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+\documentclass{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsfonts}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage{enumerate}
+\usepackage{hyperref}
+
+\begin{filecontents}[overwrite]{galois-theory-notes.bib}
+@misc{ianstewart,
+  author = {Ian Stewart},
+  title = {{Galois Theory, Third Edition}},
+  year = {2004}
+}
+\end{filecontents}
+\nocite{*}
+
+
+\theoremstyle{definition}
+
+\newtheorem{innerdefn}{Definition}
+\newenvironment{defn}[1]
+{\renewcommand\theinnerdefn{#1}\innerdefn}
+{\endinnerdefn}
+
+\newtheorem{innerthm}{Theorem}
+\newenvironment{thm}[1]
+{\renewcommand\theinnerthm{#1}\innerthm}
+{\endinnerthm}
+
+\newtheorem{innerlemma}{Lemma}
+\newenvironment{lemma}[1]
+{\renewcommand\theinnerlemma{#1}\innerlemma}
+{\endinnerlemma}
+
+\newtheorem{innercor}{Lemma}
+\newenvironment{cor}[1]
+{\renewcommand\theinnercor{#1}\innercor}
+{\endinnercor}
+
+\newtheorem{innereg}{Example}
+\newenvironment{eg}[1]
+{\renewcommand\theinnereg{#1}\innereg}
+{\endinnereg}
+
+
+\title{Galois Theory notes}
+\author{arnaucube}
+\date{2023-2024}
+
+\begin{document}
+
+\maketitle
+
+\begin{abstract}
+	Notes taken while studying Galois Theory, mostyly from Ian Stewart's book "Galois Theory" \cite{ianstewart}.
+
+	Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
+
+	The notes are not complete, don't include all the steps neither all the proofs.
+\end{abstract}
+
+\tableofcontents
+
+\section{Recap on the degree of field extensions}
+
+\begin{defn}{4.10}
+  A \emph{simple extension} is $L:K$ such that $L=K(\alpha)$ for some $\alpha \in L$.
+\end{defn}
+\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}
+  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 $m(i)=0$.
+\end{defn}
+
+\begin{lemma}{5.9}
+  Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \delta m$.
+\end{lemma}
+\begin{proof}
+  Divide $a / m$ with remainder, $a= qm +r$, with $q,r \in K[t]$ and $\delta r < \delta m$.
+  Then, $a-r=qm$, so $a \equiv r \pmod{m}$.
+
+  It remains to prove uniqueness.
+
+  Suppose $\exists~ r \equiv s \pmod{m}$, with $\delta r, \delta s < \delta m$.
+  Then, $r-s$ is divisible by $m$, but has smaller degree than $m$.
+
+  Therefore, $r-s=0$, so $r=s$, proving uniqueness.
+\end{proof}
+
+\begin{lemma}{5.14}
+  Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$.
+
+  Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
+  In particular, $[K(\alpha):K]=n$.
+\end{lemma}
+
+\begin{defn}{6.2}
+  The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$.
+
+  Equivalently, the dimension of $L$ as a vector space over $K$ is the number of terms in the expression for a general element of $L$ using coefficients from $K$.
+\end{defn}
+
+\begin{eg}{6.3}
+  \begin{enumerate}
+    \item $\mathbb{C}$ elements are 2-dimensional over $\mathbb{R}$ ($p+qi \in \mathbb{C}$, with $p,q \in \mathbb{R}$), because a basis is $\{1, i\}$, hence $[\mathbb{C}:\mathbb{R}]=2$.
+    \item $[ \mathbb{Q}(i, \sqrt{5}) : \mathbb{Q}]=4$, since the elements $\{1, \sqrt{5}, i, i\sqrt{5}\}$ form a basis for $\mathbb{Q}(i, \sqrt{5})$ over $\mathbb{Q}$.
+  \end{enumerate}
+\end{eg}
+
+\begin{thm}{6.4}\emph{(Short Tower Law)}
+  If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$.
+\end{thm}
+\begin{proof}
+  Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$,
+  let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\
+  $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$.
+  \\
+  Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$.
+  \begin{enumerate}[i.]
+    \item prove linear independence:\\
+      Suppose that
+      $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$
+      rearrange
+      $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$
+      Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$.
+      \\
+      Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$.
+      \\
+      So the elements $x_i y_j$ are linearly independent over $K$.
+
+    \item prove that $x_i y_j$ span $M$ over $K$:\\
+      Any $x \in M$ can be written $x=\sum_j \lambda_j y_j$ for $\lambda_j \in L$, because $y_j$ spans $M$ over $L$.
+      Similarly, $\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$ for  $\lambda_{ij} \in K$.\\
+      Putting the pieces together, $x=\sum_{ij} \lambda_{ij} x_i y_j$ as required.
+  \end{enumerate}
+\end{proof}
+
+\begin{cor}{6.6}\emph{(Tower Law)}\\
+  If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then
+  $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$
+\end{cor}
+
+\bibliographystyle{unsrt}
+\bibliography{galois-theory-notes.bib}
+
+\end{document}