diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf
index bae5bb7..1d9c7fe 100644
Binary files a/commutative-algebra-notes.pdf and b/commutative-algebra-notes.pdf differ
diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex
index 42a92cd..34d1460 100644
--- a/commutative-algebra-notes.tex
+++ b/commutative-algebra-notes.tex
@@ -73,7 +73,7 @@
 
 \title{Commutative Algebra notes}
 \author{arnaucube}
-\date{}
+\date{2026}
 
 \begin{document}
 
@@ -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 $K=A/\mM$:
@@ -233,7 +233,7 @@
 
 
 
-\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
 
 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 @@ 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 $M=0$.
@@ -687,7 +687,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
 
 
 \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 @@ Properties:
   \end{itemize}
 
   \vspace{0.3cm}
-  TL;DR:\\
+  Overview:\\
 
 $$
 0 \longrightarrow L
@@ -938,8 +938,7 @@ As in with rings, it is equivalent to say that
 
   $J_n \subset J_{n+1}$, 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.
@@ -991,8 +990,87 @@ As in with rings, it is equivalent to say that
   $A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
 \end{proof}
 
+\vspace{1cm}
+\section{Finite ring extensions and Noether normalisation}
+
+\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{ennumerate}
+
+  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}
+\end{proof}
 
 \newpage
 
@@ -1353,7 +1431,7 @@ Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bi
 \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 $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.