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diff --git a/semestr2/algebra2r/notatki/notatki.pdf b/semestr2/algebra2r/notatki/notatki.pdf
index b34dcad..9c9aec4 100644
--- a/semestr2/algebra2r/notatki/notatki.pdf
+++ b/semestr2/algebra2r/notatki/notatki.pdf
diff --git a/semestr2/algebra2r/notatki/sections/finite-fields.tex b/semestr2/algebra2r/notatki/sections/finite-fields.tex
new file mode 100644
index 0000000..4e623f0
--- /dev/null
+++ b/semestr2/algebra2r/notatki/sections/finite-fields.tex
@@ -0,0 +1,143 @@
+\documentclass[../notes.tex]{subfiles}
+\begin{document}
+  Non-trivial finite fields can be only of $\Char = p$ prime. Also, we can
+  describe all of them very specifically.
+
+  \begin{remark}
+    Let $K$ be a finite field $\Char = p > 0$, Then $K^{*}$ (the multiplicative
+    subgroup) is also finite and for $n = |K|$ we have $x^n = x$ for any $x\in
+    K^*$.
+  \end{remark}
+
+  \begin{theorem}
+    Let $K$ be a finite field of $\Char = p$. Then
+    \begin{enumerate}
+      \item $|K| = p^n$ for some $n > 0$,
+      \item For all $n > 0$ there exists a field of order $p^n$ and it is unique up to
+        isomorphism.
+    \end{enumerate}
+  \end{theorem}
+
+  \begin{theorem}
+    Let $K$ be a finite field of characteristics $p$.
+    \begin{enumerate}
+      \item Every irreducible polynomial $f\in K[X]$ divides the polynomial $X^n
+        -1$ for some $n$ not divisible by $p$.
+      \item Such $f$ has no multiple roots in any field $L\supseteq K$.
+        For $a\in L\supset K$, the following conditions are equivalent:
+        enumerate$a$ is algebraic over $K$,
+      \item $K[a] = K(a)$,
+      \item $[K(a):K] < \infty$.
+    \end{enumerate}
+  \end{theorem}
+
+  We can very well describe every subfields of a finite field.
+  \begin{theorem}
+    \mbox{}
+    \begin{enumerate}
+      \item If $K\subseteq L$ is are finite fields, $|K| = p^m, |L| = p^n$, then
+        $m\mid n$.
+      \item Every field of size $p^n$ elements contains a unique subfield of
+        size $p^m$, where $m\mid n$.
+    \end{enumerate}
+  \end{theorem}
+
+  Hence, we can name the finite fields with $F_{p^n}$ to be the very unique
+  subfield of order $p^n$ (as even in bigger fields this is the only subfield of
+  this size).
+
+  \begin{definition}
+    Let $K$ be a field of $\Char = p > 0$. Then the function $\phi\colon K\to K$
+    such that $\phi(a) = a^p$ is called \emph{Frobenius automorphism}.
+  \end{definition}
+
+  It's easy to check that $\phi$ is an automorphism. This gives a funny property
+  of fields of prime characteristics:
+
+  \begin{fact}
+    For any $a, b\in K$, where $\Char(K) = p > 0$, we have $(a + b)^p = a^p +
+    b^p$.
+  \end{fact}
+
+  \begin{theorem}
+    \label{theorem:degree-finite}
+    Assume that $F_p\subseteq K$ is a finite field extension. Let $a\in K$ be a
+    primitive root of unity of degree $m$. Let $n > 0$ be the smallest number such
+    that $m\mid p^n - 1$. Then $\deg(a/F_p) = n$ and $n\mid \phi(m)$.
+  \end{theorem}
+
+  \begin{proof}
+    Look at $\langle a\rangle = F_{p^n}$ for some $n$. Then $a^{p^n - 1} = 1$
+    (so this is the smallest such $n$),
+    as $p^n -1 = |F_{p^n}^*|$, but $\ord(a) = m$, hence $m\mid p^{n}-1$. Hence
+    $\deg(a/F_p) = [F_{p^n}:F_p] = n$ (\color{red}{why??} we can see $\ge n$ by
+    looking on $\phi^{k}(a)$, how the other way around?).
