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@ -5,31 +5,29 @@
Then the following sets coincide
\begin{enumerate}
\item
$\left\{ \sum_{s \in
S'} r_{s} \cdot s ~ |~ S
\subseteq S' \text{finite}, r_s \in R, \right\}$
$\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$,
\item
$\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$
$\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$,
\item
The $\subseteq$-smallest submodule of $M$ containing $S$
The $\subseteq$-smallest submodule of $M$ containing $S$.
\end{enumerate}
This subset of $N \subseteq M$ is called the \vocab[Module!
Submodule]{submodule of $M $ generated by $S$}.
If $N= M$ we say that \vocab[Module!
generated by subset $S$]{$ M$ is generated by $S$}.
$M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is
generated by $S$.
This subset of $N \subseteq M$ is called the
\vocab[Module!Submodule]{submodule of $M $ generated by $S$}.
If $N= M$ we say that
\vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}.
$M$ is finitely generated
$:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$.
\end{definition}
\begin{definition}[Noetherian $R$-module]
$M$ is a \vocab{Noetherian} $R$-module if the
following equivalent conditions hold:
$M$ is a \vocab{Noetherian} $R$-module
if the following equivalent conditions hold:
\begin{enumerate}
\item
Every submodule $N \subseteq M$ is finitely generated.
\item
Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates
Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates.
\item
Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a
$\subseteq$-largest element.
@ -37,9 +35,9 @@
\end{definition}
\begin{proposition}[Hilbert's Basissatz]
\label{basissatz}
If $R$ is a Noetherian ring, then the polynomial rings
$R[X_1,\ldots, X_n]$ in
finitely many variables are Noetherian.
If $R$ is a Noetherian ring,
then the polynomial rings $R[X_1,\ldots, X_n]$
in finitely many variables are Noetherian.
\end{proposition}
\subsubsection{Properties of finite generation and Noetherianness}
@ -61,8 +59,8 @@
By definition, $M$ is a submodule of itself.
Thus it is finitely generated.
\item
Since $M$ is finitely generated, there exists a surjective homomorphism $R^n
\to M$.
Since $M$ is finitely generated,
there exists a surjective homomorphism $R^n \to M$.
As $R$ is Noetherian, $R^n$ is Noethrian as well.
\item
trivial
@ -73,49 +71,47 @@
Let $M, M', M''$ be $R$-modules.
\begin{enumerate}
\item
Suppose $M \xrightarrow{p}
M''$ is surjective.
If $M$ is finitely generated (resp.
Noetherian), then so is $M''$.
Suppose $M \xrightarrow{p} M''$ is surjective.
If $M$ is finitely generated (resp. Noetherian),
then so is $M''$.
\item
Let $M' \xrightarrow{f}
M \xrightarrow{p} M'' \to 0$ be exact.
If $M'$ and $M ''$ are finitely generated (reps.
Noetherian), so is $M$.
Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact.
If $M'$ and $M ''$ are finitely generated (reps. Noetherian),
so is $M$.
\end{enumerate}
\end{fact}
\begin{proof}
\begin{enumerate}
\item
Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$.
Then $p^{-1} M_i''$ yields a strictly ascending
sequence.
If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$.
Then $p^{-1} M_i''$ yields a strictly ascending sequence.
If $M$ is generated by $S,
|S| < \omega$, then $M''$ is generated by $p(S)$.
\item
Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M'
\xrightarrow{f}
M \xrightarrow{p} M'' \to 0$ to be exact.
The fact about finite generation follows from EInführung in die Algebra.
If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq
f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely
generated.
Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated.
Because of 1.~we can replace $M'$ by $f(M')$
and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$
to be exact.
The fact about finite generation follows from
Einführung in die Algebra.
If $M', M''$ are Noetherian, $N \subseteq M$ a submodule,
then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$
are finitely generated.
Since $0 \to N' \to N \to N'' \to 0$ is exact,
$N$ is finitely generated.
\end{enumerate}
\end{proof}
\subsection{Ring extensions of finite type}
\begin{definition}[$R$-algebra]
Let $R$ be a ring.
An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R
\xrightarrow{\alpha} A$.
An $R$-algebra $(A, \alpha)$ is a ring $A$
with a ring homomorphism $R \xrightarrow{\alpha} A$.
$\alpha$ will usually be omitted.
In general $\alpha$ is not assumed to be injective.
\\
\\
An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq
A$.\\
An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\
A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is
a ring homomorphism with $\tilde{\alpha} = f \alpha$.
\end{definition}
@ -123,9 +119,9 @@
\begin{definition}[Generated (sub)algebra, algebra of finite type]
Let $(A, \alpha)$ be an $R$-algebra.
\begin{align*}
\alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m] \\
P = \sum_{\beta \in \N^m} p_\beta X^{\beta} & \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta)
X^{\beta}
\alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m]\\
P = \sum_{\beta \in \N^m} p_\beta X^{\beta}
& \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta}
\end{align*}
is a ring homomorphism.
We will sometimes write $P(a_1,\ldots,a_m)$ instead of
@ -134,33 +130,31 @@
Fix $a_1,\ldots,a_m \in A^m$.
Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$.
The image of this ring homomorphism is the $R$-subalgebra of $A$
\vocab[Algebra!
generated subalgebra]{generated by the $a_i$}.
$A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i
\in I$.
\vocab[Algebra!generated subalgebra]{generated by the $a_i$}.
$A$ is \vocab[Algebra!of finite type]{of finite type}
if it can be generated by finitely many $a_i \in I$.
For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the
intersection of all subalgebras containing $S$ \\ $=$ the union of subalgebras
generated by finite $S' \subseteq S$\\ $= $ the image of
$R[X_s | s \in S]$
under $P \mapsto (\alpha(P))(S)$.
intersection of all subalgebras containing $S$ \\
$=$ the union of subalgebras generated by finite $S' \subseteq S$\\
$= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$.
\end{definition}
\subsection{Finite ring extensions} % LECTURE 2
\begin{definition}[Finite ring extension]
Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a
module over itself and the ringhomomorphism $R \to A$ allows us to derive an
$R$-module structure on $A$.
$A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is
finite if $A$ is finitely generated as an $R$-module.
Let $R$ be a ring and $A$ an $R$-algebra.
$A$ is a module over itself and the ringhomomorphism $R \to A$
allows us to derive an $R$-module structure on $A$.
$A$ \vocab[Algebra!finite over]{is finite over} $R$ /
the $R$-algebra $A$ is finite / $A / R$ is finite
if $A$ is finitely generated as an $R$-module.
