@@ -1,7 +1,7 @@
 name: typos
 on:
   pull_request:
-    branches: [ main ]
+    branches: [ master ]
     types: [ready_for_review, opened, synchronize, reopened]
   push:
     branches:

@@ -73,7 +73,7 @@ \newcommand{\mM}{\mathfrak{m}}
 
 \title{Commutative Algebra notes}
 \author{arnaucube}
-\date{}
+\date{2026}
 
 \begin{document}
 
@@ -93,44 +93,44 @@ \section{Ideals}
 
 \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 $K=A/\mM$:
@@ -233,7 +233,7 @@ \end{proof}
 
 
 
-\subsection{Lemmas, propositions and corollaries}
+\subsection{Zorn's lemma and Jacobson radicals}
 
 Let $\Sigma$ be a partially orddered set. Given subset $S \subset \Sigma$, an \emph{upper bound} of $S$ is an element $u \in \Sigma$ such that $s<u \forall s \in S$.
 
@@ -241,8 +241,8 @@ A \emph{maximal element} of $\Sigma$, is $m \in \Sigma$ such that $m<s$ does not hold for any $s \in \Sigma$.
 
 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}
-  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 $\Sigma$.
+\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.
 \end{lemma}
@@ -262,7 +262,7 @@ \begin{cor}{AM.1.5} \label{1.5}
   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 @@ \end{proof}
 
 
 
-\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 $M=0$.
@@ -687,7 +687,7 @@
 
 
 \begin{prop}{AM.2.8} \label{2.8}
-  Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vector space. Then the $x_i$ generate $M$.
+  Let $x_i ~\forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vector space. Then the $x_i$ generate $M$.
 \end{prop}
 \begin{proof}
   Let $N$ submodule $M$, generated by the $x_i$.
@@ -787,7 +787,7 @@ \item[(iii to i)] similar to the previous one.
   \end{itemize}
 
   \vspace{0.3cm}
-  TL;DR:\\
+  Overview:\\
 
 $$
 0 \longrightarrow L
@@ -938,15 +938,14 @@ $J_n$ is an ideal, since $I$ is an ideal.
 
   $J_n \subset J_{n+1}$, since for $f \in I$ also $x f \in I$.
 
-  Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.
-
+  Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.\\
   Using the assumption that $A$ is Noetherian, deduce that $J_n = J_{n+1}$ for some $n$.
 
   For each $m \leq n, ~~ J_m \subset A$ is fingen, ie.
   $$J_m = (a_{m,1}, \ldots a_{m, r_m})$$
 
   By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\
-  $\exissts$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
+  $\exists$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
 
   $$\Longrightarrow~~ \{ f_{m,j} \}_{m<n; 1 \leq j \leq r_m}$$
   the set of elements of $I$.
@@ -993,6 +992,229 @@ \end{proof}
 
 
 