+
+    \color{red}{Why does $n\mid \phi(m)$?}
+  \end{proof}
+
+  \subsection{Algebraic closure of $F_p$}
+  Algebraic closure of $F_p$ is also quite well understood. We know that
+  $F_{p^n}\subset F_{p^m}$ if and only if $n\mid m$. Hence, $F_{p^{n!}}\subset
+  F_{p^{m!}}$ for any $n < m$.
+  \begin{remark}
+    Let $F_\infty = \bigcup_{n=1}^\infty F_{p^{n!}}$. $F_\infty$ is
+    algebraically closed.
+  \end{remark}
+
+  See that $F_{p^n}\subset F_\infty$ for any $n\bN$, from the construction.
+  So, this is the algebraic closure for every finite field. However, there are
+  many infinite fields contained in $F_\infty$.
+
+  \begin{remark}
+    $F_\infty$ has at least $2^{\aleph_0}$ distinct subfields, every pair having a finite
+    intersection. Also, every automorphism of $F_\infty$ does not move any of
+    these.
+  \end{remark}
+
+  \begin{proof}
+    Let $\bP = \{p_0, p_1,\ldots\}$ be the enumeration of all prime numbers. For 
+    $\eta\colon \bN\to \{0, 1\}$, let $\rho(\eta, k) = p_0^{\eta(0)}p_1^{\eta(1)}\ldots
+    p_k^{\eta(k)}$ and $$F_\eta = \bigcup^\infty_{k=0} F_{p^{\rho(\eta, k)}}$$
+
+    Intuitively, $F_\eta$ is the union of fields of some power of $p$ given by
+    some branch of the following tree:
+
+    \begin{center}
+    \begin{tikzpicture}[very thick,level/.style={sibling distance=60mm/#1}]
+    \node [vertex] (r){$1$}
+      child {
+        node [vertex] (a) {$1$}
+        child {
+          node [vertex] {$1$}
+          child {node [vertex] {$\ldots$}}
+          child {node [vertex] {$\ldots$}}
+        }
+        child {
+          node [vertex] {$p_1$}
+          child {node [vertex] {$\ldots$}}
+          child {node [vertex] {$\ldots$}}
+        }
+      }
+      child {
+        node [vertex] {$p_0$}
+        child {
+          node [vertex] {$1$}
+          child {node [vertex] {$\ldots$}}
+          child {node [vertex] {$\ldots$}}
+        }
+        child {
+          node [vertex] {$p_1$}
+          child {node [vertex] {$\ldots$}}
+          child {node [vertex] {$\ldots$}}
+        }
+      };
+    \end{tikzpicture}
+    \end{center}
+    Product of the elements on the way from root through some branch yields the
+    sizes of the subfields of $F_\eta$. It's not hard to see now that the
+    remark is true for such subfields, and it's well known that there are
+    $2^{\aleph_0}$ such functions.
+  \end{proof}
+  
+\end{document}
diff --git a/semestr2/algebra2r/notatki/sections/galois-cont.tex b/semestr2/algebra2r/notatki/sections/galois-cont.tex
new file mode 100644
index 0000000..119e31a
--- /dev/null
+++ b/semestr2/algebra2r/notatki/sections/galois-cont.tex
@@ -0,0 +1,337 @@
+\documentclass[../notes.tex]{subfiles}
+\begin{document}
+  \subsection{Galois extension and Artin's theorem}
+  So far we were focusing on the absolute Galois group $G(\hat{K}/K)$. In this
+  section we will move forward to the Galois correspondence theorem, which gives
+  a beautiful description of the algebraic extension $K\subset L$ through
+  properties of the Galois group $G(L/K)$.
