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\centerline{\Kopfont Commutative Algebra}

\smallskip
\centerline{Excercises 4}

\bigskip\noindent
{\bf 1.} {\it A silly argument using the complex numbers!}
Let ${\bbb C}$ be the complex number field. Let $V$ be a vector 
space over ${\bbb C}$. The spectrum of a linear operator
$T : V \to V$ is the set of complex numbers $\lambda \in {\bbb C}$
such that the operator $T - \lambda {\rm id}_V$ is not invertible.
\item{(a)} Show that ${\bbb C}(X)=f.f.({\bbb C}[X])$
has uncountable dimension over ${\bbb C}$.
\item{(b)} Show that any linear operator on $V$ has a
nonempty spectrum if the dimension of $V$ is finite or
countable.
\item{(c)} Show that if a finitely generated ${\bbb C}$-algebra
$R$ is a field, then the map ${\bbb C}\to R$ is an isomorphism.
\item{(d)} Show that any maximal ideal $\gm$ of ${\bbb C}[x_1,\ldots,x_n]$
is of the form $(x_1-\alpha_1,\ldots,x_n-\alpha_n)$ for some
$\alpha_i \in {\bbb C}$.

\medskip\noindent
{\bf Remark.} Let $k$ be a field. Then 
\item{($\dagger$)} for every integer $n\in {\bbb N}$ and every maximal
ideal $\gm \subset k[x_1,\ldots,x_n]$ the quotient $k[x_1,\ldots,x_n]/\gm$
is a finite field extension of $k$. 

\noindent
This will be shown later in the course. Of course (please check this)
it implies a similar statement for maximal ideals of finitely generated
$k$-algebras. The exercise above proves ($\dagger$) in the case $k={\bbb C}$. 

\medskip\noindent
{\bf 2.} Let $k$ be a field. Please use $(\dagger)$ in (b) below. 
\item{(a)} Let $R$ be a $k$-algebra. Suppose that $\dim_k R < \infty$
and that $R$ is a domain. Show that $R$ is a field.
\item{(b)} Suppose that $R$ is a finitely generated $k$-algebra, and
$f\in R$ not nilpotent. Show that there exists a maximal ideal
$\gm \subset R$ with $f\not\in \gm$.
\item{(c)} Show by an example that this statement fails when $R$
is not of finite type over a field. 
\item{(d)} Show that any radical ideal $I \subset {\bbb C}[x_1,\ldots,x_n]$
is the intersection of the maximal ideals containing it.

\medskip\noindent
{\bf Remark.} This is the Hilbert Nullstellensatz. Namely it says
that the closed subsets of $\Spec k[x_1,\ldots,x_n]$ (which correspond
to radical ideals by a previous exercise) are determined by the 
closed points contained in them.

\medskip\noindent
{\bf 3.} Let $A={\bbb C}[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12},
y_{21}, y_{22}]$. Let $I$ be the ideal of $A$ generated by the entries of the
matrix $XY$, with 
$$
X=\left(\matrix{x_{11} & x_{12}\cr x_{21} & x_{22}\cr}\right)
\quad\hbox{and}\quad
Y=\left(\matrix{y_{11} & y_{12}\cr y_{21} & y_{22}\cr}\right)
.
$$
Find the irreducible components of the closed subset $V(I)$ of $\Spec A$.
(I mean describe them and give equations for each of them. You do not have
to prove that the equations you write down define prime ideals.) Hints:
\item{(1)} You may use the Hilbert Nullstellensatz, and it suffices to find
irreducible locally closed subsets which cover the set of closed points of $V(I)$.
\item{(2)} There are two easy components. 
\item{(3)} An image of an irreducible set under a continuous map is 
irreducible.

\end