\end{definition}
\begin{fact}[Basic properties of finiteness]
\begin{enumerate}[A]
\item
Every ring is finite over itself.
\item
A field extension is finite as a ring extension iff it is finite as a field
extension.
A field extension is finite as a ring extension
iff it is finite as a field extension.
\item
$A$ finite $\implies$ $A$ of finite type.
\item
@ -177,16 +171,14 @@
Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module.
Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
\item
Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by
$b_1,\ldots,b_n$ as an $A$-module.
For every $b$ there exist $\alpha_j \in A$ such that $b =
\sum_{j=1}^{n}
\alpha_j b_j$.
We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some
$\rho_{ij} \in R$
thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij}
a_i b_j$ and the $a_ib_j$
generate $B$ as an $R$-module.
Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module
and $B$ by $b_1,\ldots,b_n$ as an $A$-module.
For every $b$ there exist $\alpha_j \in A$
such that $b = \sum_{j=1}^{n} \alpha_j b_j$.
We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$
for some $\rho_{ij} \in R$
thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$
and the $a_ib_j$ generate $B$ as an $R$-module.
\end{enumerate}
\end{proof}
@ -194,39 +186,35 @@
\subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements?
This generalizes some facts about matrices to matrices with elements from
commutative rings with $1$.
\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.}
\footnote{Most of this even works in commutative rings without $ 1$,
since $1$ simply can be adjoined.}
\begin{definition}[Determinant]
Let $A = (a_{ij})
\Mat(n,n,R)$.
Let $A = (a_{ij}) \in \Mat(n,n,R)$.
We define the determinant by the Leibniz formula
\[
\det(A) \coloneqq \sum_{\pi
\in S_n} \sgn(\pi)
\prod_{i=1}^{n} a_{i, \pi(i)}
\det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}.
\]
Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij}
\coloneqq (-1)^{i+j} \cdot
M_{ij}$, where $M_{ij}$ is the
determinant of the matrix resulting from $A$
after deleting the $i^{\text{th}}$ row and the
$j^{\text{th}}$ column.
Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$,
where $M_{ij}$ is the determinant of the matrix resulting from $A$
after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column.
\end{definition}
\begin{fact}
\begin{enumerate}
\item
$\det(AB) = \det(A)\det(B)$
$\det(AB) = \det(A)\det(B)$.
\item
Development along a row or column works.
\item
Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot
A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible
iff $\det(A)$ is a unit.
Cramer's rule:
$A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$.
$A$ is invertible iff $\det(A)$ is a unit.
\item
Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then
$P_A(A) = 0$.
Caley-Hamilton:
If $P_A = \det(T \cdot \mathbf{1}_n - A)$%
\footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$},
then $P_A(A) = 0$.
\end{enumerate}
\end{fact}
\begin{proof}
All rules hold for the image of a matrix under a ring homomorphism if they hold
@ -234,20 +222,17 @@ commutative rings with $1$.
The converse holds in the case of injective ring homomorphisms.
Caley-Hamilton was shown for algebraically closed fields in LA2 using the
Jordan normal form.
Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds
for fields.
Fields can be embedded into their algebraic closure,
thus Caley-Hamilton holds for fields.
Every domain can be embedded in its field of quotients $\implies$
Caley-Hamilton holds for domains.
In general, $A$ is the image of
$(X_{i,j})_{i,j = 1}^{n} \in
\Mat(n,n,S)$ where
$S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the
morphism $S \to A$ of evaluation defined by $X_{i,j}
\mapsto a_{i,j}$.
In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$
where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain)
under the morphism $S \to A$ of evaluation defined by
$X_{i,j} \mapsto a_{i,j}$.
Thus Caley-Hamilton holds in general.
\end{proof}
%TODO: lernen
\subsection{Integral elements and integral ring extensions} %LECTURE 2
\begin{proposition}[on integral elements]
@ -256,88 +241,77 @@ commutative rings with $1$.
Then the following are equivalent:
\begin{enumerate}[A]
\item
$\exists n \in
\N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n =
\sum_{i=0}^{n-1} r_i a^i$
$\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$.
\item
There
exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
\end{enumerate}
If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of
$A$ finite over $R$ and containing all $a_i$.
If $a_1, \ldots, a_k \in A$ satisfy these conditions,
there is a subalgebra of $A$ finite over $R$ and containing all $a_i$.
\end{proposition}
\begin{definition}
\label{intclosure}
Elements that satisfy the conditions from
\ref{propinte} are called
Elements that satisfy the conditions from \ref{propinte} are called
\vocab{integral over} $R$.
$A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$.
$A / R$ is \vocab[Algebra!integral]{integral},
if all $a \in A$ are integral over $R$.
The set of elements of $A$ integral over $R$ is called the
\vocab{integral
closure} of $R$ in $A$.
\vocab{integral closure} of $R$ in $A$.
\end{definition}
\begin{proof}
\hskip 10pt
\begin{enumerate}
{\color{gray}
\item[B $\implies$ A]
{\color{gray} \item[B $\implies$ A]
Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$
and finite over $R$.
Let $(b_i)_{i=1}^{n}$ generate $B$ as an
$R$-module.
Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module.
\begin{align*}
q: R^n & \longrightarrow B \\
q: R^n & \longrightarrow B\\
(r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
\end{align*}
is surjective.
Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$
such that
$a b_i = q(\rho_i)$.
Thus there are $\rho_{i} = \left(r_{i,j}\right)_{j=1}^n \in R^n$
such that $a b_i = q(\rho_i)$.
Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$.
By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$
for all $P
\in R[T]$.
Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using
Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$.
$P$ is monic.
Since $q$ is surjective, we find $v \in R^{n} : q(v) =
1$.
Thus $P(a) = 0$ and $a$ satisfies A.
}
By induction it follows that
$q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$.
Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$
and using Caley-Hamilton,
we obtain $P(a) \cdot q(v) = 0$. $P$ is monic.
Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$.
Thus $P(a) = 0$ and $a$ satisfies A.}
\item[B $\implies$ A]
if $R$ is Noetherian.\footnote{This suffices in the exam.}
Let $a \in A$ satisfy B.
Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$.
Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ with $0 \le
i < n$.
As a finitely generated module over the Noetherian ring $R$, $B$ is a
Noetherian $R$-module.
Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in
M_d$.
Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such
that $a^d = \sum_{i=0}^{d-1}
r_ia^i$.
Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$
with $0 \le i < n$.