+\vspace{1cm}
+\section{Finite ring extensions and Noether normalisation}
+
+\subsection{A-algebras and integral domains}
+
+\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 $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
+
+  When $A \subset B$, $B$ is an extenaion ring of $A$; denoted $\psi(A) = A' \subset B$.
+\end{defn}
+
+\begin{defn}{R.4.1}\label{R.4.1}
+  Let $B$ be an $A$-algebra.
+
+  \begin{enumerate}[i.]
+    \item $B$ is a \emph{finite} $A$-algebra (\emph{finite over $A$}) if it is finite as an $A$-module.
+    \item $y \in B$ is \emph{integral over} $A$ if $\exists$ a monic polynomial
+      $$f(Y) = Y^n + a_{n-1} Y^{n-1} + \ldots + a_0 ~\in A'[Y]$$
+      such that $f(y)=0:$
+      $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
+
+      The algebra $B$ is \emph{integral over} $A$ if $\forall~ b \in B$ is integral.
+  \end{enumerate}
+\end{defn}
+
+
+\begin{prop}{R.4.2}\label{R.4.2}
+  Let $\psi: A \longrightarrow B$ be an $A$-algebra, and $y \in B$. Three equivalent conditions:
+
+  \begin{enumerate}[i.]
+    \item $y$ is integral over $A$
+    \item subring $A'[y] \subset B$ generated by $A' = \psi(A)$ and $y$ is finite over $A$
+    \item $\exists$ an $A$-subalgebra $C \subset B$ such that $A'[y] \subset C$ and $C$ is finite over $A$
+  \end{enumerate}
+
+  Notes: $A'$ is the image of $A$ in $B$, ie. $A' = \psi(A)$.\\
+  $A'[y]$ is the smallest subring of $B$ containing both coefficients from $A$ and the element $y$.
+\end{prop}
+\begin{proof} .\\
+  \begin{itemize}
+    \item[(i to ii):]
+  since $y$ integral over $A$ $~~\Longrightarrow~$ by \ref{R.4.1} (ii), $y$ satisfies
+  $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
+
+  So any power $y^k~~ (k\geq n)$ can be expressed in terms of $\{1, y, y^2, \ldots, y^{n-1} \}$.
+
+  Thus the set $\{1, y, y^2, \ldots, y^{n-1} \}$ spans $A'[y]$ as an $A$-module.
+
+    \item[(iii to i):]
+      since $A'[y] \subset C ~~\Longrightarrow~ y \in C$\\
+      since $C$ finite over $A$, $C$ has finite generators $\{ c_1, \ldots, c_n \}$ such that $C = A \cdot c_1 + A \cdot c_2 + \ldots + A \cdot c_n$
+
+      Thus $y \cdot c_i \in C$,
+      $$y \cdot c_i = \sum_{j=1}^n a_{ij} c_j$$
+      with $a_{ij} \in A$.
+
+      By the Cayley-Hamilton theorem (\ref{2.4}),
+      $$y^n + a_{n-1} y^{n-1} + \ldots + a_1 y + a_0 = 0$$
+      Therefore, $y$ is integral (by \ref{R.4.1} (ii)).
+  \end{itemize}
+\end{proof}
+
+
+\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,\\
+      then $C$ is finite over $A$.
+    \item Finiteness of generated algebras: if $y_1, \ldots, y_m \in B$ are integral over $A$, then $A[y_1, \ldots, y_m]$ is finite over $A$.\\
+      In particular, every $f \in A[y_1, \ldots, y_m]$ is integral over $A$.
+    \item Transitivity of integrality: if $A \subset B \subset C$ with $C$ integral over $B$, and $B$ integral over $A$,\\
+      then $C$ is integral over $A$.
+    \item Integral closure as a subring: the subset
+      $$\tilde{A} = \{ y \in B ~|~ y ~\text{is integral over}~ A \} \subset B$$
+      is a subring of $B$.
+
+      Moreover, if $y \in B$ is integral over $\tilde{A}$ then $y \in \tilde{A}$, so that $\tilde{\tilde{A}} = \tilde{A}$.
+  \end{enumerate}
+\end{prop}
+\begin{proof}.\\
+  \begin{enumerate}[a.]
+    \item if $\{ \beta_1, \ldots, \beta_n \}$ generate $B$ as an $A$-module and $\{ \gamma_1, \ldots, \gamma_n \}$ generate $C$ as an $B$-module,\\
+      then the set of products $\{ \beta_i \gamma_j \}$ generates $C$ as an $A$-module.
+
+      Since there are $n \times m$ generators (ie. finite), $C$ is finite over $A$.
+
+    \item proof by induction:
+
+      base case: if $y_1$ integral over $A ~\Longrightarrow~$ it satisfies a monic polynomial.
+
+      Thus $A[y_1]$ is generated as an $A$-module by $\{1, y_1, y_1^2, \ldots, y_1^{n-1} \}$, making it a finite $A$-algebra.
+
+      inductive step: let $R_k = A[y_1, \ldots, y_k]$. Assume $R_k$ is finite over $A$.
+
+      Since $y_{k+1}$ is integral over $A~~ \Longrightarrow~$ it is also integral over $R_k$.
+
+      Thus $R_{k+1} = R_k[y_{k+1}]$ is finite over $R_k$.
+
+      Applying part (a) (transitivity of finiteness), if $R_{k+1}$ is finite over $R_k$ and $R_k$ finite over $A$, then $R_{k+1}$ is finite over $A$.
+
+      Consequence: since any $f \in A[y_1, \ldots, y_m]$ belongs to a finite $A$-algebra, $f$ must be integral over $A$ (since an element is integral iff it is contained in a finite extension).
+
+    \item let $x \in C$, since $x$ integral over $B$, it satisfies:
+      $$x^n + b_{n-1} x^{n-1} + \ldots + b_1 x + b_0 = 0,~~~~~b_i \in B$$
