%% -*- coding:utf-8 -*-
\begin{appendices}

\chapter{Course prerequisites}
There are several prerequisites for the course there. They consists of
definitions, theorems and examples mostly taken from Wikipedia.

\section{Sets}

\begin{definition}[Class]
  A class is a collection of sets (or sometimes other mathematical
  objects) that can be unambiguously defined by a property that all
  its members share.
  \label{def:class}
\end{definition}

\begin{definition}[Disjoint union]
  Let \cite{wiki:disjointunion} $\{A_i: i \in I\}$ be a family of sets
  indexed by $I$. The
  disjoint union of this family is the set
  \[
  \sqcup_{i \in I} A_i = \cup_{i \in I}\left\{
  \left(x, i\right): x \in A_i
  \right\}.
  \]
  The elements of the disjoint union are ordered pairs $(x, i)$. Here $i$
  serves as an auxiliary index that indicates which $A_i$ the element $x$
  came from.
  \label{def:disjointunion}
\end{definition}

\begin{example}[Disjoint union]
  Let \cite{wiki:disjointunion} we have 2 sets $A_0 = \{1,2,3\}$ and
  $A_1 = \{1,2\}$.
  We can construct the following sets of pairs
  \begin{eqnarray}
  A_0^\ast = \left\{
  \left(1,0\right),
  \left(2,0\right),
  \left(3,0\right)
  \right\},
  \nonumber \\
  A_1^\ast = \left\{
  \left(1,1\right),
  \left(2,1\right)
  \right\}
  \nonumber
  \end{eqnarray}
  so
  \[
  A_0 \sqcup A_1 = A_0^\ast \cup A_1^\ast =
  \left\{
  \left(1,0\right),
  \left(2,0\right),
  \left(3,0\right),
  \left(1,1\right),
  \left(2,1\right)
  \right\}
  \]
  \label{ex:disjointunion}
\end{example}

\section{Groups}

\begin{definition}[Monoid]
  \label{def:monoid}
  The set of elements $M$ with defined binary operation $\circ$ we will call
  as a monoid if the following conditions are satisfied.
  \begin{enumerate}
  \item Closure: $\forall a, b \in M$: $a \circ b \in M$
  \item Associativity: $\forall a, b, c \in M$:
    $a \circ \left( b \circ c \right) =
    \left( a \circ b \right) \circ c$
  \item Identity element: $\exists e \in M$ such that
    $\forall a \in M$: $e \circ a = a \circ e = a$
  \end{enumerate}
\end{definition}

\begin{definition}[Group]
  \label{def:group}
  Let we have a set of elements $G$ with a defined binary operation
  $\circ$ that satisfied the following properties.
  \begin{enumerate}
  \item Closure: $\forall a, b \in G$: $a \circ b \in G$
  \item Associativity: $\forall a, b, c \in G$:
    $a \circ \left( b \circ c \right) =
    \left( a \circ b \right) \circ c$
  \item Identity element: $\exists e \in G$ such that
    $\forall a \in G$: $e \circ a = a \circ e = a$
  \item Inverse element: $\forall a \in G$ $\exists a^{-1} \in G$ such that
    $a \circ a^{-1} = e$
  \end{enumerate}
  In this case $\left(G, \circ\right)$ is called as group.
\end{definition}
Therefore the group is a \mynameref{def:monoid} with inverse element
property.

\begin{example}[Group $\mathbb{Z}/2\mathbb{Z}$]
  Consider a set of 2 elements: $G = \left\{0, 1\right\}$ with the
  operation $\circ$ defined by the table \ref{tab:CayleyZ2Z}.
  \begin{table}
    \centering
    \caption{Cayley table for $\mathbb{Z}/2\mathbb{Z}$}
    \label{tab:CayleyZ2Z}
    \begin{tabular}{l|ll}
      \toprule
      $\circ$ & 0 & 1 \\
      \midrule
      0 & 0 & 1 \\
      1 & 1 & 0 \\
      \bottomrule
    \end{tabular}
  \end{table}

  The identity element is $0$ i.e. $e = 0$.
  Inverse element is the element itself
  because $\forall a \in G$: $a \circ a = 0 = e$.