As a finitely generated module over the Noetherian ring $R$,
$B$ is a Noetherian $R$-module.
Thus the ascending sequence $M_n$ stabilizes at some step $d$
and $a^d \in M_d$.
Thus there are $(r_i)_{i=0}^{d-1} \in R^d$
such that $a^d = \sum_{i=0}^{d-1} r_ia^i$.
\item[A $\implies$ B]
Let $a = (a_i)_{i=1}^n$ where all $a_i$
satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1}
r_{i,j}a_i^j$ with $r_{i,j} \in
R$.
Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha =
\prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$.
Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A,
i.e.~$a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$
with $r_{i,j} \in R$.
Let $B \subseteq A$ be the sub-$R$-module generated by
$a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$
with $0 \le \alpha_i < d_i$.
$B$ is closed under $a_1 \cdot $ since
\[
a_1a^{\alpha} =
\begin{cases}
a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1 \\
\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1
a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1, \\
\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1.
\end{cases}
\]
By symmetry, this hold for all $a_i$.
By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant
under
$a^{\alpha}\cdot $.
Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed.
By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$,
$B$ is invariant under $a^{\alpha}\cdot $.
Since these generate $B$ as an $R$-module,
$B$ is multiplicatively closed.
Thus A holds.
Furthermore we have shown the final assertion of the proposition.
\end{enumerate}
@ -348,13 +322,15 @@ commutative rings with $1$.
\item[Q]
Every finite $R$-algebra $A$ is integral.
\item[R]
The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$
The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$.
\item[S]
If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$,
If $A$ is an $R$-algebra,
$B$ an $A$-algebra and $b \in B$ integral over $R$,
then it is integral over $A$.
\item[T]
If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral
over $A$, then $b$ is integral over $R$.
If $A$ is an integral $R$-algebra and $B$ any $A$-algebra,
$b \in B$ integral over $A$,
then $b$ is integral over $R$.
\end{enumerate}
\end{corollary}
\begin{proof}
@ -362,136 +338,126 @@ commutative rings with $1$.
\item[Q]
Put $ B = A $ in B.
\item[R]
For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral
over $R$.
From B it follows, that the integral closure is closed under ring operations.
For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$,
hence integral over $R$.
From B it follows,
that the integral closure is closed under ring operations.
\item[S]
trivial
\item[T]
Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$.
Then there is a subalgebra $\tilde{A} \subseteq A$ finite over
$R$, such that
all $a_i \in \tilde{A}$.
$b$ is integral over $\tilde{A} \implies \exists
\tilde{B} \subseteq B$ finite over $\tilde{A}$ and
$b \in \tilde{B}$.
Since $\tilde{B} / \tilde{A} $ and
$\tilde{A} / R$ are finite, $\tilde{B} / R$
is finite and $b$ satisfies B.
Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$,
such that all $a_i \in \tilde{A}$.
$b$ is integral over $\tilde{A}$
Hence $\exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$.
Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite,
$\tilde{B} / R$ is finite and $b$ satisfies B.
\end{enumerate}
\end{proof}
\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality
\subsection{Finiteness, finite generation and integrality}
% some more remarks on finiteness, finite generation and integrality
\begin{fact}[Finite type and integral $\implies$ finite]
\label{ftaiimplf}
If $A$ is an integral $R$-algebra of finite type, then it is a finite
$R$-algebra.
If $A$ is an integral $R$-algebra of finite type,
then it is a finite $R$-algebra.
\end{fact}
\begin{proof}
Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra.
By the proposition on integral elements (
\ref{propinte}), there is a
finite
$R$-algebra $B \subseteq A$ such that all $a_i \in B$.
Let $A $ be generated by $\left( a_i \right)_{i=1}^{n}$ as an $R$-algebra.
By the proposition on integral elements (\ref{propinte}),
there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$.
We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra.
\end{proof}
\begin{fact}[Finite type in tower]
If $A$ is an $R$-algebra of finite type and $B$ an
$A$-algebra of finite type, then $B$ is an $R$-algebra of finite type.
If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra
of finite type, then $B$ is an $R$-algebra of finite type.
\end{fact}
\begin{proof}
If $A / R$ is generated by $(a_i)_{i=1}^m$ and
$B / A$ by $(b_j)_{j=1}^{n}$,
If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$,
then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$.
\end{proof}
{\color{red}
\begin{fact}[About integrality and fields]
\label{fintaf}
Let $B$ be a domain integral over its subring $A$.
Then $B$ is a field iff $A$ is a field.
\end{fact}
{
\color{red}
\begin{fact}[About integrality and fields]
\label{fintaf}
Let $B$ be a domain integral over its subring $A$.
Then $B$ is a field iff $A$ is a field.
\end{fact}
}
\begin{proof}
Let $B$ be a field and $a \in A \setminus \{0\} $.
Then $a^{-1} \in B$ is integral over $A$, hence
$a^{-d} = \sum_{i=0}^{d-1}
\alpha_i a^{-i}$ for some $\alpha_i \in A$.
Then $a^{-1} \in B$ is integral over $A$,
hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$
for some $\alpha_i \in A$.
Multiplication by $a^{d-1}$ yields
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i
a^{d-1-i} \in A$.
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
On the other hand, let $B$ be integral over the field $A$.
Let $b \in B \setminus \{0\}$.
As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B}
\subseteq B,
b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a
finite-dimensional
$A$-vector space.
Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot }
\tilde{B}$ is
injective, hence surjective, thus $\exists x \in \tilde{B} : b
\cdot x \cdot
1$.
As $B$ is integral over $A$,
there is a sub-$A$-algebra $\tilde{B} \subseteq B$,
$b \in \tilde{B}$ finitely generated as an $A$-module,
i.e.~a finite-dimensional $A$-vector space.
Since $B$ is a domain,
$\tilde{B} \xrightarrow{b\cdot } \tilde{B}$
is injective,
hence surjective,
thus $\exists x \in \tilde{B} : b \cdot x \cdot 1$.
\end{proof}
\subsection{Noether normalization theorem}
\begin{lemma}
\label{nntechlemma}
Let $S \subseteq \N^n$ be finite.
Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and
$w_{\vec k}(\alpha)
\neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$,
where $w_{\vec
k}(\alpha) = \sum_{i=1}^{n} k_i
\alpha_i$.
Then there exists $\vec k \in \N^n$ such that $k_1 =1$
and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$
for $\alpha \neq \beta \in S$,
where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$.
