\documentclass{amsart}
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\newcommand{\superset}{\supset}
\newcommand{\bbZ}{{\mathbb Z}}
\newcommand{\bbA}{{\mathbb A}}
\newcommand{\bbP}{{\mathbb P}}
\DeclareMathOperator{\Spec}{Spec}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Ann}{Ann}
\DeclareMathOperator{\Fin}{{\bf Fin}}
\DeclareMathOperator{\Irr}{{\bf Irr}}
\DeclareMathOperator{\Proj}{{\bf Proj}}
\usepackage{hyperref}
\begin{document}
\title[Part IV]{Projective Schemes}
\author{Kapil Hari Paranjape}
\maketitle
\section{Going Projective}
There are two ways in which the study of graded rings arises
naturally out of the study of affine schemes. Let $R$ be a quotient of
the polynomial algebra as considered earlier. Let $R_n$ be a filtration
(increasing or decreasing) on the ring $R$ such that $R_n\cdot R_m\subset
R_{n+m}$. We obtain a graded ring $\tilde{R}=\oplus R_n$.
One way to obtain an increasing filtration on a quotient of a polynomial
ring is to take $R_n$ to the subgroup (or $k$ vector subspace) generated
by polynomials of degree at most $n$. This process is called
homogenisation and the associated ring is denoted by $R^h$.
A way to obtain an increasing filtration is to take $R_n=I^n$ where $I$
is some ideal in $R$. The resulting graded ring $\tilde{R}_I=\oplus I^n$
is called (for reasons that will become clear later) the ``blow-up'' of
$\Spec(R)$ along $\Spec(R/I)$ or of $R$ along $I$ for short.
In each case the study of the graded ring allows us to formalise (and
unify) arguments by induction. However, this is not just an algebraic
trick---we will see below that the constructions have a nice geometric
interpretation as well.
We note that in both cases the rings obtained are quotients of the
appropriate polynomial rings. The ideal generated by all the graded
elements of positive degree is called the irrelevant ideal $\Irr(R)$ of
such a ring. We later encounter another ideal that has ``equal right''
to being called the irrelevant ideal. Perhaps this should be called the
``big'' irrelevant ideal.
\subsection{Projective Schemes}
If $R$ is a graded finitely generated ring, then consider a morphism
$f:R\to A$ such that image of the irrelevant ideal
generates the unit ideal in $A$. If $\lambda\in A^{\times}$ is a unit in
$A$ then we can modify $f$ by defining, for each homogeneous element $x$
of $R$, $(\lambda\cdot f)(x)=\lambda^{\deg(x)}f(x)$. Then $(\lambda\cdot
f):R\to A$ is again a homomorphism. If we write $R$ as a quotient of the
polynomial ring $k[X_1,\dots,X_n,Y_0,\dots,Y_m]$, with $\deg(X_i)=0$ and
$\deg(Y_i)>0$, by a homogeneous ideal $I$, then such a homomorphism
amounts to choosing elements $a_1$,\dots $a_n$ and $b_0$, \dots $b_m$ in
$A$ such that they satisfy the polynomials in $I$ and $(b_0,\dots,b_m)$
is the unit ideal in $A$. The action of units is given by replacing
$b_i$ by $\lambda^{\deg(Y_i)}b_i$; since the equations are generated by
homogeneous equations, this again gives a solution.
An important example is the case where there are no $X$'s and no
equations. The scheme thus obtained is called the projective space of
dimension $m$ and denoted $\bbP^m$. More generally, we can see that all
our schemes are subschemes (in fact closed subschemes) of
$\bbA^n\times\bbP^m$.
\subsection{Segre embedding}
Consider the product $\bbP^p\times\bbP^q$. As a functor of points it
assigns to each finite ring $A$ the set of pairs of tuples of the form
$((a_0,\dots,a_p),(b_0,\dots,b_q))$ where the $a_i$'s and $b_j$'s lie in
$A$ and the $a_i$'s generate the unit ideal as do the $b_j$'s. It
follows that if we define $c_{ij}=a_ib_j$, then we obtain a
$(p+1)(q+1)$-tuple of elements of $A$ which also generate the unit
ideal. Moreover, this tuple satisfies the collection of equations of the
form $Z_{ij}Z_{kl}=Z_{il}Z_{kj}$ as $i$,$k$ run from 0 to $p$ and
$j$,$l$ run from 0 to q.