+
+      Let $B''=A[b_0, b_1, \ldots, b_{n-1}]$. Since each $b_i \in B$ and $B$ is integral over $A$\\
+      \hspace*{2em} $\Longrightarrow~$ each $b_i$ is integral over $A$.
+
+      Since all $b_i$ are integral over $B'$ $~~\Longrightarrow~ B'[x]$ is a finite $B'$-algebra.
+
+      By part (a) (transitivity of finiteness), $B'[x]$ is a finite $A$-algebra.
+
+      Therefore, $x$ is integral over $A$.
+
+    \item
+      \begin{enumerate}[I.]
+        \item subring:\\
+          let $x,y \in \tilde{A}$. Want to show $x+y, xy \in \tilde{A}$:
+
+          by part (b), the algebra $A[x,y]$ is finite over $A$.
+
+          Since $x+y,xy \in A[x,y]$, they are integral over $A$.
+
+          Thus $x+y, xy \in \tilde{A}$, since $\tilde{A} = \{ b \in B ~|~ b ~\text{integral over}~ A \}$.
+
+        \item idempotence\\
+          let $z \in B$ be integral over $\tilde{A}$
+          
+          we have a chain $A \subseteq \tilde{A} \subseteq \tilde{A}[x]$.
+
+          By definition, $\tilde{A}$ is integral over $A$, and $z$ is integral over $\tilde{A}$\\
+          thus by part (c), $z$ is integral over $A$.
+
+          Therefore, $z \in \tilde{A}$.
+      \end{enumerate}
+  \end{enumerate}
+\end{proof}
+
+
+\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 $B$.
+
+  If $A=\tilde{A}$, then $A$ is \emph{integrally closed} in $B$.
+
+  An integral domain $A$ is \emph{normal} if it is \emph{integrally closed in its field of fractions}, that is if
+  $$A = \tilde{A} \subset K = Frac(A)$$
+
+  For any integral domain $A$, the integral closure of $A$ in its field of fractions $K=Frac(A)$ is also called the \emph{normalization} of $A$.
+\end{defn}
+
+
+\vspace{0.5cm}
+
+\subsection{Noether normalization}
+
+\begin{defn}{4.6}{Algebraically independent.}
+  $y_1, \ldots, y_n \in A$ 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 $K$.
+\end{defn}
+
+Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some finite set $y_1, \ldots, y_n$.
+
+\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 $F(y_1, \ldots, y_n)=0$.
+
+  Then $\exists~ y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
+  $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~~\text{and}~~ A=A^*[y_n]$$
+\end{lemma}
+\begin{proof}
+  (todo)
+\end{proof}
+
+\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
+  \begin{enumerate}[i.]
+    \item $z_1, \ldots, z_m$ are algebraically independent over $K$
+    \item $A$ is finite over $B=K[z_1, \ldots, z_m]$
+  \end{enumerate}
+
+  That is, a fingen extension $K \subset A$ can be written as a composite
+  $$K \subset B = K[z_1, \ldots, z_m] \subset A$$
+  where $K \subset B$ is a polynomial extension, and $B \subset A$ is finite.
+\end{thm}
+\begin{proof}
+  induction on $n$.
+
+  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 $K$,
+      $$\exists 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
+  \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.
+
+  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
+  $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~\text{and}~ A=A^*[y_n]$$
+
+  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}
+
+
+\begin{eg}{ }
+  $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
+
+  This corresponds to the fact that the hyperbola $XY=1$ has the line $X=0$ as an asymptotic line (so that its projection to the $X$-axis misses a root over $X=0$).
+
+  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.
+\end{eg}
+
 
 \newpage
 
@@ -1353,7 +1575,7 @@ \begin{proof}
 \begin{enumerate}[i.]
   \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
     $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
-  \item Take the canoncial homomorphism
+  \item Take the canonical homomorphism
     $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
     by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
 

@@ -517,12 +517,12 @@ \item show $HN \subseteq G$:\\
       Note that $HN = \{ hn : h\in H, n\in N \}$. Let $h_1 n_1, h_2 n_2 \in HN$.
 
       Since $N$ normal $\Longrightarrow~ h_2^{-1} n_1 h_2 \in N$, so
-      $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \in HN$$
+      $$(h_1 n_1)(h_2 n_2) = h_1 h_2 (h_2^{-1} n_1 h_2) \cdot n_2 \in HN$$
 
       [Recall: since $N \triangleleft G$, $gN=Ng ~\forall g \in G$ $\Longrightarrow gn=n'g$ for some $n' \in N$.]
 
       To see that $(hn)^{-1} \in HN$:\\
-      since $(hn)^{-1} = n^{-1} h^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
+      since $(hn)^{-1} = h^{-1} n^{-1} = h^{-1} (h n^{-1} h^{-1})$, thus $(hn)^{-1} \in HN$.
 
       Thus $HN \subseteq G$.