\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 $i<s$ and $g_r$ is the lowest degree term of $g$ which is non-zero.
Let $N$ be chosen so that $(f_0+\dots+f_{s-1})^N=0$ and consider the
binomial expansion of $(f-(f_0+\dots+f_{s-1}))^Ng$. The only term that
does not contain $fg$ as a factor is $(f_0+\dots+f_{-s})^Ng$. Thus
$(f_s+\dots)^Ng=0$; in particular, the lowest degree term $f_s^Ng_r=0$.
Now applying the homogeneous case proved above we see that $f_s$ is
nilpotent as well. By induction we thus see that $f$ is nilpotent as
required.

Now any homogeneous ideal $I$ in $R$ has a primary decomposition $I=\cap
Q_i$. Since $I$ is homogeneous we see that $I=\cap Q_i^h$ as well. So
that we have a homogeneous primary decomposition of $I$. We shall only
deal with such primary decompositions below. We note in passing that the
radical of a homogeneous ideal is also homogeneous and so the associated
and minimal primes for a homogeneous ideal are also homogeneous.

\subsection{The ``small'' irrelevant ideal}
Let $R$ be a finitely generated graded ring. Let $N_k=\Ann(\Irr(R)^k))$.
This is an increasing collection of homogeneous ideals for $k$
sufficiently large; by the Noetherian-ness condition, this sequence must
terminate. Let $N_R=N_k$ for $k$ sufficiently large.  Under any morphism
$R\to A$ where the image of the irrelevant ideal generates the unit
ideal, the annihilator of any power of this ideal must go to zero; $N_R$
must go to 0 under such a homomorphism. Thus it is truly
``irrelevant''! To summarise the ``big'' irrelevant ideal of $R$ is
irrelevant since under all homomorphisms under consideration its image
generates the full ring, while the ``small'' irrelevant ideal of $R$ is
irrelevant since its image under all homomorphisms under considertation
is 0.

Let $(0)=\cap_i Q_i$ be a primary decomposition of the zero ideal in
$R$. From this collection we drop all those primary ideals that contain
a power of $\Irr(R)$ (note that it is enough that for each generator of
$\Irr(R)$ there is some power which lies in an ideal for that ideal to
contain some power of $\Irr(R)$). Let $N'$ be the intersection of the
remaining ideals if that set is non-empty and $N'=R$ otherwise. It is
clear that $N'\subset N_R$ since $N'\Irr(R)^l\subset \cap_i Q_i = (0)$
for a suitable $l$.  Conversely, let $Q_i$ be a primary ideal in the
decomposition above that does not contain any power $\Irr(R)^l$, then
there is an i such that $Y_i^l$ does not lie in $Q_i$ for any $l$. But
$Y_i^l N_R=0$ for $l$ sufficiently large, so $N_R\subset Q_i$. This
proves that $N_R\subset N'$ as well.

We note that $N_R=R$ if and only if $\Irr(R)$ is nilpotent which happens
if and only if $R$ is a finite $R_0$ module. Moreover, these conditions
imply that $\Proj(R)(A)$ is empty for all $A$. One we prove Hilbert's
Nullstellensatz we will show that the converse is true as well.

\subsection{Homogenisation}
Let $R$ be a finitely generated ring which we think of as a quotient of
the polynomial ring with its induced filtration by degree. The
associated graded ring $R^h$ can be thought of as the subring of
$R[T]$ whose homogeneous elements of degree $n$ have the form $aT^n$,
where the degree of $a$ is at most $n$. The natural homomorphism
$R[T]\to R$ which sends $T$ to 1 induces a homomorphism $R^h\to R$ as
well; this is the dehomogenisation homomorphism.

If $I$ is an ideal in $R$, then its homogenisation is the ideal $J=I^h$
in $\tilde{R}$ that consists of elements $aT^n$ with $a$ in $I$. If
$bT^n$ is an element of $\tilde{R}$ such that $bT^nT^m$ lies in $J$ then
clearly $bT^n$ itself lies in $J$; in other words $(J:T^m)=J$ for all
$m$. Conversely, given a homogeneous ideal $J$ in $R^h$ we can take its
dehomogenisation $I$ and consider $I^h$. If $bT^n$ is an element of $J$,
then $b$ is an element of $R$ and so $bT^n$ lies in $I^h$; thus $J$ is
contained in $I^h$. In fact $(J:T^m)\subset I^h$ for all $m$ and $I^h$
is the stable union of the ideals $(J:T^m)$; the increasing sequence
$(J:T^m)$ stops after a finite stage by the Noetherian-ness of $R^h$.