+
+  As before, we work with algebraic extension $K\subset L\subset \hat{K}$. 
+
+  \begin{definition}
+    The algebraic field extension $K\subset L$ is \emph{Galois} when $\forall
+    a\in L\setminus K$ $\exists f\in G(L/K)$ $f(a)\neq a$.
+  \end{definition}
+  \begin{definition}
+    Let $G<\Aut(L)$. Then $L^G\defeq \{a\in L\mid \forall f\in G$ $f(a) = a\} =
+    \bigcup_{f\in G}\Fix(f)$ is the \emph{fixed-point field of $G$}.
+  \end{definition}
+
+  We get a nice equivalent definition for Galois extensions with the notion of
+  fixed-point field:
+  \begin{remark}
+    Extension $K\subset L$ is Galois $\Leftrightarrow$ $K = L^{G(L/K)}$.
+  \end{remark}
+
+  \begin{example}
+    \mbox{}
+    \begin{enumerate}
+      \item Let $L = K(a)$ for algebraic $a/K$. Let $a= a_1,\ldots, a_k$ be the roots of
+        the minimal polynomial $W_a(X)$ in $L$ (so not every root!). However we know
+        that any $f\in G(L/K)$ is determined by $f(a)$, which can only be one of
+        $a_1,\ldots, a_k$, hence $\left|G(L/K)\right|\le k$.
+      \item Let $L\supset K$ be the splitting field of $W(X)\in K[X]$, i.e. $L =
+        K(a_1,\ldots,a_n)$, where $a_i$ are all roots of $W(X)$. Then every
+        $f\in G(L/K)$ is determined by $\restr{f}{\{a_1,\ldots,a_n\}}\in
+        \Sym(\{a_1,\ldots,a_n\})$. We get a very natural embedding:
+        \[
+          \begin{matrix}
+          G(L/K) & \hookrightarrow & \Sym(\{a_1,\ldots, a_n\}) \\
+          \rotatebox[origin=c]{90}{$\in$} &&  \rotatebox[origin=c]{90}{$\in$} \\
+          f & \mapsto & \restr{f}{\{a_1,\ldots,a_n\}}
+          \end{matrix}
+        \]
+      \item Let $\zeta\in\bC$ be a primitive root of unity of degree $m$. Then
+        $[\bQ(\zeta)\colon \bQ] = \phi(m)$. Moreover, if $\{\zeta=\zeta_1,
+        \zeta_2,\ldots,\zeta_{\phi(m)}\}$ where all primitive roots of unity of
+        degree $m$, then $f\in G(\bQ(\zeta_1)/\bQ)$ is determined by
+        $f(\zeta_1) = \zeta_i$ for any $i=1,\ldots,\phi(m)$. This shows that not
+        every permutation of roots of a minimal polynomial extends to a valid
+        automorphism. Moreover, in this case we see that $f(\zeta_1) =
+        \zeta_1^{l_f}$ for some $0<l_f<m$ such that $\gcd(l_f,m) = 1$. Hence,
+        \[
+          \begin{matrix}
+          G(\bQ(\zeta_1)/\bQ) & \hookrightarrow & \bZ^*_m \\
+          \rotatebox[origin=c]{90}{$\in$} &&  \rotatebox[origin=c]{90}{$\in$} \\
+          f & \mapsto & l_f 
+          \end{matrix}
+        \]
+    \end{enumerate}
+  \end{example}
+
+  \begin{theorem}
+    Let $K\subset L$ be an algebraic extension. Then $K\subset L$ is Galois
+    $\Leftrightarrow$ $K\subset L$ is separable and normal.
+  \end{theorem}
+
+  Proof of this theorem gives great intuition on Galois extensions and utilizes
+  a very important observation, that polynomials of $K[X]$ are fixed by any
+  automorphism of $L$ over $K$.
