@@ -104,10 +104,6 @@ \end{enumerate}
     we call $H$ a \emph{subgroup} of $G$. The operation of $H$ is the same as the operation of $G$.
 \end{definition}
 
-\begin{theorem}
-    Every subgroup of a cyclic group is cyclic.
-\end{theorem}
-
 \section{Functions}
 \begin{definition}[Function]
     If $A$ and $B$ are sets, then a function from $A$ to $B$ is a rule which to every element $x$ in $A$ assigns a unique element $y$ in $B$.

@@ -1361,7 +1361,24 @@ It's a fingen $k$-algebra, so by Zariski's lemma, $k[X_1, \ldots, X_n]/m$ is algebraic over $k$.
 
 Since $k[X_1, \ldots, X_n]/m$ is algebraically closed, it must equal $k$; ie.
 $$\frac{k[X_1, \ldots, X_n]}{m}=k$$
-Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown at \ref{5.2}).
+
+
+Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, we have the natural map
+$$\psi: k \longrightarrow \frac{k[X_1, \ldots, X_n]}{m}$$
+so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $k[X_1, \ldots, X_n]/m$ must be an element of $k$).\\
+Then, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
+Thus,
+$$a_i = X_i +m$$
+$$\Longrightarrow X_i - a_i \in m$$
+
+  % So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
+  % $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
+  % $$\Longrightarrow x_i - a_i \in m$$
+
+The ideal generated by these terms is a subset of $m$:
+$$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
+
+Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown also at \ref{5.2}).
 
 \vspace{0.4cm}
 Then:
@@ -1410,6 +1427,7 @@ \vspace{1cm}
 
 \section{Nullstellensatz}
 
+\subsection{Corollary from Weak Nullstellensatz}
 Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue field $K=k[X_1, \ldots, X_n]/m$ satisfies the Zariski's lemma (\ref{zariski}), thus $K$ is a finite algebraic extension of $k$.
 
 \vspace{0.3cm}
@@ -1431,16 +1449,19 @@ let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.
 
   By fundamental property of maximal ideals, $K=A/m$ is a field.
 
-  Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $x_i' = x_i +m$.
+  Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $\psi(x_i) = x_i' = x_i +m$ (where $\psi: k \longrightarrow A/m$).
 
   By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$.
 
   Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\
   \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$.
   
-  So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
-  $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
-  $$\Longrightarrow x_i - a_i \in m$$
+  \vspace{0.3cm}
+  Recall that since $A$ is a $k$-algebra, we have the natural map $\psi: k \longrightarrow A/m$, so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $A/m$ must be an element of $k$).\\
+  Also, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
+  Thus,
+  $$a_i = X_i +m$$
+  $$\Longrightarrow X_i - a_i \in m$$
 
   The ideal generated by these terms is a subset of $m$:
   $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
@@ -1452,12 +1473,15 @@
   \vspace{0.3cm}
   Let
   \begin{align*}
-    \psi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
-    \psi: x_i &\longmapsto a_i
+    \phi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
+    \phi: x_i &\longmapsto a_i
   \end{align*}
 
-  Since $\psi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
-  $$\psi(f(X_1, \ldots, X_n)) = f(\psi(x_1), \ldots, \psi(x_n))= f(a_1, \ldots, a_n)$$
+  Which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
+  \vspace{0.2cm}
+
+  Since $\phi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
+  $$\phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n))= f(a_1, \ldots, a_n)$$
 
   Thus there is a one-to-one correspondence:
 
@@ -1677,7 +1701,6 @@ \end{align*}
 
   Therefore,
   $$Spec~ k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$
-\end{corollary}
 \end{cor}
 
 \begin{prop}{5.8} \label{5.8}
@@ -1692,7 +1715,7 @@ \begin{proof}
   By definition, $Spec~ A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$.
 
   By Corollary \ref{cor.5.8}:
-    $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
+    $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} \longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
 
   About varieties:
   \begin{itemize}
@@ -1726,7 +1749,7 @@
   $\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$.
 
 
-  Now, $J \susbset \mathfrak{P} = I(X)$;\\
+  Now, $J \subset \mathfrak{P} = I(X)$;\\
   by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so
   $$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$
   $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$.
@@ -1736,6 +1759,181 @@ $$\mathfrak{P} \in Spec~ A \longleftrightarrow X \subseteq V(J)$$
   where $X = V(\mathfrak{P})$.
 \end{proof}
 