  See also example \ref{ex:quotientgroup}
  \label{ex:groupZ2}
\end{example}

\begin{definition}[Order of element in group]
  Order, sometimes period, of an element a of a group is the smallest
  positive integer $m$ such that $a^m = e$ (where $e$ denotes the identity
  element of the group, and am denotes the product of $m$ copies of
  $a$). If no such m exists, a is said to have infinite order.
  \label{def:grouporder}
\end{definition}

\begin{theorem}[Lagrange]
  For any finite group $G$, the order (number of elements) of every
  subgroup $H$ of $G$ divides the order of $G$.
  \label{thm:lagrange}
\end{theorem}

\begin{theorem}[Finite group of prime order]
  Let $G$ is a finite group of prime order i.e.
  $\left|G\right| = p$, where $p$ is prime number. Then $G$ is a
  \mynameref{def:cyclicgroup}.
  \label{thm:finite_group_of_prime_order}
  \begin{proof}
    Let $g \in G$ such that $g \ne e$ then, by theorem
    \ref{thm:lagrange},
    we have
    $\left|\left<g\right>\right| \mid \left|G\right|$. Thus
    $\left|\left<g\right>\right| = p$ because
    $\left|G\right| = p$ and $p$ is a prime number. As result we have
    $G = \left<g\right>$ and therefore $G$ is a cyclic group.
  \end{proof}
\end{theorem}

\begin{definition}[Subgroup]
  Let we have a \mynameref{def:group} $\left(G, \circ\right)$. The
  subset $S \subset G$ is called as subgroup if $\left(S,
  \circ\right)$ is a \mynameref{def:group}.
  \label{def:subgroup}
\end{definition}

\begin{definition}[Proper subgroup]
  A proper subgroup of a group $G$ is a \mynameref{def:subgroup} $H$
  which is a proper subset of $G$ (i.e. $H \ne G$) \cite{wiki:group}
  \label{def:propersubgroup}
\end{definition}

\begin{definition}[Coset]
  If $G$ is a group, and $H$ is a subgroup of $G$, and $g$ is an
  element of $G$, then
  \[
  gH = \left\{ gh \vert h \in H\right\}
  \]
  is the left coset of $H$ in $G$ with respect to $g$, and
  \[
  Hg = \left\{ hg \vert h \in H\right\}
  \]
  is the right coset of $H$ in $G$ with respect to $g$.
  \label{def:coset}
\end{definition}

\begin{definition}[Normal subgroup]
  A subgroup, $N$, of a group $G$, is called a normal subgroup if it
  is invariant under conjugation i.e.
  \[
  N \triangleleft G \Leftrightarrow
  \forall n \in N, \forall g \in G, g n g^{-1} \in N
  \]
  The definition taken from \cite{wiki:normalsubgroup}

  We also can write normality in the \mynameref{def:coset} notation as
  follows.
  $N \triangleleft G$ if $\forall g \in G: gN = Ng$.
  \label{def:normalsubgroup}
\end{definition}

\begin{definition}[Simple group]
   A simple group is a nontrivial group whose only
   \mynameref{def:normalsubgroup}s are the trivial group and the group
   itself.

   The definition taken from \cite{wiki:simplegroup}
   \label{def:simplegroup}
\end{definition}

\begin{definition}[Quotient group]
  A quotient group or factor group is a mathematical group obtained by
  aggregating similar elements of a larger group using an equivalence
  relation that preserves the group structure. For example, the
  \mynameref{def:cyclicgroup} of addition modulo n can be obtained from
  the integers by
  identifying elements that differ by a multiple of n and defining a
  group structure that operates on each such class (known as a
  congruence class) as a single entity. It is part of the mathematical
  field known as group theory \cite{wiki:quotientgroup}.

  In a quotient of a group, the equivalence class of the identity
  element is always a normal subgroup of the original group, and the
  other equivalence classes are precisely the cosets of that normal
  subgroup. The resulting quotient is written $G / N$, where $G$ is the
  original group and $N$ is the normal subgroup.

