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\centerline{\Kopfont Commutative Algebra}

\smallskip
\centerline{Excercises 3}

\bigskip\noindent
{\bf 1.} Find a flat but not free module over ${\bbb Z}_{(2)}$.

\medskip\noindent
{\bf 2.} Let $k$ be any field. Suppose that $A=k[[x,y]]/(f)$ and
$B=k[[u,v]]/(g)$, where $f=xy$ and $g=uv+\delta$ with $\delta \in (u,v)^3$.
Show that $A$ and $B$ are isomorphic rings.

\medskip\noindent
{\bf Remark.} A singularity on a curve over a field $k$ is called an
ordinary double point if the complete local ring of the curve at the
point is of the form $k'[[x,y]]/(f)$, where (a) $k'$ is a finite separable
extension of $k$, (b) the initial term of $f$ has degree two, i.e., it
looks like $q=ax^2+bxy+cy^2$ for some $a,b,c\in k'$ not all zero, and (c) 
$q$ is a nondegenerate quadratic form over $k'$ (in char 2 this means that
$b$ is not zero). In general there is one isomorphism class of such rings for
each isomophism class of pairs $(k',q)$.

\medskip\noindent
{\bf 3.} Suppose that $A$ is a ring and $M$ is an $A$-module.
Let $f_i$ be a collection of elements of $A$ such that
$$
\Spec(A) = \bigcup D(f_i).
$$
\item{(a)} Show that if $M_{f_i}$ is a finitely generated $A_{f_i}$-module,
then $M$ is a finitely generated $A$-module.
\item{(b)} Show that if $M_{f_i}$ is a flat $A_{f_i}$-module,
then $M$ is a flat $A$-module. (This is kind of silly if you think about
it right.)

\medskip\noindent
{\bf Remark.} In algebraic geometric language this means that the property 
of ``being finitely generated'' or ``being flat'' is local for the Zariski
topology (in a suitable sense). You can also show this for the property
``being of finite presentation''.

\medskip\noindent
{\bf 4.} Suppose that $(A,\gm,k)$ is a Noetherian local ring. For any
finite $A$-module $M$ define $r(M)$ to be the minimum number
of generators of $M$ as an $A$-module. This number equals
$\dim_k M/\gm M = \dim_k M\otimes_A k$ by NAK.
\item{(a)} Show that $r(M\otimes_AN) = r(M)r(N)$.
\item{(b)} Let $I\subset A $ be an ideal with $r(I)>1$. Show that
$r(I^2) < r(I)^2$.
\item{(c)} Conclude that if every ideal in $A$ is a flat module, then
$A$ is a PID (or a field).


\medskip\noindent
{\bf 5.} Flat deformations. 
\item{(a)} Suppose that $k$ is a field and $k[\epsilon]$ is the ring of 
dual numbers $k[\epsilon]=k[x]/(x^2)$ and $\epsilon = \bar x$. Show that for
any $k$-algebra $A$ there is a flat $k[\epsilon]$-algebra $B$ such that
$A$ is isomorphic to $B/\epsilon B$.
\item{(b)} Suppose that $k={\bbb F}_p = {\bbb Z}/p{\bbb Z}$ and 
$A = k[x_1,x_2,x_3,x_4,x_5,x_6]/(x_1^p,x_2^p,x_3^p,x_4^p,x_5^p,x_6^p)$.
Show that there exists a flat ${\bbb Z}/p^2{\bbb Z}$-algebra $B$ such
that $B/pB$ is isomorphic to $A$. (So here $p$ plays the role of $\epsilon$.)
\item{(c)} Now let $p=2$ and consider the same question for
$k={\bbb F}_2 = {\bbb Z}/2{\bbb Z}$ and 
$A = k[x_1,x_2,x_3,x_4,x_5,x_6]/(x_1^2,x_2^2,x_3^2,x_4^2,x_5^2,x_6^2,
x_1x_2+x_3x_4+x_5x_6)$. 
However, in this case show that there does {\it not} exist a flat
${\bbb Z}/4{\bbb Z}$-algebra $B$ such that $B/2B$ is isomorphic to
$A$. (Find the trick! The same example works in arbitrary characteristic
$p>0$, except that the computation is more difficult.)

\end