+  \begin{proof}
+    $\Rightarrow$ Let $a\in L\setminus K$ and $a_1,\ldots,a_k\in L$ be pairwise
+    disjoint roots of
+    $W_a(X)$ that reside in $L$. Look at $V(X) =
+    (X-a_1)\cdot\ldots\cdot(X-a_k)\in L[X]$, of course $V(X)\mid W_a(X)$ in
+    $L[X]$, but as any $f\in G(L/K)$ permutes $\{a_1,\ldots,a_k\}$ we get that
+    $f(V) = V$, hence $V(X)\in L^{G(L/K)[X} = K[X]$ as $K\subset L$ is Galois.
+    But $W_a$ is minimal, hence $V = W_a$, so all roots of $W_a$ are in $L$ and
+    every root is simple, so $L$ is normal and separable over $K$.
+
+    $\Leftarrow$ Normal extension says that for a given minimal polynomial $W_a(X)$ of $a\in
+    L$ all roots of this polynomial are inside $L$. Combined with separability,
+    we get that $a$ can be moved to some other element by $f\in G(L/K)$ if $\deg
+    W_a > 1$, but if $\deg W_a = 1$ then $a\in K$, so it's clear that $K =
+    L^{G(L/K)}$. 
+  \end{proof}
+
+  \begin{corollary}
+    \label{corollary:galois-exts}
+    Let $K\subset L\subset M$ be field extensions. If $K\subset M$ is Galois
+    then $L\subset M$ is Galois.
+  \end{corollary}
+
+  \begin{corollary}
+    In $Char K = 0$, let $L \supset K$ be a splitting field of some polynomial $f\in K[X]$.
+    Then $K\subset L$ is a Galois extension.
+  \end{corollary}
+
+  \begin{proof}
+    Because $\Char K = 0$, the extension is separable. It's also easy to see
+    that $L$ is normal over $K$, hence it's Galois.
+  \end{proof}
+
+  The following theorem is also extremely important for studying examples of
+  Galois extensions. It is a corollary from Abel's primitive element
+  theorem\ref{theorem:abel}.
+  \begin{theorem}[Artin's]
+    \label{theorem:artins}
+    Let $G<\Aut(L)$ be a finite subgroup. Then $L^G\subset L$ is Galois
+    extension with $G = G(L/L^G)$ and $[L\colon L^G] = \left|G\right|$.
+  \end{theorem}
+
+  \begin{proof}
+    Now it shouldn't be very hard, try doing it yourself!
+  \end{proof}
+
+  \begin{example}
+    Every finite group $G$ is isomorphic to some Galois group of some Galois
+    extension. By Cayley's theorem we may assume $G < \Sym(\{X_1,\ldots,
+    X_n\})$. Let $K$ be any field, let $L = K(X_1, \ldots, X_n)$ and $H = \{f\in
+    G(L/K)\mid \restr{f}{\{X_1,\ldots,X_n\}}\in G\}$, then $G\cong
+    G(L/L^H)$ by Artin's theorem. 
+  \end{example}
+
+  \begin{corollary}
+    Let $K\subset L$ be Galois extension. Then $[L\colon K] =
+    \left|G(L/K)\right|$.
+  \end{corollary}
+
+  \subsection{Fundamental theorem of Galois theory}
+  Let $K\subset L$ be an algebraic extension. Let $\cL =\{L'\mid K\subseteq
+  L'\subseteq L\}$ be the family of intermediate fields. Let $\cG = \{H\mid
+  H<G(L/K)\}$ be the family of subgroups of the Galois group of $L$ over $K$.
