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\centerline{\Kopfont Commutative Algebra}

\smallskip
\centerline{Excercises 8}

\bigskip\noindent
Let $R$ be a graded ring. A {\it homogeneous} ideal is simply an ideal
$I \subset R$ which is also a graded submodule of $R$. Equivalently, 
it is an ideal generated by homogeneous elements. Equivalently, if
$f \in I$ and 
$$
f = f_0 + f_1 + \ldots + f_n
$$
is the decomposition of $f$ into homogenous pieces in $R$ then $f_i \in I$
for each $i$. We define $\Proj(R)$ to be the set of homogenous, prime ideals
$\gp$ of $R$ such that $R_{+} \not \subset \gp$. Note that $\Proj(R)$
is a subset of $\Spec(R)$ and hence has a natural induced topology.

\medskip\noindent
Let $R = \oplus_{d \geq 0} R_d$ be a graded ring, let $f\in R_d$ and
assume that $d \geq 1$. We define $R_{(f)}$ to be the subring of
$R_f$ consisting of elements of the form $r/f^n$ with $r$ homogenous and
$\deg(r) = nd$. Furthermore, we define
$$
D_{+}(f) = \{ \gp \in \Proj(R) | f \not\in \gp \}.
$$
Finally, for a homogenous ideal $I \subset R$ we define 
$V_{+}(I) = V(I) \cap \Proj(R)$.

\medskip\noindent
{\bf 1.} Topology on $\Proj(R)$. With notations as above:
\item{(a)} Show that $D_{+}(f)$ is open in $\Proj(R)$, show that
$D_{+}(ff') = D_{+}(f) \cap D_{+}(f')$.
\item{(b)} Let $g = g_0 + \ldots + g_m$ be an element
of $R$ with $g_i \in R_i$. Express $D(g) \cap \Proj(R)$
in terms of $D_{+}(g_i)$, $i \geq 1$ and $D(g_0) \cap \Proj(R)$.
No proof necessary.
\item{(c)} Let $g\in R_0$ be a homogenous element of degree $0$.
Express $D(g) \cap \Proj(R)$ in terms of $D_{+}(f_\alpha)$ for a suitable
family $f_\alpha \in R$ of homogenous elements of positive degree.
\item{(d)} Show that the collection $\{D_{+}(f)\}$ of opens forms a
basis for the topology of $\Proj(R)$.
\item{(e)} Show that there is a canonical bijection
$D_{+}(f) \to \Spec(R_{(f)})$.
\item{(f)} Show that the map from (e) is a homeomorphism.
\item{(g)} Give an example of an $R$ such that $\Proj(R)$ is not
quasi-compact. No proof necessary.
\item{(i)} Show that any closed subset $T \subset \Proj(R)$ is of
the form $V_{+}(I)$ for some homogenous ideal $I \subset R$.

\medskip\noindent
{\bf Remark.} There is a continuous map
$ \Proj(R) \longrightarrow \Spec(R_0) $.

\medskip\noindent
{\bf 2.} If $R=A[X]$ with $\deg(X)=1$, show that the natural map
$\Proj(R) \to \Spec(A)$ is a bijection and in fact a homeomorphism.

\medskip\noindent
{\bf 3.} Blowing up: part I.
In this exercise $R = Bl_I(A) = A \oplus I \oplus I^2 \oplus \ldots$.
Consider the natural map $b : \Proj(R) \to \Spec(A)$. 
Set $U = \Spec(A) - V(I)$. Show that
$$
b : b^{-1}(U) \longrightarrow U
$$
is a homeomorphism.

\medskip\noindent
Thus we may think of $U$ as an open subset of $\Proj(R)$.
Let $Z \subset \Spec(A)$ be an irreducible closed subscheme
with generic point $\xi \in Z$. Assume that $\xi \not\in V(I)$,
in other words $Z \not\subset V(I)$, in other words
$\xi \in U$, in other words $Z\cap U \not=\emptyset$. We define
the {\it strict transform} $Z'$ of $Z$ to be the closure of the unique
point $\xi'$ lying above $\xi$. Another way to say this is that
$Z'$ is the closure in $\Proj(R)$ of the locally closed subset
$Z\cap U \subset U \subset \Proj(R)$.

\medskip\noindent
{\bf 4.} Blowing up: Part II.
Let $A = k[x,y]$ where $k$ is a field, and let $I=(x,y)$.
Let $R$ be the blow up algebra for $A$ and $I$.
\item{(a)} Show that the strict transforms of $Z_1=V(\{x\})$ and
$Z_2=V(\{y\})$ are disjoint.
\item{(b)} Show that the strict transforms of $Z_1=V(\{x\})$ and
$Z_2=V(\{x-y^2\})$ are not disjoint.
\item{(c)} Find an ideal $J \subset A$ such that $V(J) = V(I)$
and such that the strict transforms of $Z_1=V(\{x\})$ and
$Z_2=V(\{x-y^2\})$ are disjoint.

\medskip\noindent
{\bf 5.} Let $R$ be a graded ring.
\item{(a)} Show that $\Proj(R)$ is empty if $R_n=(0)$ for all $n >> 0$.
\item{(b)} Show that $\Proj(R)$ is an irreducible topological space
if $R$ is a domain and $R_{+}$ is not zero. (Recall that the empty
topological space is not irreducible.)

\medskip\noindent
{\bf 6.} Blowing up: Part III.
Consider $A$, $I$ and $U$, $Z$ as in the definition of strict transform.
Let $Z = V(\gp)$ for some prime ideal $\gp$. Let $\bar A = A/\gp$ and let
$\bar I$ be the image of $I$ in $\bar A$.
\item{(a)} Show that there exists a surjective ring map
$R:=Bl_I(A) \to \bar R:=Bl_{\bar I}(\bar A)$.
\item{(b)} Show that the ring map above induces a bijective map
from $\Proj(\bar R)$ onto the strict transform $Z'$ of $Z$. (This
is not so easy. Hint: Use 5(b) above.)
\item{(c)} Conclude that the strict transform $Z' = V_{+}(P)$ where
$P \subset R$ is the homogenous ideal defined by $P_d = I^d \cap \gp$.
\item{(d)} Suppose that $Z_1=V(\gp)$ and $Z_2=V(\gq)$ are irreducible
closed subsets defined by prime ideals such that  $Z_1 \not \subset Z_2$,
and $Z_2 \not \subset Z_1$. Show that blowing up the ideal
$I = \gp + \gq$ separates the strict transforms of $Z_1$ and $Z_2$,
i.e., $Z_1' \cap Z_2' = \emptyset$. (Hint: Consider the homogenous
ideal $P$ and $Q$ from part (c) and consider $V(P + Q)$.)

\end