  In other words the quotient group can be defined as a set of all
  left \mynameref{def:coset}s (they also equal to the right cosets as
  soon as $N$ is a \mynameref{def:normalsubgroup}):
  \[
  G/N = \left\{a N: a \in G\right\}
  \]

  See also example \mynameref{ex:s3a3quotientgroup}.
  \label{def:quotientgroup}
\end{definition}

\begin{theorem}[Quotient group]
  If $G$ is a group and $H \triangleleft G$ then the operation $aH
  \cdot bH = a \cdot H$ makes $G/H$ a group with identity $H$ and
  inverse element $\left(a H\right)^{-1} = a^{-1} H$.
  \begin{proof}
    See \cite{book:kostrikin} p. 33.
  \end{proof}
  \label{thm:quotientgroup}
\end{theorem}

\begin{example}[Quotient group]
  Consider \cite{wiki:quotientgroup} a group of integers $\mathbb{Z}$
  (under addition) and the
  subgroup $2\mathbb{Z}$ of all even integers. This is a normal
  subgroup, because $\mathbb{Z}$ is \mynameref{def:abeliangroup}. There
  are only two \mynameref{def:coset}s: the set
  of even integers and the set of odd integers; therefore, the
  quotient group $\mathbb{Z}/2\mathbb{Z}$ is the
  \mynameref{def:cyclicgroup} with two
  elements. This
  quotient group is isomorphic with the set $\left\{ 0, 1 \right\}$
  with addition modulo 2; informally, it is sometimes said that
  $\mathbb{Z}/2\mathbb{Z}$
  equals the set $\left\{ 0, 1 \right\}$ with addition modulo 2.

  See also example \ref{ex:groupZ2}.
  \label{ex:quotientgroup}
\end{example}

\begin{theorem}[Correspondence theorem]
  The correspondence theorem, sometimes referred to as the fourth
  isomorphism theorem or the lattice theorem, states that if
  $N$ is a \mynameref{def:normalsubgroup} of a group  $G$, then there
  exists a bijection from the set of all subgroups $A$ of $G$
  containing $N$, onto the set of all subgroups of the quotient group
  $G/N$. The structure of the subgroups of $G/N$ is exactly the same
  as the structure of the subgroups of $G$ containing $N$, with
  $N$ collapsed to the identity element \cite{wiki:correspondence}.
  \label{thm:correspondence}
\end{theorem}

\begin{definition}[Commutator]
  The commutator of two elements, $g$ and $h$, of a group $G$, is the
  element \cite{wiki:commutator}
  \[
  \left[g, h\right] = g^{-1} h^{-1} g h
  \]
  \label{def:commutator}
\end{definition}

\begin{definition}[Commutator subgroup]
  The commutator subgroup or derived subgroup of a group is the
  subgroup generated by all the \mynameref{def:commutator}s of the group
  \cite{wiki:commutatorsubgroup}.
  \label{def:commutatorsubgroup}
\end{definition}

\begin{theorem}[About quotient group and commutator subgroup]
  Given a group $G$ a \mynameref{def:quotientgroup} $G/N$ is an
  \mynameref{def:abeliangroup} if and only if $N \supseteq \left[G,
    G\right]$
  \cite{wiki:commutatorsubgroup}
  \label{thm:about_quotient_group_and_commutatorsubgroup}
\end{theorem}

\begin{definition}[Abelianization]
  The quotient $G/[G,G]$ is an abelian group
  (as it follows from theorem \ref{thm:about_quotient_group_and_commutatorsubgroup})
  called the abelianization of $G$ or $G$ made abelian.
  \cite{wiki:commutatorsubgroup}
  \label{def:abelianization}
\end{definition}