\end{lemma}
\begin{proof}
Intuitive: For $\alpha \neq \beta$ the equation
$w_{(1, \vec \kappa)}(\alpha) =
w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$)
defines a codimension $1$
Intuitive:
For $\alpha \neq \beta$ the equation
$w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$
($\kappa \in \R^{n-1}$) defines a codimension $1$
affine hyperplane in $\R^{n-1}$.
It is possible to choose $\kappa$ such that all $\kappa_i$ are $>
\frac{1}{2}$
and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these
hyperplanes.
By choosing the closest $\kappa'$ with integral coordinates, each coordinate
will be disturbed by at most $\frac{1}{2}$, thus at Euclidean
distance $\le
\frac{\sqrt{n-1} }{2}$.
It is possible to choose $\kappa$ such that all $\kappa_i$ are
$> \frac{1}{2}$
and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$
from the union of these hyperplanes.
By choosing the closest $\kappa'$ with integral coordinates,
each coordinate will be disturbed by at most $\frac{1}{2}$,
thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$.
More formally:\footnote{The intuitive version suffices in the exam.
}
Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $.
More formally:\footnote{The intuitive version suffices in the exam.}
Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\}$.
We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
Suppose $\alpha \neq \beta$.
Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$.
Then the contributions of $\alpha_j$ (resp.
$\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$
(resp. $w_{\vec k}(\beta)$) cannot undo the difference
$k_i(\alpha_i - \beta_i)$.
Then the contributions of $\alpha_j$
(resp.~$\beta_j$)
with $1 \le j < i$ to $w_{\vec k}(\alpha)$
(resp. $w_{\vec k}(\beta)$)
cannot undo the difference $k_i(\alpha_i - \beta_i)$.
\end{proof}
\begin{theorem}[Noether normalization]
\label{noenort}
Let $K$ be a field and $A$ a $K$-algebra of finite type.
Then there are $a = (a_i)_{i=1}^{n} \in A$ which
are algebraically independent
over $K$, i.e. the ring homomorphism
are algebraically independent over $K$,
i.e.~the ring homomorphism
\begin{align*}
\ev_a: K[X_1,\ldots,X_n] &
\longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n)
\ev_a: K[X_1,\ldots,X_n] & \longrightarrow A\\
P & \longmapsto P(a_1,\ldots,a_n)
\end{align*}
is
injective.
is injective.
$n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
$\ev_a$.
\end{theorem}
\begin{proof}
Let $(a_i)_{i=1}^n$ be a minimal number of
elements such that $A$ is integral
over its $K$-subalgebra generated by $a_1, \ldots, a_n$.
Let $(a_i)_{i=1}^n$ be a minimal number of elements
such that $A$ is integral over its $K$-subalgebra
generated by $a_1, \ldots, a_n$.
(Such $a_i$ exist, since $A$ is of finite type).
Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
If suffices to show that the $a_i$ are algebraically independent.
@ -547,5 +513,4 @@ commutative rings with $1$.
This contradicts the minimality of $n$, as $B$ can be generated by $< n$
elements $b_i$.
\end{proof}

9
inputs/preamble.tex
View File

@ -5,13 +5,12 @@
\end{warning}
\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the
MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}.
% TODO
MIT-License and can be obtained from
\url{https://gitlab.com/latexci/LatexPackages}.
\newline
\noindent $\mathfrak{k}$ is {\color{red} always} an
algebraically closed field and $\mathfrak{k}^n$ is equipped with the
Zariski-topology.
\noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field
and $\mathfrak{k}^n$ is equipped with the Zariski-topology.
Fields which are not assumed to be algebraically closed have been renamed
(usually to $\mathfrak{l}$).

543
inputs/projective_spaces.tex
View File

@ -1,64 +1,50 @@
Let $\mathfrak{l}$ be any field.
\begin{definition}
For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be
the set of
For a $\mathfrak{l}$-vector space $V$,
let $\mathbb{P}(V)$ be the set of
one-dimensional subspaces of $V$.
Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq
\mathbb{P}(\mathfrak{l}^{n+1})$, the
\vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$,
the \vocab[Projective space]%
{$n$-dimensional projective space over $\mathfrak{l}$}.
If $\mathfrak{l}$ is kept fixed, we will often write
$\mathbb{P}^n$ for
$\mathbb{P}^n(\mathfrak{l})$.
If $\mathfrak{l}$ is kept fixed,
we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$.
When dealing with $\mathbb{P}^n$, the usual convention is to use
$0$ as the
index of the first coordinate.
When dealing with $\mathbb{P}^n$,
the usual convention is to use $0$ as the index of the first coordinate.
We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in
\mathfrak{k}^{n+1} \setminus \{0\}$ by
We denote the one-dimensional subspace generated by
$(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by
$[x_0,\ldots,x_n] \in \mathbb{P}^n$.
If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the
$(x_{i})_{i=0}^n$ are
called
\vocab{homogeneous coordinates} of $x$.
If $x = [x_0,
ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called
\vocab{homogeneous coordinates} of $x$.
At least one of the $x_{i}$ must be $\neq 0$.
\end{definition}
\begin{remark}
There are points $[1,0],
[0,1] \in \mathbb{P}^1$ but there
is no point $[0,0]
\in \mathbb{P}^1$.
There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there
is no point $[0,0] \in \mathbb{P}^1$.
\end{remark}
\begin{definition}[Infinite hyperplane]
For $0 \le i \le n$ let $U_i \subseteq
\mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with
$x_{i}\neq 0$.
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and
$[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
same point $x \in
\mathbb{P}^n$ differ by scaling with a $\lambda \in
\mathfrak{l}^{\times}$,
For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$
denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$
and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
same point $x \in \mathbb{P}^n$ differ by scaling with a
$\lambda \in \mathfrak{l}^{\times}$,
$x_i = \lambda \xi_i$.
Since not all $x_i$ may be $0$, $\mathbb{P}^n =
\bigcup_{i=0}^n U_i$.
We identify $\mathbb{A}^n =
\mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
with
$U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$.
We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
$[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $
where $\infty=[0,1]$.
More generally, when $n > 0$ $\mathbb{P}^n \setminus
\mathbb{A}^n$ can be
More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be
identified with $\mathbb{P}^{n-1}$ identifying
$[0,x_1,\ldots,x_n] \in
\mathbb{P}^n \setminus \mathbb{A}^n$ with
$[x_1,\ldots,x_n] \in
\mathbb{P}^{n-1}$.
$[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with
$[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$.
Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong
\mathfrak{l}^n$ with a copy of
Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of
$\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
\end{definition}
@ -67,40 +53,37 @@ Let $\mathfrak{l}$ be any field.
Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$.
\end{notation}
\begin{definition}
By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a
ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
subgroups of the
additive group $(A, +)$ such that $A_a \cdot A_b \subseteq
A_{a + b}$ for $a,b
\in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
the
sense that every $r \in A$ has a unique decomposition $r =
\sum_{d \in
\mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d
\neq 0$.
By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$
we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
subgroups of the additive group $(A, +)$
such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$
and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
the sense that every $r \in A$ has a unique decomposition
$r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many
$r_d \neq 0$.
We call the $r_d$ the \vocab{homogeneous components} of $r$.
An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I
\implies
\forall d \in \mathbb{I} ~ r_d \in I_d$ where $I_d \coloneqq I
\cap A_d$.
An ideal $I \subseteq A$ is called \vocab{homogeneous} if
$r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$
where $I_d \coloneqq I \cap A_d$.
By a \vocab{graded ring} we understand an $\N$-graded ring.
Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty}
A_d = \{r \in A | r_0
= 0\} $ is called the \vocab{augmentation ideal} of $A$.
Tin this case,
$A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $
is called the \vocab{augmentation ideal} of $A$.
\end{definition}
\begin{remark}[Decomposition of $1$]
If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the
decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot
\varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with
$\varepsilon_a\varepsilon_b \in A_{a+b}$.
If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$
is the decomposition into homogeneous components,
then $\varepsilon_a = 1 \cdot \varepsilon_a
= \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$
with $\varepsilon_a\varepsilon_b \in A_{a+b}$.
By the uniqueness of the decomposition into homogeneous components,
$\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies
\varepsilon_a \varepsilon_b = 0$.
Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b =
\varepsilon_0 \varepsilon _b = 0$.
$\varepsilon_a \varepsilon_0 = \varepsilon_a$
and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$.
Applying the last equation with $a = 0$ gives
$b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$.
Thus $1 = \varepsilon_0 \in A_0$.
\end{remark}
\begin{remark}
@ -123,44 +106,41 @@ Let $\mathfrak{l}$ be any field.
\end{proposition}
\begin{proof}
Most assertions are trivial.
We only show that $J$ homogeneous $\implies \sqrt{J} $
homogeneous.
Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and
$f = \sum_{d \in
\mathbb{I}} f_d$ the decomposition.
To show that all $f_d \in \sqrt{J} $, we use induction on $N_f
\coloneqq \# \{d
\in \mathbb{I} | f_d \neq 0\}$.
We only show that $J$ homogeneous $\implies \sqrt{J}$ homogeneous.
Let $A$ be $\mathbb{I}$-graded,
$f \in \sqrt{J} $ and
$f = \sum_{d \in \mathbb{I}} f_d$ the decomposition.
To show that all $f_d \in \sqrt{J} $,
we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$.
$N_f = 0$ is trivial.
Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq
0$.
Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$.
For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$.
Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of
$J$, we find $f_e \in \sqrt{J}$.
As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in
\sqrt{J} $.
As $N_{\tilde f} = N_f -1$, the induction assumption may be
applied to $\tilde
f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$.
As $\sqrt{J} $ is an ideal,
$\tilde f \coloneqq f - f_e \in \sqrt{J} $.
As $N_{\tilde f} = N_f -1$,
the induction assumption may be applied to $\tilde f$
and shows $f_d \in \sqrt{J} $ for $d \neq e$.
\end{proof}
\begin{fact}
A homogeneous ideal is finitely generated iff it can be generated by finitely
many of its homogeneous elements.
A homogeneous ideal is finitely generated
iff it can be generated by finitely many of its homogeneous elements.
In particular, this is always the case when $A$ is a Noetherian ring.
\end{fact}
\subsubsection{The Zariski topology on $\mathbb{P}^n$}
\begin{notation}
Recall that for $\alpha \in \N^{n+1}$ $|\alpha| =
\sum_{i=0}^{n} \alpha_i$ and
$x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
Recall that for
$\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$
and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
\end{notation}
\begin{definition}[Homogeneous polynomials]
Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}}
f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d
\implies f_\alpha = 0$ .
Let $R$ be any ring and
$f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
We say that $f$ is \vocab{homogeneous of degree $d$}
if $|\alpha| \neq d \implies f_\alpha = 0$ .
We denote the subset of homogeneous polynomials of degree $d$ by
$R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
\end{definition}
@ -169,57 +149,50 @@ Let $\mathfrak{l}$ be any field.
\end{remark}
\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]
\label{ztoppn}
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.%
\footnote{As always, $\mathfrak{k}$ is algebraically closed}
For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation
$f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous
coordinates, as
For $f \in A_d = \mathfrak{k}[X_0,\ldots,
_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$
does not depend on the choice of homogeneous coordinates, as
\[
f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d
f(x_0,\ldots,x_n)
f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n).
\]
Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x)
= 0\}$.
Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$.
We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it
can be
represented as
can be represented as
\[
X = \bigcap_{i=1}^k \Vp(f_i)
\]
where the $f_i \in A_{d_i}$
are homogeneous polynomials.
where the $f_i \in A_{d_i}$ are homogeneous polynomials.
\end{definition}
\pagebreak
\begin{fact}
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq
\mathbb{P}^n$ is closed, then $Y
= X \cap \mathbb{A}^n$ can be identified with the closed subset
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$ is closed,
then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset
\[
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1
\le i \le
k\} \subseteq \mathfrak{k}^n
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\}
\subseteq \mathfrak{k}^n.
\]
Conversely, if $Y \subseteq \mathfrak{k}^n$ is
closed it has the form
Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form
\[
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
\]
and can thus be identified with $X
\cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k
\Vp(f_i)$ is given by
and can thus be identified with $X \cap \mathbb{A}^n$
where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by
\[
f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i
\ge \deg(g_i)
\ge \deg(g_i).
\]
Thus, the Zariski topology on $\mathfrak{k}^n$ can be
identified with the topology induced by the Zariski topology on
$\mathbb{A}^n =
U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
$\mathbb{A}^n = U_0$,
and the same holds for $U_i$ with $0 \le i \le n$.
In this sense, the Zariski topology on $\mathbb{P}^n$ can be
thought of as
In this sense,
the Zariski topology on $\mathbb{P}^n$ can be thought of as
gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
\end{fact}
@ -227,18 +200,16 @@ Let $\mathfrak{l}$ be any field.
\begin{definition}
Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
Let $\Vp(I) \coloneqq
\{[x_0,\ldots,_n] \in \mathbb{P}^n |
\forall f \in I ~
f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this
condition for the homogeneous elements $f \in I$.
Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements
$(f_i)_{i=1}^k$ and
$\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as
in
\ref{ztoppn}.
Conversely, if the homogeneous $f_i$ are given, then $I = \langle
f_1,\ldots,f_k \rangle_A$ is homogeneous.
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~%
f(x_0,\ldots,x_n) = 0\}$
As $I$ is homogeneous,
it is sufficient to impose this condition for the homogeneous elements
$f \in I$.
Because $A$ is Noetherian,
$I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and
$\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}.
Conversely, if the homogeneous $f_i$ are given,
then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous.
\end{definition}
\begin{remark}
Note that $V(A) = V(A_+) = \emptyset$.
@ -247,52 +218,47 @@ Let $\mathfrak{l}$ be any field.
For homogeneous ideals in $A$ and $m \in \N$, we have:
\begin{itemize}
\item
$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda}
\Vp(I_\lambda)$
$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$.
\item
$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) =
\bigcup_{k=1}^m \Vp(I_k)$
$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$.
\item
$\Vp(\sqrt{I}) = \Vp(I)$
$\Vp(\sqrt{I}) = \Vp(I)$.
\end{itemize}
\end{fact}
\begin{fact}
If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an
open covering of a topological space then $X$ is Noetherian iff there is a
finite subcovering and all $U_\lambda$ are Noetherian.
If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering
of a topological space then $X$ is Noetherian
iff there is a finite subcovering and all $U_\lambda$ are Noetherian.
\end{fact}
\begin{proof}
By definition, a topological space is Noetherian $\iff$ all open subsets are
quasi-compact.
By definition, a topological space is Noetherian
$\iff$ all open subsets are quasi-compact.
\end{proof}
\begin{corollary}
The Zariski topology on $\mathbb{P}^n$ is indeed a topology.
The induced topology on the open set $\mathbb{A}^n =
\mathbb{P}^n \setminus
\Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski
topology on $\mathfrak{k}^n$.
The same holds for all $U_i = \mathbb{P}^n \setminus
\Vp(X_i) \cong
\mathfrak{k}^n$.
The induced topology on the open set
$\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$
is the Zariski topology on $\mathfrak{k}^n$.
The same holds for all
$U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
Moreover, the topological space $\mathbb{P}^n$ is Noetherian.
\end{corollary}
\subsection{Noetherianness of graded rings}
\begin{proposition}
For a graded ring $R_{\bullet}$, the following conditions
are equivalent:
For a graded ring $R_{\bullet}$,
the following conditions are equivalent:
\begin{enumerate}[A]
\item
$R$ is Noetherian.
\item
Every homogeneous ideal of $R_{\bullet}$ is finitely
generated.
Every homogeneous ideal of $R_{\bullet}$ is finitely generated.
\item
Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals
terminates.
Every chain $I_0\subseteq I_1 \subseteq \ldots$ of
homogeneous ideals terminates.
\item
Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a
$\subseteq$-maximal element.
Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals
has a $\subseteq$-maximal element.
\item
$R_0$ is Noetherian and the ideal $R_+$ is finitely generated.
\item
@ -316,27 +282,27 @@ Let $\mathfrak{l}$ be any field.
The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$.
\end{claim}
\begin{subproof}
It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde
R$.
It is sufficient to show that every homogeneous $f \in R_d$
belongs to $\tilde R$.
We use induction on $d$.
The case of $d = 0$ is trivial.
Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$.
as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where
$g_i = \sum_{b=0}^{\infty}
g_{i,b}$ is the decomposition into homogeneous
components.
As $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
Let $f_a = \sum_{i=1}^{k} g_{i,
a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$
is the decomposition into homogeneous components.
Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into
homogeneous
components, hence $a \neq d \implies f_a = 0 $.
homogeneous components,
hence $a \neq d \implies f_a = 0 $.
Thus we may assume $g_i \in R_{d-d_i}$.
As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i
\in \tilde R$, hence $f \in \tilde R$.
As $d_i > 0$,
the induction assumption may now be applied to $g_i$,
hence $g_i \in \tilde R$,
hence $f \in \tilde R$.
\end{subproof}
\noindent\textbf{F $\implies$ A}
Hilbert's Basissatz (
\ref{basissatz})
Hilbert's Basissatz (\ref{basissatz})
\end{proof}
@ -348,148 +314,130 @@ Let $\mathfrak{l}$ be any field.
% Lecture 12
\begin{proposition}[Projective form of the Nullstellensatz]
\label{hnsp}
If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then
$\Vp(I) \subseteq \Vp(f) \iff f \in
\sqrt{I}$.
If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$,
then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$.
\end{proposition}
\begin{proof}
$\impliedby$ is clear.
Let $\Vp(I) \subseteq \Vp(f)$.
If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in
which case $f(x) =
0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in
\mathbb{P}^n$ is
well-defined and belongs to $\Vp(I) \subseteq
\Vp(f)$, hence $f(x) = 0$.
Thus $\Va(I) \subseteq \Va(f)$ and $f \in
\sqrt{I}$ be the Nullstellensatz
(
\ref{hns3}).
If $x = (x_0,
ldots,x_n) \in \Va(I)$, then either $x = 0$ in
which case $f(x) = 0$ since $d > 0$
or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is
well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$,
hence $f(x) = 0$.
Thus $\Va(I) \subseteq \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz
(\ref{hns3}).
\end{proof}
\begin{definition}
\footnote{This definition is not too important, the characterization in the following remark suffices.}.
For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of
$\fp \in
\Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
For a graded ring $R_\bullet$,
let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$
such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
\end{definition}
\begin{remark}
\label{proja}
As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for
every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ |
\fp
\text{ is homogeneous}\} $.
As the elements of $A_0 \setminus \{0\}$ are units in $A$
it follows that for every homogeneous ideal $I$
we have $I \subseteq A_+$ or $I = A$.
In particular,
$\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | \fp \text{ is homogeneous}\} $.
\end{remark}
\begin{proposition}
\label{bijproj}
There is a bijection
\begin{align*}
f: \{I \subseteq A_+ | I \text{ homogeneous
ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{
closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X
\subseteq \Vp(f)\} \rangle & \longmapsfrom X
f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &
\longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\}\\
I &
\longmapsto \Vp(I)\\
\langle \{f \in A_d | d > 0, X \subseteq \Vp(f)\} \rangle &
\longmapsfrom X
\end{align*}
Under this bijection,
the irreducible subsets correspond to the elements of
$\Proj(A_\bullet)$.