+  We get a very natural maps from one family to another:
+
+  \[
+    \begin{matrix}
+      \Gamma\colon & \cL & \lhook\joinrel\longrightarrow & \cG & &
+      \Lambda\colon & \cG & \longrightarrow & \cL \\
+      & \rotatebox[origin=c]{90}{$\in$} & &  \rotatebox[origin=c]{90}{$\in$} & &
+      & \rotatebox[origin=c]{90}{$\in$} & &  \rotatebox[origin=c]{90}{$\in$} & \\
+      & L' &\overset{\Gamma}{\longmapsto}& G(L/L') & & & H &\overset{\Lambda}{\longmapsto}& L^H 
+    \end{matrix}
+  \]
+  \[
+    \begin{matrix}
+    \end{matrix}
+  \]
+
+  It's obvious that $\Gamma$ is 1-1 and that $\Lambda$ is onto. Maybe it's an
+  isomorphism?
+
+  \begin{theorem}[Fundamental theorem of Galois theory]
+    Let $K\subset L$ be a finite Galois extension. Then $\Gamma$ is a bijection
+    and $\Gamma^{-1} = \Lambda$.
+  \end{theorem}
+
+  \begin{proof}
+    For any $L'\in\cL$, $L\overset{\Gamma}{\longmapsto} G(L/L')
+    \overset{\Lambda}{\longmapsto} L^{G(L/L')}$, but by corollary
+    \ref{corollary:galois-exts} $L'\subset L$ is also Galois, hence $L^{G(L/L')}
+    = L'$. So, $\Lambda\circ\Gamma =\id_\cL$.
+
+    The other way, let $H\in\cG$, then
+    $H\overset{\Lambda}{\longmapsto}L^H\overset{\Gamma}{\longmapsto}G(L/L^H) =
+    H$, where the last equality follows from the Artin's theorem
+    \ref{theorem:artins}. Hence $\gamma\circ\Lambda = \id_\cG$.
+  \end{proof}
+
+  We will look on example usages of the theorem a little bit later. Now we'll
+  show one more corollary, that gives a wonderful connection between normal
+  extensions and normal subgroups. First, we need this easy lemma:
+
+  \begin{lemma}
+    \label{lemma:normal}
+    Let $K\subset L'\subset L$ such that $K\subset L$ is a normal finite
+    extension. Then $K\subset L'$ is normal $\Leftrightarrow$ $\forall
+    f\in G(L/K)$ $f[L'] = L'$.
+  \end{lemma}
+
+  \begin{theorem}
+    Let $K\subset L$ be Galois (i.e. normal and separable). Then, for any
+    $H<G(L/K)$ we have that $H\triangleleft G(L/K) \Leftrightarrow K\subset L^H$ is
+    Galois and $G(L^H/K) \cong G(L/K)/H$.
+  \end{theorem}
+
+  \begin{proof}
+    $L$ is a Galois extension of $K$, so it suffices to show that $K\subset L^H$
+    normal (as it is separable).
+
+    Every $f\in\Aut(L)$ induces an inner automorphism $\hat{f}$ of $\Aut(L)$ by
+    conjugation, i.e. $\hat{f}(g) = f\circ \phi\circ f^{-1}$. But, as $L$ is
+    normal, this is also an automorphism of $G(L/K)$. Hence, for any $H<G(L/K)$
+    we have that $\hat{f}[H] = H^f$ and
+    \begin{align*}
+      L^H &= \bigcap_{g\in H}\Fix(g) \\
+      & \Rightarrow f[L^H] = \bigcap_{g\in H^f}\Fix(g) = L^{H^f}
+    \end{align*}
+    
+    Hence, because $\Gamma\colon\cG\to\cL$ is a bijection, we have the following chain of equivalences:
+    \begin{align*}
+      H\triangleleft G(L/K) &\Leftrightarrow \forall f\in G(L/K) \; H^f = H
+      \\
+      &\Leftrightarrow \forall f\in G(L/K) \; L^{H^f} = L^H \\
+      &\Leftrightarrow \forall f\in G(L/K) \; f[L^H] = L^H \\
+      &\overset{\ref{lemma:normal}}{\Leftrightarrow} K\subset L^H \text{ is
+      normal}
+    \end{align*}
+
+    The fact that $\Gamma$ is a bijection is used in the second equivalence.