\subsection{Cyclic group}

\begin{definition}[Cyclic group]
  A cyclic group or monogenous group is a group that is generated by a
  single element.
  Note that \mynameref{ex:groupZ2} is a cyclic group.
  See example \ref{ex:multiplicativegroup}.
  \label{def:cyclicgroup}
\end{definition}

\begin{theorem}[Fundamental theorem of cyclic groups]
  In abstract algebra, every subgroup of a \mynameref{def:cyclicgroup} is
  cyclic. Moreover, for a finite cyclic group of order $n$, every
  \mynameref{def:subgroup}'s order is a divisor of $n$, and there is
  exactly one subgroup for each divisor.This result has been called the
  fundamental theorem of cyclic groups
  \cite{wiki:subgroups_of_cyclic_groups}
  \label{thm:fundamentaltheoremofcyclicgroup}
\end{theorem}

\begin{theorem}[About subgroups of a cyclic group]
  Let \cite{mathstackexchange:cyclicgroupssubgroup}
    $G=\left<a\right>$ be a cyclic group.
  \begin{enumerate}
  \item Every subgroup $S$ of $G$ is cyclic
  \item If $\left|G\right| = n$, then $G$ has a unique subgroup of
    order $d$ for each divisor $d$ of $n$
  \end{enumerate}
  See example \ref{ex:multiplicativegroup}
  \label{thm:subgroupofcyclicgroup}
\end{theorem}

\subsection{Group action}

\begin{definition}[Action]
  An action of a group is a way of
  interpreting the elements of the
  group as "acting" on some space in a way that preserves the structure
  of that space. See also \cite{wiki:groupaction}.
  \label{def:action}
\end{definition}

\begin{definition}[Orbit]
  Consider \cite{wiki:groupaction} a group $G$ acting on a set
  $X$. The orbit of an element $x \in X$
  is the set of elements in $X$ to which $x$ can be moved by the elements
  of $G$:
  \[
  Orb\left(x\right) = \left\{y \in X: \exists g \in G: y = g \cdot x \right\}
  \]
  The orbit of element $x$  is also denoted as $G\left(x\right)$.
  \label{def:orbit}
\end{definition}

\begin{definition}[Fixed point]
  The set of points of $X$ fixed by a group action are called the
  group's set of fixed points, defined by
  \[
  \left\{
  x: g x = x, \forall g \in G
  \right\}.
  \]
  see also \cite{mathworld:groupfixedpoint}.
  \label{def:fixedpoint}
\end{definition}

\begin{definition}[Stabilizer subgroup]
  For every $x$ in $X$, we define \cite{wiki:groupaction} the
  stabilizer subgroup of $G$ with
  respect to $x$ (also called the isotropy group) as the set of all
  elements in $G$ that fix $x$:
  \[
  G_{x}=\{g\in G\mid g\cdot x=x\}
  \]
  \label{def:stabilizersubgroup}
\end{definition}

\begin{theorem}[Orbit-stabilizer theorem]
  If group $G$ and the set the group acting $X$ are finite then

  \[
  \left|G\right| = \left|Orb\left(x\right)\right|\left|G_x\right|
  \]
  where $x \in X$, $G\left(x\right)$ - is the \mynameref{def:orbit},
  $G_x$ - \mynameref{def:stabilizersubgroup}.

  Note: the result was got from \cite{wiki:groupaction} as
  orbit-stabilizer theorem + \mynameref{thm:lagrange}
  \label{thm:orbitstabilizertheorem}
\end{theorem}

\begin{definition}[Transitive group action]
  The action of $G$ on $X$ is called \cite{wiki:groupaction}
  transitive if $X$ is non-empty and if for each pair $x, y \in X$ there
  exists a $g \in G$ such that $gx = y$.
  \label{def:transitive}
\end{definition}

\begin{definition}[Free group action]
  The action of $G$ on $X$ is called \cite{wiki:groupaction}
  free if, given $g, h \in G$, the existence of an $x \in X$ with $g(x)
  = h(x)$ implies $g = h$.
  \label{def:freeaction}
\end{definition}

\begin{definition}[Quotient of the group action]
  Consider \cite{wiki:groupaction} a group $G$ acting on a set
  $X$. The set of all \mynameref{def:orbit}s of $X$ under the action of
  $G$ is written as $X/G$, and is called the quotient of the action.
  \label{def:quotientofgroupaction}
\end{definition}

\subsection{Direct product}

\begin{definition}[Direct product]
  Given groups $G$ and $H$, the direct product $G \times H$ is defined as follows:

  The underlying set is the Cartesian product, $G \times H$. That is, the
  ordered pairs $\left(g,h\right)$, where $g \in G$ and $h \in H$.
  The binary operation on $G \times H$ is defined component-wise:
  \[
  \left(g_1,h_1\right)\left(g_2,h_2\right) =
  \left(g_1 \cdot g_2, h_1 \circ h_2\right)
  \]
  The resulting algebraic object satisfies the axioms for a group.
  See \cite{wiki:directproduct}.