+\subsection{Zariski, Weak Nullstellensatz, Nullstllensatz}
+This subsection proves again Zariski's lemma, Weak Nullstellensatz and Nullstellensatz, in a more interconnected way than the proofs described previously.
+
+\begin{lemma}{}[Zariski's lemma]
+  $k$ a field, $K$ a fingen $k$-algebra and a field.
+
+  Then $K$ is a finite algebraic extension of $k$.
+
+  If $k$ algebraically closed, then $K=k$.
+\end{lemma}
+\begin{proof}
+  $K$ a fingen $k$-algebra and a field $\Longleftrightarrow$ $K=k[X_1, \ldots, X_n]$ a field.\\
+  Want to show that $K$ is algebraic over $k$.
+
+  Since $K$ is a fingen $k$-algebra, by Noether normalization \ref{noether-normalization},\\
+  $\exists~ \{y_1, ldots, y_r\} \subset K$ such that
+  \begin{itemize}
+    \item the set $\{y_1, \ldots, y_r\}$ is algebraically independent over $k$
+    \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \sbuset K$.
+  \end{itemize}
+
+  By \ref{R.4.9}:\\
+  if $A \subset B$ integral extension of integral domain,\\
+  then $A$ is a field $\Longleftrightarrow~ B$ is a field.
+
+  which applied to our case:
+  $$B:=K ~\text{(a field)},~~ A:= R=k[y_1, \ldots, y_r]$$
+  since $K$ is integral over $R$ and $K$ a field $\Longrightarrow$ by \ref{R.4.9} $R$ must be a field.
+
+  \vspace{0.3cm}
+  Since $y_i \in K$ are algebraically independent\\
+  $\Longrightarrow~ R=k[y_1, \ldots, y_r]$ is a polynomial ring in $r$ indeterminates.
+
+  Then, if $r \geq 1$, we have at least one variable ($k[y_1]$);\\
+  in $k[y_1]$, the element $y_1$ is not invertible, since $1/y_1$ is a rational function, not a polynomial.\\
+  (alternatively: the only invertible elements in a polynomial ring over a field are the non-zero constants of the field $k$).
+
+  Therefore, $R=k[y_1, \ldots, y_r]$ can be a field iff $r=0$, and we know that is a field, therefore $r=0$ and $R=k$.
+
+  By Noether normalization, at the previous step we knew that $K$ is integral over $R$ ($=k$).
+
+  \vspace{0.3cm}
+  By \ref{integral-implies-finite}, a fingen algebra that is integral over its base field is a finite vector space over that field.
+
+  Thus $[K:k]<\infty$, and $K$ is algebraic over $k$.
+\end{proof}
+
+\vspace{0.5cm}
+This brings us to the \emph{Weak Nullstellensatz}:
+
+\begin{thm}{4.10}[Weak Nullstellensatz]\label{weak-nullstellensatz-again}
+  $k$ a field, $K$ a $k$-algebra with
+  \begin{itemize}
+    \item $K$ fingen as a $k$-algebra
+    \item $K$ a field
+  \end{itemize}
+  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}
+  proven at previous proof.
+\end{proof}
+
+
+\vspace{0.8cm}
+Now, let $m\subset k[X_1, \ldots, X_n]$ a maximal ideal. Let $K=k[X_1, \ldots, X_n]$.
+
+Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{K}{m}$ is a field (by basic ideal quotient theory).
+
+If it's a fingen $k$-algebra $\Longrightarrow$ by Zariski's lemma, $\frac{K}{m}$ is algebraic over $k$.
+
+And since $k$ is algebraically closed $\Longrightarrow~ \frac{K}{m}=k$.
+
+By definition of $K$ being a $k$-algebra,
+$$\exists~ \psi: k \longrightarrow \frac{K}{m} ~~~\Longrightarrow~ k \stackrel{iso}{\cong} \frac{K}{m}$$
+
+Thus, $\forall x_i \in k,~ \psi(x_i)=a_i \in k$.
+And by definition of a quotient ring,
+
+$psi(x_i)$ is the coset
+\begin{align*}
+  \psi(x_i)&=x_i+m\\
+           &\Longrightarrow a_i = x_i +m\\
+           &x_i - a_i \in m
+\end{align*}
+
+Since $x_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
+
+since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal.
+
+(Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
+
+\vspace{0.4cm}
+We had $k \stackrel{iso}{\cong} \frac{K}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
+
+We have the quotient map
+\begin{align*}
+  \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
+  x_i &\mapsto a_i \in k
+\end{align*}
+which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
+
+Since $\phi$ is a $k$-algebra homomorphism, then
+$$\forall~ f \in K,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n)) = f(a_1, \ldots, a_n)$$
+
+Thus, $\exists$ a one-to-one correspondence:
+\begin{align*}
+  \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~ A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\
+  (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)
+\end{align*}
+
+\vspace{0.5cm}
+Observe: $x_i - a_i \in ker(\psi) =m~ \forall~ i \in [n]$.
+
+We also have seen that $m=(X_1-a_1, \ldots, X_n-a_n)$.
+
+So for $P=(a_1, \ldots, a_n) \in k^n$, it is a zero for every polynomial in $m$, ie.
+$$\forall~ f \in m ~\Longrightarrow~ f(P)=0$$
+
+Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$.
+
+$\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states:
+\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
+  \begin{enumerate}[a.]
+    \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
+    \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
+      $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
+  \end{enumerate}
+\end{thm}
+\begin{proof}
+  \begin{enumerate}[a.]
+  \item already proven in the previous steps.
+    
+  \item $I(V(J)) = rad J$\\
+  ie. vanishing ideal of a variety = radical of the ideal definingh the variety
+
+  where $rad~J = \{ f \in R | f^n \in J ~\text{for some}~ n>0 \}$.
+
+  Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
+
+  Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
+
+  Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
+  $$\exists~ \underbrace{(a_1, \ldots, a_n, a_{n+1})}_{a} \in V(J')$$
+
+  Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
+  $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
+  a contradiction.
+
+  Therefore, $V(J') = \emptyset$.
+
+  \vspace{0.4cm}
+
+  By the Weak Nullstellensatz (\ref{weak-nullstellensatz-again}),\\
+  \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
+
+  Every element in an ideal is a linear combination of its generators:\\
+  $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
+
+  $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
+
+  which, since $1 \in J'$,
+  $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
+
+  substitute $Y=1/f$,
+  $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
+  thus
+  $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
+  multiply by $f^n$,
+  $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
+
+  thus $f^n$ is a linear combination of $g_i$.\\
+  Therefore $f^n \in J$, so $f \in rad~J$.
+  \end{enumerate}
+\end{proof}
+
 