+    It's not necessary from the left to right implication, but it gives us the
+    implication in other direction.
+
+    Lastly, when $H\triangleleft G(L/K)$, the $\phi\colon G(L/K)\to G(L^H/K)$
+    with $\phi(f) = \restr{f}{L^H}$ is an epimorphism and $\ker{\phi} = G(L/L^H)
+    = H$ by Artin's theorem \ref{theorem:artins}. Hence:
+
+    \begin{center}
+      \begin{tikzcd}
+        G(L/K) \arrow[r, "\phi"] \arrow[d, "j"] & G(L^H/K) \\
+        G(L/K)/H \arrow[ur, dashed, "\cong"] & 
+	  \end{tikzcd}
+    \end{center}
+  \end{proof}
+
+  \subsection{Fundamental theorem applied}
+  
+  We'll work in $\Char = 0$.
+  \begin{example}
+    Look at $W(X) = X^3 - 2\in \bQ$. It's roots are $\sqrt[3]{2},
+    \zeta\sqrt[3]{2}$ and $\zeta^2\sqrt[3]{2}$, where $\zeta\in \bC$ is the
+    primitive root of unity of degree $3$. So, the splitting field of $W$ is
+    $\bQ(\sqrt[3]{2}, \zeta)$, which has degree $6$ over $\bQ$. By Abel's
+    theorem we know that there is $a\in\bQ(\sqrt[3]{2}, \zeta)$ such that
+    $\bQ(a) = \bQ(\sqrt[3]{2}, \zeta)$, moreover, it is a linear combination of
+    $\sqrt[3]{2}$ and $\zeta$. So, let's try $a = \sqrt[3]{2} + \zeta$ and see
+    if it works.
+
+    From Artin's theorem \ref{theorem:artins} we know that $6 = [\bQ(\sqrt[3]{2},
+    \zeta)\colon\bQ] = \left|G(\bQ(\sqrt[3]{2},\zeta)/\bQ)\right|$ We know for
+    sure that $\bQ(\sqrt[3]{2} + \zeta)\subseteq \bQ(\sqrt[3]{2}, \zeta)$, so
+    it's sufficient to show that it's degree over $\bQ$ is also $6$, or in other
+    words, that it's Galois group is the same as $G(\bQ(\sqrt[3]{2},\zeta)/\bQ)$. 
+    So, if we find $6$ distinct automorphisms of $\bQ(\sqrt[3]{2} + \zeta)$ over $\bQ$
+    then we're done (as there can't be more).
+
+    Let $r, s\in G(\bQ(\sqrt[3]{2},\zeta)/\bQ)$ such that:
+    \begin{equation*}
+      \begin{split}
+        \begin{cases}
+          r\colon \sqrt[3]{2}&\longmapsto \zeta\sqrt[3]{2} \\
+          r\colon \zeta&\longmapsto \zeta
+        \end{cases}
+      \end{split}
+      \quad\quad\quad
+      \begin{split}
+        \begin{cases}
+          s\colon \sqrt[3]{2} &\longmapsto \sqrt[3]{2} \\
+          s\colon \zeta &\longmapsto \zeta^2
+        \end{cases}
+      \end{split}
+    \end{equation*}
+
+    It's clear that $r$ and $s$ are correctly determined by this maps. Also, see
+    that any automorphism has to map $\sqrt[3]{2}$ to some other root of $X^3 -
+    2$ and can map $\zeta$ only to itself or $\zeta^2$ (which are the roots of
+    irreducible polynomial $X^2 + X + 1$). Moreover, we get $6$ different
+    automorphisms from $r$ and $s$, namely $\id, r, r^2, s, rs, r^2s$. So, the
+    Galois group $G(\bQ(\sqrt[3]{2}, \zeta)/\bQ)$ is generated by $r, s$. We
+    need to check that $\sqrt[3]{2} + \zeta$ goes to distinct elements through
+    every of these automorphisms which is quite easy to check.
