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  9\begin{filecontents}[overwrite]{galois-theory-notes.bib}
 10@misc{ianstewart,
 11  author = {Ian Stewart},
 12  title = {{Galois Theory, Third Edition}},
 13  year = {2004}
 14}
 15\end{filecontents}
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 46
 47\title{Galois Theory notes}
 48\author{arnaucube}
 49\date{2023-2024}
 50
 51\begin{document}
 52
 53\maketitle
 54
 55\begin{abstract}
 56	Notes taken while studying Galois Theory, mostyly from Ian Stewart's book "Galois Theory" \cite{ianstewart}.
 57
 58	Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
 59
 60	The notes are not complete, don't include all the steps neither all the proofs.
 61\end{abstract}
 62
 63\tableofcontents
 64
 65\section{Recap on the degree of field extensions}
 66
 67\begin{defn}{4.10}
 68  A \emph{simple extension} is $L:K$ such that $L=K(\alpha)$ for some $\alpha \in L$.
 69\end{defn}
 70\begin{eg}{4.11}
 71  Beware, $L=\mathbb{Q}(i, -i, \sqrt{5}, -\sqrt{5}) = \mathbb{Q}(i, \sqrt{5}) = \mathbb{Q}(i+\sqrt{5})$.
 72\end{eg}
 73
 74\begin{defn}{5.5}
 75  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 $m(\alpha)=0$.
 76  \\
 77  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 $m(i)=0$.
 78\end{defn}
 79
 80\begin{lemma}{5.9}
 81  Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \delta m$.
 82\end{lemma}
 83\begin{proof}
 84  Divide $a / m$ with remainder, $a= qm +r$, with $q,r \in K[t]$ and $\delta r < \delta m$.
 85  Then, $a-r=qm$, so $a \equiv r \pmod{m}$.
 86
 87  It remains to prove uniqueness.
 88
 89  Suppose $\exists~ r \equiv s \pmod{m}$, with $\delta r, \delta s < \delta m$.
 90  Then, $r-s$ is divisible by $m$, but has smaller degree than $m$.
 91
 92  Therefore, $r-s=0$, so $r=s$, proving uniqueness.
 93\end{proof}
 94
 95\begin{lemma}{5.14}
 96  Let $K(\alpha):K$ be a simple algebraic extension, let $m$ be the minimal polynomial of $\alpha$ over $K$, let $\delta m =n$.
 97
 98  Then $\{1, \alpha, \alpha^2, \ldots, \alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
 99  In particular, $[K(\alpha):K]=n$.
100\end{lemma}
101
102\begin{defn}{6.2}
103  The degree $[L:K]$ of a field extension $L:K$ is the dimension of L considered as a vector space over $K$.
104
105  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 $K$.
106\end{defn}
107
108\begin{eg}{6.3}
109  \begin{enumerate}
110    \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 $[\mathbb{C}:\mathbb{R}]=2$.
111    \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 $\mathbb{Q}$.
112  \end{enumerate}
113\end{eg}
114
115\begin{thm}{6.4}\emph{(Short Tower Law)}
116  If $K, L, M \subseteq \mathbb{C}$, and $K \subseteq L \subseteq M$, then $[M:K]=[M:L]\cdot [L:K]$.
117\end{thm}
118\begin{proof}
119  Let $(x_i)_{i \in I}$ be a basis for $L$ over $K$,
120  let $(y_j)_{j \in J}$ be a basis for $M$ over $L$.\\
121  $\forall i \in I, j \in J$, we have $x_i \in L, u_j \in M$.
122  \\
123  Want to show that $(x_i y_j)_{i\in I, j\in J}$ is a basis for $M$ over $K$.
124  \begin{enumerate}[i.]
125    \item prove linear independence:\\
126      Suppose that
127      $$\sum_{ij} k_{ij} x_i y_j = 0 ~(k_{ij} \in K)$$
128      rearrange
129      $$\sum_j (\underbrace{\sum_i k_{ij} x_i}_{\in L}) y_j = 0 ~(k_{ij} \in K)$$
130      Since $\sum_i k_{ij} x_i \in L$, and the $y_j \in M$ are linearly independent over $L$, then $\sum_i k_{ij} x_i = 0$.
131      \\
132      Repeating the argument inside $L$ $\longrightarrow$ $k_{ij}=0 ~~\forall i\in I, j\in J$.
133      \\
134      So the elements $x_i y_j$ are linearly independent over $K$.
135
136    \item prove that $x_i y_j$ span $M$ over $K$:\\
137      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 $L$.
138      Similarly, $\forall j\in J,~ \lambda_j = \sum_i \lambda_{ij} x_i y_j$ for  $\lambda_{ij} \in K$.\\
139      Putting the pieces together, $x=\sum_{ij} \lambda_{ij} x_i y_j$ as required.
140  \end{enumerate}
141\end{proof}
142
143\begin{cor}{6.6}\emph{(Tower Law)}\\
144  If $K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$ are subfields of $\mathbb{C}$, then
145  $$[K_n:K_0] = [K_n:K_{n-1}] \cdot [K_{n-1}:K_{n-2}] \cdot \ldots \cdot [K_1: K_0]$$
146\end{cor}
147
148\bibliographystyle{unsrt}
149\bibliography{galois-theory-notes.bib}
150
151\end{document}