 
 \section{Rings of fractions $S^{-1}A$ and localization}
@@ -1747,7 +1945,7 @@ let $A$ a ring, $S \subset A$ a multiplicative set ($1 \in S$, and $st \in S
   ~\forall s, t \in S$).
 
   Introduce th following relation $\sim$ on $A \times S$:
-  $$(a, s) \sim (b, t) \Longleftrightarrow \exist y \in S ~\text{such that}~ u(at - bs) = 0$$
+  $$(a, s) \sim (b, t) \Longleftrightarrow \exists y \in S ~\text{such that}~ u(at - bs) = 0$$
 
   (write $a/s$ for the equivalence class of $(a, s)$.
 
@@ -1920,14 +2118,13 @@ $a/s \in A_P$ is a unit.
 
     \item[$\Longrightarrow$] (if $a/s$ unit $~\Longrightarrow~ a \not\in P$)\\
 
-      Suppose $a/s$ a unit; then $\exist~ \frac{b}{t} \in A_P$ such that
+      Suppose $a/s$ a unit; then $\exists~ \frac{b}{t} \in A_P$ such that
       $\frac{a}{s} \cdot \frac{b}{t} = 1$.
 
-      By definition of equality in localization, $\frac{ab}{st}=1$ means $\exist
-      u \in S$ such that
+      By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that
       \begin{equation}
         u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust
-        \tag{eq.6.4}
+        \label{eq:6.4}
       \end{equation}
       with $u,s,t \in S$.
 
@@ -1935,7 +2132,7 @@ Since $S=A \setminus P$ and $P$ is a prime ideal,\\
       the products of elements outside $P$ must also be outside of $P$.
       $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$
 
-      At \eq{eq.6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
+      At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
 
       If $uab \not\in P$, then $a \not\in P$.
 
@@ -1996,10 +2193,11 @@ the quotient ring $A/I$ can be viewed as an $A$-module and
   $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong
   \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$
 
-  The \ref{6.6}.i, gives an isomorphism of modules
+  The \ref{cor.6.6}.i, gives an isomorphism of modules
   $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
   it's easy to see that this is a ring homomorphism.
 \end{proof}
+% TODO 6.6 (proposition and corollary)
 
 
 

@@ -123,7 +123,7 @@ \begin{eg}{4.11}
   Beware, $L=\mathbb{Q}(i, -i, \sqrt{5}, -\sqrt{5}) = \mathbb{Q}(i, \sqrt{5}) = \mathbb{Q}(i+\sqrt{5})$.
 \end{eg}
 
-\begin{defn}{5.5}
+\begin{defn}{5.5}[Minimal polynomial]
   Let $L:K$, suppose $\alpha \in L$ is algebraic over $K$. Then, the \emph{minimal polynomial} of $\alpha$ over $K$ is the unique monic polynomial $m$ over $K$, $m(t) \in K[t]$, of smallest degree such that $m(\alpha)=0$.
   \\
   eg.: $i \in \mathbb{C}$ is algebraic over $\mathbb{R}$. The minimal polynomial of $i$ over $\mathbb{R}$ is $m(t)=t^2 +1$, so that $m(i)=0$.