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add weak nullstellensatz theorem and proof

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.github/workflows/typos.toml vendored
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@@ -16,4 +16,6 @@ nd = "nd"
# names
Strang = "Strang"
Pinter = "Pinter"
Bootle = "Bootle"
Groth = "Groth"

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commutative-algebra-notes.pdf
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commutative-algebra-notes.tex
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@@ -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.
@@ -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.
@@ -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 $M=0$.
@@ -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
@@ -889,7 +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.
@@ -926,7 +926,7 @@ As in with rings, it is equivalent to say that
\vspace{0.5cm}
\subsection{Hilbert basis}
\begin{thm}{R.3.6}{Hilbert basis theorem.} \label{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}
@@ -1060,7 +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,\\
@@ -1136,38 +1136,42 @@ As in with rings, it is equivalent to say that
\end{proof}
\begin{lemma}{4.3.Aux}[Integrality implies finiteness] \label{integral-implies-finite}
If $y_n$ integral over $A$ then $A[y_n]$ is finite over $A$
If $y$ integral over $A$ then $A[y]$ is finite over $A$.
This extends on point (b) from the previous proposition \ref{R.4.3}:
This extends on point (b) from the previous proposition \ref{R.4.3}.
\end{lemma}
\begin{proof}
Suppose $y_n$ is integral over $A$. By definition $\exists~~ f \in A[T]$, with $f$ monic, such that $f(y_n)=0$.
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 $f(y_n)=0$ we have
$$y_n^d + a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0 ~~~~ a_i \in A$$
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_n^d = -(a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0)
y^d = -(a_{d-1} y^{d-1} + \ldots + a_1 y + a_0)
\label{eq:yn}
\end{equation}
Thus $y_n^d$ can be written using lower powers of $y_n$ with coefficients in $A$.
Thus $y^d$ can be written using lower powers of $y$ with coefficients in $A$.
\vspace{0.5cm}
Consider any element $p \in A[y_n]$, $p = c_m y_n^m + c_{m-1} y_n^{m-1} + \ldots + c_0$.
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 $y_n^d$ with lower powers.
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 $y_n$ down to a linear combination of $\{1, y_n, y_n^2, \ldots, y_n^{d-1} \}$.
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 elemen in $A[y_n]$ can be expressed as
$$\lambda_{d-1} y_n^{d-1} + \ldots + \lambda_2 y_n^2 + \lambda_1 y_n + \lambda_0 \cdot 1~~~~ \lambda_i \in A$$
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 $\{1, y_n, y_n^2, \ldots, y_n^{d-1} \}$ generates $A[y_n]$ as a finite $A$-module.
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 $B$.
@@ -1185,7 +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 $K$.
@@ -1226,7 +1230,7 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
\vspace{0.5cm}
\begin{thm}{R.4.6}{Noether normalization lemma.} \label{noether-normalization}
\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
@@ -1240,43 +1244,47 @@ 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}
(by induction on the number of generators ($n$) of $A$).
% (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, with $m=n$.
\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}
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 $K[y_1, \ldots, y_{n-1}]$.
$$y_n^d+ a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0~~~~~~~a_i \in K[y_1^*,
\ldots, y_{n-1}^*]$$
\hspace*{2em} ie. monic, so that by definition (\ref{R.4.1}), $y_n$ is
integral over $K[y_1^*, \ldots, y_{n-1}^*]$.
$~\longrightarrow~$ 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
$$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~\text{and}~ A=A^*[y_n]$$
$$A^* = K[y^*_1, \ldots, y^*_{n-1}], ~~\text{and}~ A=A^*[y_n]$$
Setting $y_i^* = y_i - y_n^{r_i}$, so that $y_i = y_i^* + y_i^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.
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 $f(y_1, \ldots, y_n)=0$:
$$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_n^* + y_n^{r_n}, y_n) = 0$$
$$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_{n-1}^* + y_n^{r_{n-1}}, y_n) = 0$$
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}
Now, $A$ is a finite module over $A^*=K[y_1^*, \ldots, y_{n-1}^*]$, so that $A^*$ is generated by $n-1$ elements.
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 $A^*$ (by \ref{integral-implies-finite}).\\
Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required.
\end{proof}
@@ -1292,6 +1300,69 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
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