $\kbar[x]$.  As this is an integral domain, $Y$
is irreducible.

\item[] (b) Let $Z$ be the plane curve $xy=1$ (in $\A^2(\kb)$).   Show that $A(Z)$ is {\em not} isomorphic to $\kbar[t]$.
Prove that $Z$ is irreducible.  

\item[] (c)  Let $Z$ be the plane curve $y^2 = x(x-1)(x-\lambda)$ (where $\lambda \in \kb$).  (Some of you
will recognize this as an elliptic curve minus the origin.)  Prove that $Z$ is irreducible.

\item Prove that the morphism $f$ from $\A^1$ (with co-ordinate $t$) to the curve $C$ given by $y^2 = x^3$ in the
$xy$-plane $\A^2$, given by $t \mapsto (t^2, t^3)$ is not an
isomorphism, by showing directly that the induced map on rings is not
an isomorphism.

\item  (Practice with morphisms and algebra.)
The plane curve $C \subset \A^2(\com)$ given by $y^2 = x^3 + x$ is actually
isomorphic to $\com$ modulo a lattice (the Gaussian integers, i.e. $\{
a + bi | a, b \in \zed \}$), minus one point.  The symmetry group of
$\com$ preserving the lattice has order 4 {\em (that's too vague,
and technically incorrect)}, 
and as a result $C$ has an
automorphism group of order 4.  Find the group.

\item Let $R$ be the ring whose elements
are finite sums of terms of the form $a x^i y^j$ where $a \in \kbar$,
and $i$ and $j$ are non-negative integers such that $i+j \neq 1$.
This is a finitely generated $\kbar$-algebra with no nilpotents
(generated by $x^2$, $xy$, and $y^2$ {\em --- that's not right}), so we know that it is the
ring of regular functions for some affine algebraic set.  Find
such a set.  (Give equations cutting out the set.  Don't bother
proving that your answer is correct.)

\item  Let $F$ be the sheaf of differentiable real-valued
functions on the unit disc $\{ x^2 + y^2 < 1\}$ (in the classical
topology).  Show that the stalk $F_{(0,0)}$ of $F$ at the origin is a local ring
(i.e. that it has one maximal ideal).  (Hint:  first find the maximal ideal $\fm$,
by describing a surjection from the stalk $F_{(0,0)}$ onto some field.  Then show
that anything not in this maximal ideal is invertible.  Finally, show that if $I$ is any
other proper ideal of $F_{(0,0)}$, then it must be contained in $\fm$.)


\item  {\em Images of morphisms.}  If $f: X \rightarrow Y$ is a morphism of algebraic sets, then
the image of $f(X)$ in $Y$ need not be an algebraic set (although it can still
be ``described in terms of polynomials'').  To see some pathologies that can
come up, find the image of $f:  \A^2 \rightarrow \A^2$ given by $(x,y) \mapsto (xy,y)$ (as a set).

\item {\em Fibers of morphisms.}  Each fiber of a morphism $f:  X \rightarrow Y$ (i.e. the set $f^{-1}(y)$ for 
some $y \in Y$) can be easily understood algebraically.  For example,
suppose let $C$ be the curve $y^2 = x^3$ in the plane, and $f: C
\rightarrow \A^2$ the morphism $(x,y) \rightarrow (x^2, y^2)$.  Then
the fiber $f^{-1}(1,1)$ can be found as follows.  (i) Informally:
Solve $y^2 = x^3$, $x^2=1$, $y^2=1$ to find $(x,y) = (1,1)$ or
$(1,-1)$.  (ii) More correctly: Consider the ideal $I = (y^2-x^3,
x^2-1, y^2-1)$, and show that its radical is $\fm_{(1,1)} \cap
\fm_{(1,-1)}$.

(a)  Consider the morphism $f:  \A^1 \rightarrow \A^1$, given by $t \mapsto t^2$.
Calculate the fiber $f^{-1}(y)$ for all $y$ in $\kb$ (using (ii) above).

(b)  Consider the morphism described in the previous question.  Calculate
the fibers $f^{-1}(0,0)$, $f^{-1}(0,1)$, and $f^{-1}(1,0)$ (using (ii) above).

In the future, feel free to use (i) above; by the Nullstellensatz, they are the same. 

\end{enumerate}


} % end of parskip; it started just before the introduction
\end{document}

