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Noether normalization & weak nullstellensatz thm (#4)

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* add lemma 4.6 proof, polish Noether normalization lemma's proof, add aux lemma on integrality implies finiteness

* add weak nullstellensatz theorem and proof
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commutative-algebra-notes.tex
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@@ -93,44 +93,44 @@
\subsection{Definitions}
\begin{defn}{}{ideal}
\begin{defn}{}[ideal]
$I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\
\hspace*{2em} ie. $I$ absorbs products in $R$.
\end{defn}
\begin{defn}{}{prime ideal}
\begin{defn}{}[prime ideal]
if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$.
\end{defn}
\begin{defn}{}{principal ideal}
\begin{defn}{}[principal ideal]
generated by a single element, $(a)$.
$(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
\end{defn}
\begin{defn}{}{maximal ideal}
\begin{defn}{}[maximal ideal]
$\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$.
\end{defn}
\begin{defn}{}{unit}
\begin{defn}{}[unit]
$x \in A$ such that $xy=1$ for some $y \in A$. ie. element \emph{which divides 1}.
\end{defn}
\begin{defn}{}{zerodivisor}
\begin{defn}{}[zerodivisor]
$x \in A$ such that $\exists 0 \neq y \in A$ such that $xy=0 \in A$. ie. $x$ \emph{divides 0}..
If a ring does not have zerodivisors is an integral domain.
\end{defn}
\begin{defn}{}{prime spectrum - $Spec(A)$}
\begin{defn}{}[prime spectrum - $Spec(A)$]
set of prime ideals of $A$. ie.
$$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
\end{defn}
\begin{defn}{}{integral domain}
\begin{defn}{}[integral domain]
Ring in which the product of any two nonzero elements is nonzero.
ie. no zerodivisors.
@@ -140,15 +140,15 @@
Every field is an integral domain, not the converse.
\end{defn}
\begin{defn}{}{principal ideal domain - PID}
\begin{defn}{}[principal ideal domain - PID]
integral domain in which every ideal is principal. ie.
ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
\end{defn}
\begin{defn}{}{nilpotent}
\begin{defn}{}[nilpotent]
$a \in A$ such that $a^n=0$ for some $n>0$.
\end{defn}
\begin{defn}{}{nilrad A}
\begin{defn}{}[nilrad A]
set of all nilpotent elements of $A$; is an ideal of $A$.
if $nilrad A = 0 ~\Longrightarrow$ $A$ has no nonzero nilpotents.
@@ -157,11 +157,11 @@
$$nilrad A = \bigcap_{P \in Spec(A)} P$$
\end{defn}
\begin{defn}{}{idempotent}
\begin{defn}{}[idempotent]
$e \in A$ such that $e^2=e$.
\end{defn}
\begin{defn}{}{radical of an ideal}
\begin{defn}{}[radical of an ideal]
$$rad I = \{ f \in A | f^n \in I~ \text{for some} n \}$$
$rad I$ is an ideal.
@@ -171,7 +171,7 @@
$rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$
\end{defn}
\begin{defn}{}{local ring}
\begin{defn}{}[local ring]
A \emph{local ring} has a unique maximal ideal.
Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
@@ -241,7 +241,7 @@ A \emph{maximal element} of $\Sigma$, is $m \in \Sigma$ such that $m<s$ does not
A subset $S \subset \Sigma$ is \emph{totally ordered} if for every pair $s_1,s_2 \in S$, either $s_1 \leq s_2$ or $s_2 \leq s_1$.
\begin{lemma}{R.1.7}{Zorn's lemma.} \label{zorn}
\begin{lemma}{R.1.7}[Zorn's lemma] \label{zorn}
Suppose $\Sigma$ a nonempty partially ordered set (ie. we are given a relation $x \leq y$ on $\Sigma$), and that any totally ordered subset $S \subset \Sigma$ has an upper bound in $\Sigma$.
Then $\Sigma$ has a maximal element.
@@ -262,7 +262,7 @@ A subset $S \subset \Sigma$ is \emph{totally ordered} if for every pair $s_1,s_2
Every non-unit of $A$ is contained in a maximal ideal.
\end{cor}
\begin{defn}{}{Jacobson radical}
\begin{defn}{}[Jacobson radical]
The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$.
Denoted $Jac(A)$.