\end{proposition}
\begin{proof}
From the projective form of the Nullstellensatz it follows that $f$ is
injective and that $f^{-1}(\Vp\left( I \right))
= \sqrt{I} = I$.
If $X \subseteq \mathbb{P}^n$ is closed, then $X =
\Vp(J)$ for some homogeneous
ideal $J \subseteq A$.
From the projective form of the Nullstellensatz it follows
that $f$ is injective
and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$.
If $X \subseteq \mathbb{P}^n$ is closed,
then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$.
Without loss of generality loss of generality $J = \sqrt{J}$.
If $J \not\subseteq A_+$, then $J = A$ (
\ref{proja}), hence $X =
\Vp(J) =
\emptyset = \Vp(A_+)$.
If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}),
hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
Suppose $\fp \in \Proj(A_\bullet)$.
Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
proven part of
the proposition.
Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where
$X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k =
\sqrt{I_k}$.
Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
proven part of the proposition.
Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets,
where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$.
Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq
\Vp(I_k)$ hence $\Vp(f_1f_2)
\supseteq \Vp(I_1) \cup \Vp(I_2) = X =
\Vp(\fp)$ and it follows that $f_1f_2\in
\sqrt{\fp} = \fp \lightning$.
We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$
hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$
and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$.
Assume $X = \Vp(\fp)$ is irreducible, where $\fp =
\sqrt{\fp} \in A_+$ is
homogeneous.
Assume $X = \Vp(\fp)$ is irreducible,
where $\fp = \sqrt{\fp} \in A_+$ is homogeneous.
The $\fp \neq A_+$ as $X = \emptyset$ otherwise.
Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i}
\setminus \fp$.
Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \setminus \fp$.
Then $X \not \subseteq \Vp(f_i)$ by the projective
Nullstellensatz when $d_i >
0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$.
Nullstellensatz when $d_i > 0$
and because $\Vp(1) = \emptyset$ when $d_i = 0$.
Thus $X = (X \cap \Vp\left( f_1 \right)) \cup (X \cap \Vp(f_2))$ is a
proper
decomposition $\lightning$.
By lemma
\ref{homprime}, $\fp$ is a prime ideal.
proper decomposition $\lightning$.
By lemma \ref{homprime}, $\fp$ is a prime ideal.
\end{proof}
\begin{remark}
It is important that $I \subseteq A_{\color{red} +}$, since
$\Vp(A) = \Vp(A_+)
= \emptyset$ would be a counterexample.
It is important that $I \subseteq A_{\color{red} +}$,
since $\Vp(A) = \Vp(A_+) = \emptyset$ would be a counterexample.
\end{remark}
\begin{corollary}
$\mathbb{P}^n$ is irreducible.
\end{corollary}
\begin{proof}
Apply
\ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$.
Apply \ref{bijproj} to $\{0\} \in \Proj(A_\bullet)$.
\end{proof}
\subsection{Some remarks on homogeneous prime ideals}
\begin{lemma}
\label{homprime}
Let $R_\bullet$ be an $\mathbb{I}$ graded ring
($\mathbb{I} = \N$ or
$\mathbb{I} = \Z$).
A homogeneous ideal $I \subseteq R$ is a prime ideal iff $1 \not\in I$ and for
homogeneous elements $f, g \in R , fg \in I \implies f \in I \lor g \in I$.
($\mathbb{I} = \N$ or $\mathbb{I} = \Z$).
A homogeneous ideal $I \subseteq R$ is a prime ideal
iff $1 \not\in I$ and for all homogeneous elements $f, g \in R$
\[fg \in I \implies f \in I \lor g \in I.\]
\end{lemma}
\begin{proof}
$\implies$ is trivial.
It suffices to show that for arbitrary $f,g \in R fg \in I \implies f \in I
\lor g \in I$.
It suffices to show that for arbitrary $f,g \in R$
we have that $fg \in I \implies f \in I \lor g \in I$.
Let $f = \sum_{d \in \mathbb{I}} f_d, g = \sum_{d \in \mathbb{I}} g_d $ be the
decompositions into homogeneous components.
If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I, g_e
\in I$, and they may assumed to be maximal with this property.
As $I$ is homogeneous and $fg \in I$, we have
$(fg)_{d+e} \in I$ but
If $f \not\in I$ and $g \not\in I$ there are $d,e \in I$ with $f_d \in I$,
$g_e \in I$, and they may assumed to be maximal with this property.
As $I$ is homogeneous and $fg \in I$, we have $(fg)_{d+e} \in I$ but
\[
(fg)_{d+e} = f_dg_e + \sum_{\delta = 1}^{\infty} (f_{d + \delta} g_{e - \delta}
+ f_{d - \delta} g_{e + \delta})
\]
where $f_dg_e \not\in I$ by our assumption
on $I$ and all other summands on the right hand side are $\in I$ (as
$f_{d+
\delta} \in I$ and $g_{e + \delta} \in
I$ by the maximality of $d$ and $e$), a
contradiction.
on $I$ and all other summands on the right hand side are $\in I$
(as $f_{d+ \delta} \in I$ and $g_{e + \delta} \in I$ by the maximality
of $d$ and $e$),
a contradiction.
\end{proof}
\begin{remark}
If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$, then $\fp \oplus R_+ =
\{r \in R | r_0 \in \fp\} $ is a homogeneous prime ideal of $R$.
If $R_\bullet$ is $\N$-graded and $\fp \in \Spec R_0$,
then $\fp \oplus R_+ = \{r \in R | r_0 \in \fp\} $
is a homogeneous prime ideal of $R$.
\[
\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of }
R_\bullet\} = \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec
R_0\}
\{\fp \in \Spec R | \fp \text{ is a homogeneous ideal of } R_\bullet\}
= \Proj(R_\bullet) \sqcup \{\fp \oplus R_+ | \fp \in \Spec R_0\}.
\]
\end{remark}
@ -500,38 +448,35 @@ Let $\mathfrak{l}$ be any field.
$\mathbb{P}^n$ is catenary.
\item
$\dim(\mathbb{P}^n) = n$.
Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in
\mathbb{P}^n$.
Moreover, $\codim(\{x\} ,\mathbb{P}^n) = n$ for every $x \in \mathbb{P}^n$.