+  \end{example}
+
+  \begin{example}
+    Let's find all intermediate fields of $\bQ\subset \bQ(\sqrt[3]{2},
+    \zeta)$. We already know that $\Gal(\bQ(\sqrt[3]{2},\zeta)/\bQ)\cong D_3$,
+    i.e. the dihedral group of triangle. It has four subgroups, namely $\{\id, r,
+    r^2\}$, $\{\id, s\}$, $\{\id, rs\}$ and $\{\id, r^2s\}$. So, we have four 
+    intermediate fields corresponding two these subgroups. 
+
+    Now we need to find the intermediate fields that correspond to the subgroups
+    of the Galois group. In this case, I like to think of roots of the $X^3-2$
+    as the vertices of a triangle and each automorphism is some permutation of
+    the vertices (i.e. some symmetry of the triangle).
+
+    \begin{center}
+      \begin{tikzcd}
+        & \sqrt[3]{2} &\\
+        & & \\
+        \zeta\sqrt[3]{2} \arrow[dash, uur] & & \zeta^2\sqrt[3]{2} \arrow[dash,
+        ll] \arrow[dash, uul]
+      \end{tikzcd}
+    \end{center}
+
+    The intuition to finding the intermediate fields is given by the fundamental
+    theorem. For every subgroup of the Galois group look for some elements that
+    are fixed by this group (or by the automorphism that generates this group)
+    and check if this is the correct element. In our case, it's easy to find all
+    of them thanks to the triangle.
+
+    \begin{center}
+        \begin{tikzcd}
+          & &  \bQ(\sqrt[3]{2}, \zeta) & \\
+          & &  & \bQ(\sqrt[3]{2}) \arrow[ul, dash] \\
+          \bQ(\zeta) \arrow[uurr, dash] & \bQ(\zeta\sqrt[3]{2}) \arrow[uur,
+          dash] &
+          \bQ(\zeta^2\sqrt[3]{2}) \arrow[uu, dash] & \\
+          & &  \bQ \arrow[ull, dash, "2"]\arrow[u, dash, "2"] \arrow[ul,
+          dash, "2"'] \arrow[uur, dash, "3"'] &\\
+        \end{tikzcd}
+    \end{center}
+  \end{example}
+
+  \newcommand{\bQQ}{\bQ(\sqrt[3]{2}, \sqrt{3})}
+  \begin{example}
+    Let's find all intermediate fields between $\bQ$ and $\bQQ$. In this case, 
+    it's not a splitting field of any polynomial over $\bQ$ so it's not normal.  
+    A friend of mine told me a nice metaphor, that sometimes it's nice to go up
+    to the skies and get a bird's eye view to a broader picture, 
+    as it allows to understand the details better.
+    In this case, it's
+    better to find a splitting field of some polynomial that contains $\bQQ$,
+    because then we will be able to use the fundamental theorem.
+
+    In our case a natural candidate is the splitting field of the polynomial
+    $(X^3-2)(X^2-3)$, which is $\bQQ(\zeta) = \bQ(\sqrt[3]{2}, \sqrt{3},
+    \zeta)$. What is the Galois group of this? This is a group of size $12$
+    thanks to Artin's theorem, and we see that the only possibilities for the
+    automorphisms is to move the roots of respective minimal polynomials to
+    other roots. In our case, we have at the automorphisms that move
+    $\sqrt[3]{2}$ and $\zeta$, but fix $\sqrt{3}$ and the analogous ones that
+    map $\sqrt{3}$ to $-\sqrt{3}$ and we get $12$ cases so it must be it! It's
+    clear that $G(\bQ(\sqrt[3]{2}, \sqrt{3}, \zeta)/\bQ) = D_3\times \bZ_2$.
+
+    There's a lot more subgroups of this group which I'm not going to point out,
+    try to do it as an exercise and find the respective intermediate fields.
+  \end{example}
+\end{document}