diff --git a/.github/workflows/typos.toml b/.github/workflows/typos.toml
index de9c696..acadd66 100644
--- a/.github/workflows/typos.toml
+++ b/.github/workflows/typos.toml
@@ -1,3 +1,7 @@
+# usage:
+# install `typos`: https://github.com/crate-ci/typos
+# run: typos -c .github/workflows/typos.toml
+
 [default.extend-words]
 iddeal = "ideal"
 iddeals = "ideals"
diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf
index 034d87c..a45cec5 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 4a7ec0f..84686ae 100644
--- a/commutative-algebra-notes.tex
+++ b/commutative-algebra-notes.tex
@@ -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.
@@ -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 $K=A/\mM$:
@@ -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 $Jac(A)$.
@@ -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 $N=M / L$.
 
@@ -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
   $$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
@@ -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 $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
@@ -1132,8 +1135,40 @@ 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 $y_n$ integral over $A$ then $A[y_n]$ 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 $y_n$ is integral over $A$. By definition $\exists~~ f \in A[T]$, with $f$ monic, such that $f(y_n)=0$.
+
+  Let $deg(f)=d$, so that for $f(y_n)=0$ we have
+  $$y_n^d + a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0 ~~~~ a_i \in A$$
+
+  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)
+    \label{eq:yn}
+  \end{equation}
+  Thus $y_n^d$ can be written using lower powers of $y_n$ with coefficients in $A$.
+
+  \vspace{0.5cm}
+  Consider any element $p \in A[y_n]$, $p = c_m y_n^m + c_{m-1} y_n^{m-1} + \ldots + c_0$.
+
+  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.
+
+  Repeating this process, can reduce any power of $y_n$ down to a linear combination of $\{1, y_n, y_n^2, \ldots, y_n^{d-1} \}$.
+
+  Thus every elemen in $A[y_n]$ can be expressed as
+  $$\lambda_{d-1} y_n^{d-1} + \ldots + \lambda_2 y_n^2 + \lambda_1 y_n + \lambda_0 \cdot 1~~~~ \lambda_i \in A$$
+
+  Henceforth, the set $\{1, y_n, y_n^2, \ldots, y_n^{d-1} \}$ generates $A[y_n]$ as a finite $A$-module.
+\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 $B$.
 
@@ -1150,7 +1185,7 @@ 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]
   $y_1, \ldots, y_n \in A$ 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 +1193,8 @@ 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 $F(y_1, \ldots, y_n)=0$.
 
@@ -1165,9 +1202,30 @@ 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}
 
+
+\vspace{0.5cm}
 \begin{thm}{R.4.6}{Noether normalization lemma.} \label{noether-normalization}
   Let $K$ a field, $A$ a fingen $K$-algebra.
 
@@ -1182,29 +1240,48 @@ 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 $A$).
 
   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 $A \cong K[y_1, \ldots, y_n]$, 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
+  $~\longrightarrow~$ 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 $y_i = y_i^* + y_i^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.
+
+  Use those new variables at $f(y_1, \ldots, y_n)=0$:
+  $$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_n^* + y_n^{r_n}, y_n) = 0$$
+
+  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.
+
   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 $A^*$.\\
+  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}
 
 
+\vspace{0.5cm}
 \begin{eg}{ }
   $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
 
@@ -1212,7 +1289,7 @@ 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}
 
 
diff --git a/typos.toml b/typos.toml
deleted file mode 100644
index 0788b15..0000000
--- a/typos.toml
+++ /dev/null
@@ -1,7 +0,0 @@
-# usage:
-# install `typos`: https://github.com/crate-ci/typos
-# run: typos --config typos.toml
-
-[default.extend-words]
-groth = "groth"
-pinter = "pinter"