\item
If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$, then
$\codim(\{x\}, X) = \dim(X) = n -
\codim(X, \mathbb{P}^n)$.
If $X \subseteq \mathbb{P}^n$ is irreducible and $x \in X$,
then $\codim(\{x\}, X) = \dim(X) = n - \codim(X, \mathbb{P}^n)$.
\item
If $X \subseteq Y \subseteq \mathbb{P}^n$ are irreducible subsets,
then $\codim(X,Y) = \dim(Y) -
\dim(X)$.
then $\codim(X,Y) = \dim(Y) - \dim(X)$.
\end{itemize}
\end{proposition}
\begin{proof}
Let $X \subseteq \mathbb{P}^n$ be irreducible.
If $x \in X$, there is an integer $0 \le i \le n$ and $X \in U_i =
\mathbb{P}^n \setminus \Vp(X_i)$.
If $x \in X$, there is an integer $0 \le i \le n$ and
$X \in U_i = \mathbb{P}^n \setminus \Vp(X_i)$.
Without loss of generality loss of generality $i = 0$.
Then $\codim(X, \mathbb{P}^n) = \codim(X \cap \mathbb{A}^n, \mathbb{A}^n)$ by
the locality of Krull codimension (
\ref{lockrullcodim}).
Applying this with $X = \{x\}$ and our results about the affine case gives the
second assertion.
If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$, then
$\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$, $\codim(X,Z)
= \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$ and $\codim(Y,Z) =
\codim(Y
\cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
the locality of Krull codimension (\ref{lockrullcodim}).
Applying this with $X = \{x\}$
and our results about the affine case gives the second assertion.
If $Y$ and $Z$ are also irreducible with $X \subseteq Y \subseteq Z$,
then
$\codim(X,Y) = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n)$,
$\codim(X,Z) = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$
and $\codim(Y,Z) = \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
Thus
\begin{align*}
\codim(X,Y) + \codim(Y,Z) & = \codim(X \cap \mathbb{A}^n, Y
\cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & =
\codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = \codim(X, Z)
\codim(X,Y) + \codim(Y,Z)
& = \codim(X \cap \mathbb{A}^n, Y \cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\
& = \codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n)\\
& = \codim(X, Z)
\end{align*}
because $\mathfrak{k}^n$ is catenary and the first point follows.
The remaining assertions can easily be derived from the first two.
@ -540,15 +485,12 @@ Let $\mathfrak{l}$ be any field.
\subsection{The cone $C(X)$}
\begin{definition}
If $X \subseteq \mathbb{P}^n$ is closed, we define the
\vocab{affine cone over
$X$}
\vocab{affine cone over $X$}
\[
C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus
\{0\} | [x_0,\ldots,x_n] \in X\}
C(X) = \{0\} \cup \{(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\} | [x_0,\ldots,x_n] \in X\}
\]
If $X = \Vp(I)$ where $I \subseteq A_+ =
\mathfrak{k}[X_0,\ldots,X_n]_+$ is homogeneous, then $C(X) =
\Va(I)$.
If $X = \Vp(I)$ where $I \subseteq A_+ = \mathfrak{k}[X_0,\ldots,X_n]_+$
is homogeneous, then $C(X) = \Va(I)$.
\end{definition}
\begin{proposition}
\label{conedim}
@ -564,17 +506,14 @@ Let $\mathfrak{l}$ be any field.
\end{itemize}
\end{proposition}
\begin{proof}
The first assertion follows from
\ref{bijproj} and
\ref{bijiredprim}
The first assertion follows from \ref{bijproj} and \ref{bijiredprim}
(bijection of irreducible subsets and prime ideals in the projective
and affine case).
Let $d = \dim(X)$ and
\[
X_0 \subsetneq \ldots \subsetneq X_d = X \subsetneq
X_{d+1} \subsetneq \ldots \subsetneq X_n =
\mathbb{P}^n
X_0 \subsetneq \ldots \subsetneq X_d = X
\subsetneq X_{d+1} \subsetneq \ldots \subsetneq X_n = \mathbb{P}^n
\]
be a chain of irreducible subsets of $\mathbb{P}^n$.
Then
@ -586,8 +525,7 @@ Let $\mathfrak{l}$ be any field.
Hence $\dim(C(X)) \ge 1 + d$ and
$\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
Since
$\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1})
= n+1$,
$\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1}) = n+1$,
the two inequalities must be equalities.
\end{proof}
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
@ -605,13 +543,11 @@ Let $\mathfrak{l}$ be any field.
\end{corollary}
\begin{proof}
If $H = \Vp(P)$ then $C(H) = \Va(P)$
is a hypersurface in $\mathfrak{k}^{n+1}$
by \ref{irredcodimone}.
is a hypersurface in $\mathfrak{k}^{n+1}$ by \ref{irredcodimone}.
By \ref{conedim}, $H$ is irreducible and of codimension $1$.
Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$.
By \ref{conedim}, $C(H)$ is a hypersurface in
$\mathfrak{k}^{n+1}$,
By \ref{conedim}, $C(H)$ is a hypersurface in $\mathfrak{k}^{n+1}$,
hence $C(H) = \Vp(P)$ for some prime element $P \in A$
(again by \ref{irredcodimone}).
We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$.
@ -636,8 +572,7 @@ Let $\mathfrak{l}$ be any field.
\begin{corollary}
Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be
irreducible subsets of dimensions $a$ and $b$.
If $a+ b \ge n$,
then $A \cap B \neq \emptyset$
If $a+ b \ge n$, then $A \cap B \neq \emptyset$
and every irreducible component of $A \cap B$
has dimension $\ge a + b - n$.
\end{corollary}
@ -656,17 +591,17 @@ Let $\mathfrak{l}$ be any field.
From the definition of the affine cone it follows that
$C(A \cap B) = C(A) \cap C(B)$.
We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by
\ref{conedim}.
If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an
irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for
We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by \ref{conedim}.
If $A \cap B = \emptyset$,
then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an irreducible component,
contradicting the lower bound $a + b + 1 - n > 0$ for
the dimension of irreducible components of $C(A) \cap C(B)$
(again \ref{codimintersection}).
\end{proof}
\begin{remark}[Bezout's theorem]
If $A \neq B$ are hypersurfaces of degree $a$ and $b$
in $\mathbb{P}^2$, then $A \cap B$ has $ab$ points counted by
(suitably defined) multiplicity.
in $\mathbb{P}^2$,
then $A \cap B$ has $ab$ points counted by (suitably defined) multiplicity.
\end{remark}

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