@@ -591,7 +591,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
\begin{prop}{AM.2.6}{Nakayama's lemma.} \label{2.6}
\begin{prop}{AM.2.6}[Nakayama's lemma] \label{2.6}
Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
Then $\aA M = M$ implies $M=0$.
@@ -698,7 +698,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
\subsection{Sequences}
\begin{defn}{R.2.9.a}{Exact Sequence}
\begin{defn}{R.2.9.a}[Exact Sequence]
Let a sequence of homomorphisms
$$L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N$$
It is \emph{exact} at $M$ if $im(\alpha)=ker(\beta)$.
@@ -707,7 +707,7 @@ ie. $\beta \circ \alpha = 0$ and $\alpha$ maps surjectively to
$ker(\beta)$.
\end{defn}
\begin{defn}{R.2.9.b}{Short Exact Sequence (s.e.s.)} \label{2.9}
\begin{defn}{R.2.9.b}[Short Exact Sequence (s.e.s.)] \label{2.9}
$$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$$
is exact $\Longleftrightarrow~ L \subset M$ and $N=M / L$.
@@ -720,7 +720,7 @@ Properties:
\end{itemize}
\end{defn}
\begin{prop}{R.2.10}{Split exact sequence} \label{2.10}
\begin{prop}{R.2.10}[Split exact sequence] \label{2.10}
For the previous s.e.s., 3 equivalent conditions:
\begin{enumerate}[i.]
\item $\exists$ isomorphism $M \cong L \oplus N$, with
@@ -815,7 +815,7 @@ $$
\section{Noetherian rings (and modules)}
\begin{defn}{}{Ascending Chain Condition}
\begin{defn}{}[Ascending Chain Condition]
A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
$$s_1 \leq s_2 \leq \ldots \leq s_k \leq \ldots$$
eventually breaks off, that is, $s_k = s_{k+1} = \ldots$ for some $k$.
@@ -824,7 +824,8 @@ $$
$\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \Sigma$ has a maximal element.\\
\hspace*{2em} if $\empty \neq S \subset \Sigma$ does not have a maximal element, choose $s_1 \in S$, and for each $s_k$, an element $s_{k+1}$ with $s_k < s_{k+1}$, thus contradicting the a.c.c.
\begin{defn}{R.3.2}{Noetherian ring}
\subsection{Noetherian rings and modules}
\begin{defn}{R.3.2}[Noetherian ring]\\
Let $A$ a ring; 3 equivalent conditions:
\begin{enumerate}[i.]
\item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
@@ -839,7 +840,7 @@ $\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \
TODO
\end{proof}
\begin{defn}{R.3.4.D}{Noetherian modules}
\begin{defn}{R.3.4.D}[Noetherian modules]\\
An $A$-module $M$ is Noetherian if the submoles of $M$ have the a.c.c.,\\
that is, ay increasing chain
$$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
@@ -888,7 +889,7 @@ As in with rings, it is equivalent to say that
\begin{cor}{R.3.5}{Properties of Noetherian modules.}\label{R.3.5}
\begin{cor}{R.3.5}[Properties of Noetherian modules]\label{R.3.5}
\begin{enumerate}[i.]
\item if $\forall i \in [r],~~M_i$ are Noetherian modules, then
$\bigoplus_{i=1}^r M_i$ is Noetherian.
@@ -924,8 +925,9 @@ As in with rings, it is equivalent to say that
\vspace{0.5cm}
\begin{thm}{R.3.6}{Hilbert basis theorem} \label{hilbert-basis}
if $A$ a Noetherian ring, then so is the polynomial ring $A[x]$.
\subsection{Hilbert basis}
\begin{thm}{R.3.6}[Hilbert basis theorem] \label{hilbert-basis}
If $A$ a Noetherian ring, then so is the polynomial ring $A[x]$.
\end{thm}
\begin{proof}
Prove that any ideal $I \subset A[x]$ is fingen.
@@ -973,6 +975,7 @@ As in with rings, it is equivalent to say that
Thus, any ideal of $A[x]$ is finitely generated.
\end{proof}
\vspace{0.5cm}
\begin{cor}{R.3.6.C}
if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a fingen extension ring of $\psi(A)$, then $B$ is Noetherian.