  The direct product of 2 \mynameref{def:abeliangroup}s is also called
  \mynameref{def:directsum}

  \label{def:directproduct}
\end{definition}

\begin{property}[Direct product of groups]
  Let $G_1$, $G_2$ - \mynameref{def:group}s and $G = G_1 \times G_2$ -
  \mynameref{def:directproduct} of the groups. Then
  \begin{enumerate}
  \item $G_1 \cong \left(G_1, 1_{G_2}\right)$ and
    $G_2 \cong \left(1_{G_1}, G_2\right)$
  \item $G_1$ and $G_2$ are \mynameref{def:normalsubgroup}s in $G$
  \end{enumerate}
  See \cite{wiki:directproduct}.
  \label{property:directproduct}
\end{property}

\begin{property}[Quotient of direct product]
  Let $G_1$, $G_2$ - \mynameref{def:group}s and $G = G_1 \times G_2$ -
  \mynameref{def:directproduct} of the groups.
  Then
  \[
  G/G_1 \cong G_2
  \]
  and
  \[
  G/G_2 \cong G_1
  \]
  \label{property:directproductquotient}
  \begin{proof}
    Lets prove the first claim:
    $G/G_1 \cong G_2$ (the second one is analogous).
    Consider projection
    $\pi: G \xrightarrow[(g_1, g_2) \to (1_{G_1}, g_2)]{} G_2$. The
    $\pi$ is \mynameref{def:homomorphism}. Really
    let $a = (a_1, a_2), b = (b_1, b_2) \in G$ where
    $a_{1}, b_1 \in G_1$, $a_2, b_{2} \in G_2$:
    \begin{eqnarray}
      \pi\left(a \cdot b\right) = \pi\left(
      (a_1, a_2) \cdot (b_1, b_2)
      \right) =
      \nonumber \\
      = \pi\left(
      (a_1 b_1, a_2 b_2)
      \right) = a_2 b_2 =
      \pi\left(a\right) \pi\left(b\right)
      \nonumber
    \end{eqnarray}
    As result $\pi$ is a \mynameref{def:homomorphism}.

    The $\pi$ is also a \mynameref{def:surjection} because $\forall a_2
    \in G_2$ $\exists a \in G$ such that $\pi(a) = a_2$. Really we can
    use $a = (1_{G_1}, a_2)$.

    Therefore by \mynameref{thm:firstisomorphism} one can get
    \[
    G/\ker{\pi} \cong G_2,
    \]
    but $\forall g_1 \in G_1$ we have $(g_1, 1_{G_2}) \in \ker{\pi}$.
    And conversely $\forall k \in \ker \pi$ we have $k = (k_1, k_2)$
    where $k_1 \in G_1$ and $k_2 = 1_{G_2}$.
    As result (see property \ref{property:directproduct})
    \[
    \ker \pi = \left(G_1, 1_{G_2}\right) \cong G_1.
    \]
    Therefore
    \[
    G/G_1 \cong G_2.
    \]
  \end{proof}
\end{property}

\subsection{Sylow theorems}

\begin{corollary}[Sylow]
  Given a finite group $G$ and a prime number $p$ dividing the order of $G$,
  then there exists an element (and hence a subgroup) of order $p$ in
  $G$ \cite{wiki:sylow}
  \label{cor:sylow}
\end{corollary}

