josia-notes/s21-algebra-1
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

fixed some wrong linebreaks (introduced by misuse of latexindent?)

Browse Source
This commit is contained in:
Josia Pietsch 2023-07-31 02:25:22 +02:00
parent 13a7e012a9
commit 487aa3c400
Signed by: josia
GPG Key ID: E70B571D66986A2D
4 changed files with 927 additions and 1123 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

437
inputs/finiteness_conditions.tex
View File

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

9
inputs/preamble.tex
View File

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

543
inputs/projective_spaces.tex
View File

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

1061
inputs/varieties.tex
View File

File diff suppressed because it is too large Load Diff