@@ -993,11 +996,11 @@ As in with rings, it is equivalent to say that
\vspace{1cm}
\section{Finite ring extensions and Noether normalisation}
\section{Finite ring extensions and Noether normalization}
\subsection{A-algebras and integral domains}
\begin{defn}{}{A-algebra.}
\begin{defn}{}[A-algebra]
An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
$B$ is an $A$-module with multiplication defined by $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
@@ -1057,7 +1060,7 @@ As in with rings, it is equivalent to say that
\end{proof}
\begin{prop}{R.4.3}{Tower Laws.}\label{R.4.3}
\begin{prop}{R.4.3}[Tower Laws]\label{R.4.3}
\\Let $B$ be an $A$-algebra.
\begin{enumerate}[a.]
\item Transitivity of finiteness: if $A \subset B \subset C$ are extension rings such that $C$ is a finite $B$-algebra and $B$ a finite $A$-algebra,\\
@@ -1132,8 +1135,44 @@ As in with rings, it is equivalent to say that
\end{enumerate}
\end{proof}
\begin{lemma}{4.3.Aux}[Integrality implies finiteness] \label{integral-implies-finite}
If $y$ integral over $A$ then $A[y]$ is finite over $A$.
\begin{defn}{4.4}{Integral closure.}
This extends on point (b) from the previous proposition \ref{R.4.3}.
\end{lemma}
\begin{proof}
Suppose $y$ is integral over $A$. By definition $\exists~~ f \in A[T]$, with
$f$ monic, such that $f(y)=0$.
Let $deg(f)=d$, so that for $f(y)=0$ we have
$$y^d + a_{d-1} y^{d-1} + \ldots + a_1 y + a_0 = 0 ~~~~ a_i \in A$$
Since it is monic (leading coefficient is $1$), we can rearrange it to isolate the highest power:
\begin{equation}
y^d = -(a_{d-1} y^{d-1} + \ldots + a_1 y + a_0)
\label{eq:yn}
\end{equation}
Thus $y^d$ can be written using lower powers of $y$ with coefficients in $A$.
\vspace{0.5cm}
Consider any element $p \in A[y]$, $p = c_m y^m + c_{m-1} y^{m-1} + \ldots + c_0$.
if $m<d$, leave it as it is.\\
if $m \geq d$, use the monic equation \eqref{eq:yn} to replace $y^d$ with lower powers.
Repeating this process, can reduce any power of $y$ down to a linear
combination of $\{1, y, y^2, \ldots, y^{d-1} \}$.
Thus every element in $A[y]$ can be expressed as
$$\lambda_{d-1} y^{d-1} + \ldots + \lambda_2 y^2 + \lambda_1 y + \lambda_0 \cdot 1~~~~ \lambda_i \in A$$
Henceforth, the set $\{1, y, y^2, \ldots, y^{d-1} \}$ generates $A[y]$ as a finite $A$-module.
\end{proof}
\vspace{0.5cm}
\begin{defn}{4.4}[Integral closure]
Given the ring $\tilde{A}$ from \ref{R.4.3}.(d), ie. $\tilde{A} = \{ y \in B ~|~ y ~\text{integral over}~ A \} \subset B$,
$\tilde{A}$ is the \emph{integral closure} of $A$ in $B$.
@@ -1150,7 +1189,7 @@ As in with rings, it is equivalent to say that
\subsection{Noether normalization}
\begin{defn}{4.6}{Algebraically independent.}
\begin{defn}{4.6}[Algebraically independent] \label{R.4.6.D}
$y_1, \ldots, y_n \in A$ are \emph{algebraically independent} over $K$ if the natural surjection $K[Y_1, \ldots, Y_n] \longrightarrow K[y_1, \ldots, y_n]$ is an isomorphism.
$\Longrightarrow~~ \nexists~ F(y_1, \ldots, y_n)=0$ ($F$ nonzero) with coefficients in $K$.
@@ -1158,6 +1197,8 @@ As in with rings, it is equivalent to say that
Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some finite set $y_1, \ldots, y_n$.
\vspace{0.5cm}
\begin{lemma}{R.4.6.L} \label{R.4.6.L}
Let $A = K[y_1, \ldots, y_n]$ and $0 \neq F \in K[Y_1, \ldots, Y_n]$ such that $F(y_1, \ldots, y_n)=0$.