\subsection{Abelian group}

\begin{definition}[Abelian group]
  Let we have a \mynameref{def:group} $\left(G, \circ\right)$.
  The group is called an Abelian or commutative if
  $\forall a, b \in G$ it holds $a \circ b = b \circ a$.
  \label{def:abeliangroup}
\end{definition}

\begin{theorem}[About order of element of an Abelian group]
  If $G$ is a finite \mynameref{def:abeliangroup} and $m$ is the maximal
  order of the elements of $G$ then the order of every element of $G$
  divides $m$
  \label{thm:abelianelementorder}
\end{theorem}

\begin{theorem}
  Let $G$ is an \mynameref{def:abeliangroup} and $n = \left|G\right|$
  the group order (number of elements) then $\forall g \in G$ the
  following statement holds
  \[
  g^n = e,
  \]
  there $e$ is the group identity.
  \begin{proof}
    Let $m$ is the maximal order of group $G$. In this case by
    \mynameref{thm:lagrange} $m \mid n$ i. e. $n = k_1 m$ where $k_1 \in
    \mathbb{Z}$. Let $l$ is the order of $g$ i.e. $g^l = e$. By the
    theorem \ref{thm:abelianelementorder} $l \mid m$ i.e.
    $m = k_2 l$. Thus
    \[
    g^n = \left(g^m\right)^{k_1} =
    \left(g^l\right)^{k_2 k_1} = e.
    \]
  \end{proof}
  \label{thm:abelianelement}
\end{theorem}

\begin{theorem}[Cyclic Group is Abelian]
  Let $G$ be a \mynameref{def:cyclicgroup} then $G$ is
  \mynameref{def:abeliangroup}
  \label{thm:cyclic_group_is_abelian}
  \begin{proof}
    As soon as $G = \left<g\right>$ then
    $\forall x,y \in G, \exists n,m \in \mathbb{N}$ such that
    $x = g^n, y = g^m$. In the case
    \[
    xy = g^n g^m = g^{n+m} = g^m g^n = yx,
    \]
    i.e. $G$ is abelian.
  \end{proof}
\end{theorem}

\begin{remark}[Not every Abelian group is cyclic]
  The theorem statement cannot be reversed i.e. there exist abelian
  group that are not cyclic. For example well known \mynameref{def:v4}
  $V_4$  that is an
  \mynameref{def:abeliangroup} but not \mynameref{def:cyclicgroup}
\end{remark}

\begin{definition}[Klein four group]
  \label{def:v4}
  The Klein four group $V_4$ is a group of order 4
  \footnote{there are only 2 groups of order 4: $V_4$ and
    $\mathbb{Z}/4\mathbb{Z}$ (the only cyclic group of order 4)}
  The Cayley table for the group is on the table \ref{tab:CayleyV4}
  \cite{wiki:klein4group}.
  \begin{table}
    \centering
    \caption{Cayley table for $V_4$}
    \label{tab:CayleyV4}
    \begin{tabular}{l|llll}
      \toprule
      $\circ$ & $e$ & $a$ & $b$ & $c$\\
      \midrule
      $e$ & $e$ & $a$ & $b$ & $c$ \\
      $a$ & $a$ & $e$ & $c$ & $b$ \\
      $b$ & $b$ & $c$ & $e$ & $a$ \\
      $c$ & $c$ & $b$ & $a$ & $e$ \\
      \bottomrule
    \end{tabular}
    \end{table}
\end{definition}

\begin{example}[Klein four group]
  There are several examples of $V_4$:
  \begin{itemize}
  \item $V_4 \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$
  \item $V_4 \cong \left(\mathbb{Z}/8\mathbb{Z}\right)^\times \cong \left(\mathbb{Z}/12\mathbb{Z}\right)^\times$
  \end{itemize}
\end{example}

\begin{definition}[Direct sum]
  The direct sum of two abelian groups $A$ and
  $B$ is another abelian group $A \oplus B$ consisting of the ordered
  pairs
  $\left(a, b\right)$ where $a \in A$ and $b \in B$.
  \cite{wiki:directsum}
  See also \mynameref{def:directproduct}.

  \label{def:directsum}
\end{definition}

\begin{definition}[Finitely generated abelian group]
  An \mynameref{def:abeliangroup} $(G, +)$ is called finitely generated
  \cite{wiki:fgagroup}
  if there exist finitely many elements $x_1, \dots, x_s$ in $G$ such that
  $\forall x \in G$:
  \begin{equation}
    x = n_1 x_1 + \dots + n_s x_s
    \label{eq:fgagroup}
  \end{equation}
  with $n_i \in \mathbb{Z}$. In this case we say that $\{x_1, \dots,
  x_s\}$ is a generating set of $G$.

