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\def\Cat#1{\hbox{${\cal #1}$}}
\def\Shf#1{\hbox{${\cal #1}$}}
\def\Hom{\hbox{Hom}}
\def\Top#1{\hbox{$\hbox{${\cal T\!\scriptstyle\cal OP}$}(#1)$}}
\def\Set{\hbox{${\cal S\!\scriptstyle\cal ET}$}}
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\hbox to\hsize{\hfil \bf Homework 2: Math 953 Spring 2005\hfil}
\vskip\baselineskip
\hbox to\hsize{\hfil Due January 28, 2005\hfil}
(1) Let $V$ be a finite dimensional vector space over a field $k$.
Let \Cat{C} be the category of subspaces of $V$, where arrows
are arbitrary linear transformations.
Let \Cat{D} be the category of subspaces of $V^{**}$, where arrows
are arbitrary linear transformations.
Show that these are isomorphic categories.
(2) Let \Cat{C} be the category of
finite dimensional real vector spaces, where arrows
are arbitrary linear transformations.
Let \Cat{D} be the category whose objects
are the standard real vector spaces ${\bf R}^n$,
$n\ge0$, where arrows
are arbitrary linear transformations.
Show that these categories are not isomorphic but are
equivalent.
(3) Let $R$ be an integral domain. For any prime ideal $p\subset R$,
let $R_p$ denote the localization $S_p^{-1}R$, where $S_p=R-p$.
We always have $R_p\subset R_{(0)}$; $R_{(0)}$ is the field of
fractions of $R$. Thus it makes sense to take the intersection
$\cap_{m\in M}R_m$, where $M$ is the set of maximal ideals of $R$.
Show that $\cap_{m\in M}R_m=R$.
(4) Let $X$ be an affine variety contained in ${\bf A}_k^n$,
where $k$ is algebraically closed. Let $R=k[X]$ be the
coordinate ring $k[{\bf A}_k^n]/I(X)$ of $X$.
The sheaf $\Shf{O}_X$ of regular functions on $X$ is
defined, for any (Zariski)
open subset $U$ of $X$, by $\Shf{O}_X(U)= \cap_{p\in U}R_{I(p)}$.
(Intuitively, a regular function $f$ on $U$ is a function which locally
can be represented as a ratio of elements of the coordinate ring,
as long as the representation in a neighborhood of each point
$p\in U$ doesn't have a denominator that vanishes at $p$.
You don't care what happens at points not in $U$.)
And for any inclusion $U\subset V$ of open subsets of $X$,
the functorial map $\Shf{O}_X(V)\to\Shf{O}_X(U)$ is just the obvious inclusion
$\cap_{p\in V}R_{I(p)}\subset \cap_{p\in U}R_{I(p)}$.
Also consider the presheaf $\Shf{P}_X$ defined by
$\Shf{P}_X(U) = S_U^{-1}R$, where
$S_U=\{f\in R: Z(f)\cap U = \emptyset\}$.
The map $\Shf{P}_X(V)\to \Shf{P}_X(U)$ for an inclusion
$U\subset V$, is again the obvious inclusion
(note that $S_V\subset S_U$).
\hskip.2in(a) Show that $\Shf{P}_X(U)\subset\Shf{O}_X(U)$,
and that this defines a morphism
$\Shf{P}_X\to\Shf{O}_X$ of presheaves.
\hskip.2in(b) Show that $\Shf{P}_X(U_f)=\Shf{O}_X(U_f)$,
where, for any $f\in R$, $U_f=X-Z(f)$. (Note: open sets of the form
$U_f$ form a basis for the Zariski topology on $X$, so
(b) shows locally [i.e., on a basis of open sets]
that $\Shf{P}_X$ is the same as $\Shf{O}_X$, and that therefore $\Shf{O}_X$
is the sheafification of $\Shf{P}_X$. In general, as the next problem shows,
$\Shf{P}_X\to\Shf{O}_X$ need not be an isomorphism,
so $\Shf{P}_X$ need not be a sheaf.)
(5) Let $X=Z(xy-zw)\subset {\bf A}_k^4$ (assume $k$
algebraically closed). For any $f\in k[x,y,z,w]$, let
$U_f$ be the open subset of ${\bf A}_k^4$ as defined above,
and let $V_f=U_f\cap X$. Let $V=V_y\cap V_z$
and let $W=V_y\cup V_z$.
\hskip.2in(a) Show that $S_W=k$. (Here's one way.
First figure out what it means to be in $S_W$.
Then define an
appropriate injective homomorphism $R\to k[a,b,c]$,
and avail yourself of unique factorization.
What is going on here is that there is
a polynomial map $F:{\bf A}_k^3\to X$
which gives an isomorphism from a dense open subset
of ${\bf A}_k^3$ to a dense open subset of $X$,
inducing an injection $R\to k[a,b,c]$ of coordinate rings.
Let $C=X-W$; since $W$ is open, $C$ is closed.
For $f\in S_W$ to hold, we need $Z(f)\subset C$,
or equivalently, we need $Z(f\circ F)\subset F^{-1}(C)$,
which is to say we need $I(F^{-1}(C))\subset \sqrt{(f\circ F)}$.
But $\sqrt{(f\circ F)}$ is a principal ideal, so it is fairly
small as long as $f$ is not a nonzero constant.
With the fact that you're working in a UFD, this
gives you hope of explicitly showing
$I(F^{-1}(C))\subset \sqrt{(f\circ F)}$ does not hold
unless $f\in k$,
and hence that $S_W=k$.)
Conclude that $\Shf{P}_X(X)\to\Shf{P}_X(W)$
is the identity.
\hskip.2in(b) Show that $x/z$ is not in the image of
$\Shf{P}_X(W)\to \Shf{P}_X(V_z)$, but that
$x/z\in\Shf{P}_X(V_z)$ and $w/y\in\Shf{P}_X(V_y)$,
and that $x/z$ under $\Shf{P}_X(V_z)\to \Shf{P}_X(V)$
has the same image as $w/y$ under
$\Shf{P}_X(V_y)\to \Shf{P}_X(V)$
(i.e., $x/z=w/y$ on $V$). Conclude that
$\Shf{P}_X$ fails to have the sheaf property;
i.e., $x/z$ is a function regular
on $V_z$ and $w/y$ is regular on $V_y$, and they patch together
to give a function regular on $W$, but this function
is not contained in $\Shf{P}_X(W)$.
\bye