@@ -1165,10 +1206,31 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
$$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~~\text{and}~~ A=A^*[y_n]$$
\end{lemma}
\begin{proof}
(todo)
Set $y_i^* = y_i - y_n^{r_i}$ for $i \in [n-1]$ and $r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.\\
\hspace*{3em}(ie. $y_i=y_i^* + y_n^{r_i}$)
Define $G \in A$ by
$$G(y_1^*, \ldots, y_{n-1}^*, y_n) = F(y_i^* + y_n^{r_i}, y_n)=0$$
viewed as a relation for $y_n$ over $K[y_1^*, \ldots, y_{n-1}^*]$.
Since $F$ is a polynomial in $y_1, \ldots, y_{n-1}^*$, can write it as a sum of monomials
$$F= \sum_m a_m y^m = \sum_m a_m \prod y_i^{m_i}$$
where $m=(m_1, \ldots, m_n)$ and each $a_m \neq 0$.
Therefore,
$$G= \sum a_m \prod (y_i^* + y_n^{r_i})^{m_i}$$
which when expanding out, each summand $a_m \prod (y_i^* + y_n^{r_i})^{m_i}$ has a unique term of highest order in $y_n$, namely $a_m y_n^{(\sum r_i m_i)}$.
Suppose we can arrange so that
$$m \neq m' ~\Longrightarrow~ \sum r_i m_i \neq \sum r_i m_i'$$
Then $max \{ \sum r_i m_i ~|~ m ~\text{s.th.}~ a_m \neq 0 \}$ is achieved in only one summand, so that here is no cancellation; thus the highest order term in $G$ is $a_m y_n^{(\sum r_i m_i)}$ (ie. $a_m$ times a pure power of $y_n$).
\end{proof}
\begin{thm}{R.4.6}{Noether normalization lemma.} \label{noether-normalization}
\vspace{0.5cm}
\begin{thm}{R.4.6}[Noether normalization lemma] \label{noether-normalization}
Let $K$ a field, $A$ a fingen $K$-algebra.
Then $\exists~ z_1, \ldots, z_m \in A$ such that
@@ -1182,29 +1244,52 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
where $K \subset B$ is a polynomial extension, and $B \subset A$ is finite.
\end{thm}
\begin{proof}
induction on $n$.
% (by induction on the number of generators ($n$) of $A$).
Let $y_1, \ldots, y_n$ be generators of $A = K[y_1, \ldots, y_n]$.
if $n=0$, nothing to prove since $A$ is generated by $0$ elements $~\Longrightarrow~ A=K$, and $K$ is finite.
if $n>0$ we have two cases:
\begin{itemize}
\item $y_1, \ldots, y_n$ are algebraically independent over $K$, then $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself.
\item $y_1, \ldots, y_n$ are algebraically dependent over $K$,
$$\exists 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
\item[-] $y_1, \ldots, y_n$ are algebraically independent over $K$, then by
definition \ref{R.4.6.D} $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself, with $m=n$.
\item[-] $y_1, \ldots, y_n$ are algebraically dependent over $K$,
$$\exists~ 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
\end{itemize}
Goal: is to change variables so that $f$ becomes monic in one of the variables; this allows to express one generator as an integral element over the others.
Want $f$ to be \emph{monic}, so that $y_n$ is integral over new defined variables $y_1^*, \ldots, y_{n-1}^*$. In other words, want some polynomial like
$$y_n^d+ a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0~~~~~~~a_i \in K[y_1^*,
\ldots, y_{n-1}^*]$$
\hspace*{2em} ie. monic, so that by definition (\ref{R.4.1}), $y_n$ is
integral over $K[y_1^*, \ldots, y_{n-1}^*]$.
Following from Lemma \ref{R.4.6.L}, define new variables $y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
$$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~\text{and}~ A=A^*[y_n]$$
$~\longrightarrow~$ Change variables so that $f$ becomes monic in one of the variables ($y_n$); this allows to express one generator ($y_n$) as an integral element over the others.
\vspace{0.3cm}
Following from Lemma \ref{R.4.6.L}, define the new variables $y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
$$A^* = K[y^*_1, \ldots, y^*_{n-1}], ~~\text{and}~ A=A^*[y_n]$$
Setting $y_i^* = y_i - y_n^{r_i}$, so that $y_i = y_i^* + y_n^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.