From this description it is clear that the homogenisation of the primary
decomposition of an ideal is a primary decomposition of the
homogenisation.

Let $f:R^h\to A$ be a morphism which maps the irrelevant ideal to the unit
ideal. $T$ is an element of degree 1 in the irrelevant ideal. If $T$ is itself
mapped to a unit $\lambda$ in $A^{\times}$, then we can take the
equivalent homomorphism $\lambda^{-1}\cdot f$ which maps $T$ to 1 and
thus factors through $R$. In other words, the elements of $\Proj(R)(A)$
that are represented by homomorphisms $R^h\to A$ that map $T$ to a unit
correspond naturally to elements of $\Spec(R)(A)$. Thus $\Spec(R)$ can be
identifed as a subscheme of $\Proj(R)$.

At the other extreme we can consider a homomorphism $f:R^h\to A$ that
sends $T$ to 0 and yet maps the irrelevant ideal to the unit ideal. The
multiples of $T$ in $R^h$ have the form $aT^n$ where the degree of $a$
is strictly smaller than $n$ (so that $aT^n=T(aT^{n-1})$). All such
elements must also go to 0. It follows that $f$ actually gives a
homomorphism $\oplus (R_n/R_{n-1}) \to A$. When $R$ is considered as a
quotient of a polynomial ring by an ideal $I$ and $R_n$ is the induced
filtration, this amounts to considering the quotient of the same
polynomial ring by the ideal generated by the leading homogeneous
components of elements of $I$. These leading homogeneous components
define the {\em asymptotes at infinity} of $\Spec(R)$. Thus, we can think of
$\Proj(R)$ as the union of $\Spec(R)$ with its asymptotes at infinity.

\subsection{Blow-up}
Let $R$ be a finitely generated ring and $I$ an ideal. Let us write
$R$ as a quotient of the polynomial ring $k[X_1,\dots,X_n,]$ where the $X_i$ 
generate $R$ as a $k$-algebra and $X_k$, \dots, $X_n$ generate the ideal
$I$. The graded ring $\tilde{R}_I=\oplus I^n$ (where
$I^0=R$) can then be naturally thought of as a quotient of the graded
polynomial ring $k[X_1,\dots, X_n, Y_k, \dots, Y_n]$ where $\deg(X_i)=0$
and $\deg(Y_i)=1$. Moreover, we have the obvious relations
$X_iY_j=X_jY_i$. Let $K$ be the ideal generated by these relations and
$B$ be $\Proj(k[X_1,\dots,X_n,Y_k,\dots,Y_k]/K)$; the blow-up scheme
$\Proj(\tilde{R}_I)$  is a closed subscheme of $B$. Note that $B$ is itself
the blow-up of the polynomial ring with respect to the ideal generated
by $X_k$, \dots, $X_n$.

An element $B(A)$ is given by taking elements $a_1$,\dots, $a_n$ and
$b_k$, \dots $b_n$ of $A$ such that $(b_k,\dots, b_n)$ generate the
unit ideal in $A$ and $a_ib_j=a_jb_i$; in other words $(a_k,\dots, a_n)$
and $(b_k,\dots,b_n)$ are proportional. When $A$ has only one maximal
ideal this amounts to saying that $(a_k,\dots,a_n)$ is a multiple of
$(b_k,\dots,b_n)$. Now, if $(a_k,\dots,a_n)$ generate the unit ideal
then multiplying factor is a unit in $A$ and so unto equivalence we can
determine the $b$'s uniquely. In other words, each element of $B(A)$
gives an element of $\bbA^n$ and is point is ``outside the linear subspace
defined by $X_k=\dots=X_n=0$'' then the element of $B(A)$ is uniquely
determined. Here we interpret the complement of the vanishing of the
ideal as the locus where the image of the ideal generates the unit
ideal; this is {\em not} the set-theoretic complement.  Similarly, the
blow-up $\Proj(\tilde{R}_I)$ of $R$ along $I$ has a natural map to
$\Spec(R)$ which is a bijection on the locus of maps $R\to A$ under
which the image of $I$ generates the unit ideal.

The points of $\Proj(\tilde{R}_I)$ that map to points of
$\Spec(R/I)\subset\Spec(R)$ correspond to the asymptotes of $\Spec(R)$
along $\Spec(R/I)$. In terms of rings we see that these are points of
$\Proj(\oplus (I^n/I^{n-1}))$.

\end{document}