Conversely, let $c_{ij}$ be a $(p+1)(q+1)$-tuple that generates the unit
ideal in a finite ring $A$ and satisfies the above equations. We can
check that each such tuple is a ``product'' of a pair of tuples as
above.
Generalising this we see that if $\Proj(R)$ and $\Proj(S)$ are two
projective schemes then we can give a natural scheme structure to
$\Proj(R)\times\Proj(S)$.
\subsection{Morphisms of graded rings}
If $R\to S$ is a graded homomorphism of graded rings, then it need not
give a natural trasformation $\Proj(S)\to\Proj(R)$. If $S\to A$ is a
homomorphism such that the image of the irrelevant ideal $\Irr(S)$
generates the unit ideal in $A$, then it need not follow that the same
is true for the image of $\Irr(R)$ under the composite homomorphism.
Suppose that $S$ is finite as an $R$ module. The each of the generators
of $\Irr(S)$ satisfies a monic polynomial with coefficients in $R$.
Taking homogeneous components everywhere we can assume that the
polynomial has the form
\[ T^n + r_1 T^{n-1} + \dots + r_n \]
where the degree of $r_k$ is $d(n-k)$ where $d$ is the degree of the
generator under consideration. In particular, it follows that the degree
of $r_k$ is positive. Thus, some power of each generator of $\Irr(S)$
lies in the ideal $\Irr(R)S$; hence some power of the ideal $\Irr(S)$
lies in the ideal $\Irr(R)S$. As a consequence, if $S\to A$ is a
homomorphism such that the image of $\Irr(S)$ generates the unit ideal
then so does the image of $\Irr(R)$ under the composite homomorphism.
We have thus shown that a graded homomorphism $R\to S$ of graded rings
gives a morphism $\Proj(S)\to\Proj(R)$ if the homomorphism is finite.
Conversely, suppose that $R\to S$ is a graded homomorphism of graded
rings such that $R_0\to S_0$ is finite and for which there is a positive
integer $N$ such that $\Irr(S)^N$ is contained in $\Irr(R)S$. There is a
a finite collection $w_{d1}$, \dots, $w_{dk_d}$ of homogeneous elements
of degree $d$ in $S$ which generate all homogeneous elements of degree
$d$ in $S$ over $S_0$. Expanding this collection as necessary and using
the finite-ness of $S_0$ over $R_0$, we can assume that the same
generators work over $R_0$ as well. Let $e_1$, \dots, $e_r$ denote the
union of such collections for all degrees less than or equal to $N$. Any
homogeneous element $a$ of $S$ of degree greater than $N$ can be written as
a linear combination of elements of $\Irr(R)$ with coefficients from
$S$; such coefficents must have degree smaller than the degree of $a$
since elements of $\Irr(R)$ have positive degree. By induction, we can
thus write any element of $S$ as a linear combination over $R$ of the
$e_1$,\dots, $e_r$. Hence, $S$ is finite as an $R$ module.
Thus, in order that a graded homomorphism $R\to S$ be finite it is
necessary and sufficient that $R_0\to S_0$ is finite and that the image
of the irrelevant ideal of $R$ generates an ideal in $S$ that contains a
power of the irrelevant ideal of $S$.
\subsection{Homogeneous primary decomposition}
Let $R$ be a finitely generated graded ring and $Q$ a primary ideal in
$R$ that is not necessarily homogeneous. Let $Q'$ be the sub-ideal of
$Q$ generated by its homogeneous elements. We assert that $Q'$ is
primary as well.
To prove this we can go modulo $Q'$; in this case we are assuming that
$Q$ is a primary ideal such that it has no non-zero homogeneous
elements. We must then show that if $fg=0$ and $g$ is not 0,
then $f$ is nilpotent. Now if $f$ and $g$ are homogeneous and $g$ is not
zero then $g$ does not lie in $Q$ either; by the primariness of $Q$, if
follows that $f^n$ must lie in it for some $n$ and hence $f^n=0$.
In the general case, we write $f=f_0+\dots+f_{s-1}+f_s+\dots$ and
$g=g_r+\dots$ where $f_i$ and $g_i$ are homogeneous of degree $i$.
Further suppose that $s$ and $r$ are chosen so that $f_i$ are nilpotent
for $i