Use those new variables at $f(y_1, \ldots, y_n)=0$:
$$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_{n-1}^* + y_n^{r_{n-1}}, y_n) = 0$$
Then the highest power of $y_n$ in each term of $f$ will look like $y_n^{(\sum a_i r_i)}$, and with $r_i$ growing fast enough we ensure that each monomial in $f$ produces a unique power of $y_n$.
Then we have $c \cdot y_n^D + \text{(terms with lower powers of $y_n$)} = 0$ with $c \in K \setminus \{0\}$. So that dividing by $c$ we get the shape $y_n^D + \ldots =0$, thus $y_n$ is integral over $A^* = K[y_1^*, \ldots, y_{n-1}^*]$.
\vspace{0.3cm}
Induction:\\
Since $y_n$ integral over $A^* ~~\Longrightarrow~ A=A^*[y_n]$ is finite over $A^*=K[y_1^*, \ldots, y_{n-1}^*]$ (by \ref{integral-implies-finite}).
By inductive hypothesis on $A^*,~~ \exists~ z_1, \ldots, z_m \in A^*$ algebraically independent over $K$ and with $A^*$ finite over $B=K[z_1, \ldots, z_m]$.
Since $y_n$ integral over $A^* ~~\Longrightarrow~ A^*[y_n]$ is finite over $A^*$.\\
Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required.
\end{proof}
\vspace{0.5cm}
\begin{eg}{ }
$A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
@@ -1212,9 +1297,72 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
Take $X' = X- \epsilon Y$ as the element of $A$ instead of $X$; then the relation becomes $(X' + \epsilon Y) Y=1$, monic in $Y$ if $\epsilon \neq 0$.
This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymtotic line.
This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymptotic line.
\end{eg}
\vspace{1cm}
\subsection{Weak Nullstellensatz}
\begin{prop}{R.4.9}\label{R.4.9}
let $A \subset B$ be an integral extension of integral domain,\\
then $A$ is a field $\Longleftrightarrow~ B$ is a field.
\end{prop}
\begin{proof}
$\Longrightarrow$:\\
let $0 \neq x \in B$, then $\exists~~ x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0 ~~~~ a_i \in A$, monic.
Since $A$ is a field, $\exists$ inverse, observe that:
\begin{align*}
x^n &+ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\\
x(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = - a_0\\
-a_0^{-1}(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = x^{-1} \in B
\end{align*}
thus there exists inverse in $B$, so $B$ is a field too.
$\Longleftarrow$:\\
if $B$ is a field and $0 \neq x \in A$, then $x^{-1} \in B$, so $x^{-1}$ is integral over $A$.
So there is a relation of the form
$$(x^{-1})^n + a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 =0$$
Therefore
\begin{align*}
(x^{-1})^n &+ a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 = 0\\
(x^{-1})^n &= -a_{n-1} (x^{-1})^{n-1} - \ldots - a_0\\
x^{-n} &= -a_{n-1} x^{-n+1} - \ldots - a_0 ~~\text{(mult by $x^{n-1}$)}\\
x^{-n+(n-1)} &= -a_{n-1} x^{-n+1+(n-1)} - \ldots - a_0 x^{n-1}\\
x^{-1} &= -a_{n-1} - \ldots - a_0 x^{n-1} \in A
\end{align*}
thus there exists inverse in $A$, so $A$ is a field too.
\end{proof}
\begin{thm}{R.4.10}[Weak Nullstellensatz]
let $k$ a field, $K$ a $k$-algebra which
\begin{enumerate}
\item is finitely generated as a $k$-algebra
\item is a field
\end{enumerate}
Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field
extension. That is, $[K:k] < \infty$.
\end{thm}
\begin{proof}
by Noether normalization \ref{noether-normalization}, $\exists~ z_1, \ldots,
z_m \in K$ which are algebraically independent, and such that $K$ is finite
over $A=k[z_1, \ldots, z_m]$.
\\
Now we're at the situation of \ref{R.4.9}:
$A \subset K$ is integral, $K$ is a field $~~\Longrightarrow~$ therefore $A$ is a field.
Since $z_1, \ldots, z_m \in K$ are algebraically independent,\\
\hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$, and $K$ is finite over $k$.
\end{proof}
\newpage