  In the (\ref{eq:fgagroup}) we have the following:
  \[
  n_i x_i = \underbrace{x_i + \dots x_i}_{n_i \text{ times}}
  \]
  \label{def:fgagroup}
\end{definition}

\begin{theorem}[The fundamental theorem of finitely generated abelian
    groups]
  Every \mynameref{def:fgagroup} $G$ is isomorphic to a
  \mynameref{def:directsum} of primary cyclic groups and infinite cyclic
  groups. A primary cyclic group is one whose order is a power of a
  prime. That is, every finitely generated abelian group is isomorphic
  to a group of the form
  \[
  \mathbb{Z}^n \oplus \mathbb{Z}_{q_1} \oplus
  \dots \oplus \mathbb{Z}_{q_t}
  \]
  where the rank $n \ge 0$, and the numbers $q_1, \dots , q_t$ are
  powers of (not
  necessarily distinct) prime numbers. In particular, $G$ is finite if
  and only if $n = 0$. The values of $n, q_1, \dots , q_t$ are (up to
  rearranging the indices) uniquely determined by $G$.
  The statement was took from \cite{wiki:fgagroup}.
  \label{thm:fgagroup}
\end{theorem}

\begin{theorem}[Simple subgroup of an abelian group]
  Every non-simple abelian group has a simple normal subgroup.
  ??? add link ???
  \label{thm:simple_subgroup_of_abelian}
\end{theorem}

\section{Permutations}

\begin{example}[Permutation]
  The following permutation
  \[ \pi =
  \begin{array}{c}
    1 \to 2 \\
    2 \to 5 \\
    3 \to 4 \\
    4 \to 3 \\
    5 \to 1
  \end{array}
  \]
  can be also written in different forms. The most common one is the following:
  \[
  \pi = \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    2 & 5 & 4 & 3 & 1
  \end{pmatrix}.
  \]
  In the permutation we can see 2 cycles:
  $1 \to 2 \to 5 \to 1$ and $3 \to 4 \to 3$. The first cycle can be
  written as $(1,2,5)$ (or $(5,1,2)$ or $(2,5,1)$) and the second
  one as $(3,4)$ (or $(4,3)$). The cycles gives us the shortest form
  of writing the permutation:
  \[
  \pi = (1,2,5)(3,4) = (3,4)(5,1,2).
  \]

  If we have 2 permutations
  \[
  \pi_1 = \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    3 & 1 & 5 & 2 & 4
  \end{pmatrix}.
  \]
  and
  \[
  \pi_2 = \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    2 & 5 & 4 & 3 & 1
  \end{pmatrix}.
  \]
  then we can combine them into the new one (via a multiplication)
  \[ \pi = \pi_1 \pi_2 =
  \begin{array}{c}
    1 \to 2 \to 1 \\
    2 \to 5 \to 4\\
    3 \to 4 \to 2\\
    4 \to 3 \to 5\\
    5 \to 1 \to 3
  \end{array} =
  \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    1 & 4 & 2 & 5 & 3
  \end{pmatrix} = (1)(2,4,5,3) = (2,4,5,3)
  \]

  We have identity element
  \[
  e =  \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    1 & 2 & 3 & 4 & 5
  \end{pmatrix}
  \]
  such that $\forall \pi: \pi e = e \pi = \pi$.

  For every $\pi$ we can define $\pi^{-1}$ such that
  $\pi \pi^{-1} = \pi^{-1} \pi = e$.

  For our example
  \[
  \pi = \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    2 & 5 & 4 & 3 & 1
  \end{pmatrix} = (1,2,5)(3,4)
  \]
  we have
  \[
  \pi^{-1} = \begin{pmatrix}
    2 & 5 & 4 & 3 & 1 \\
    1 & 2 & 3 & 4 & 5
  \end{pmatrix} =
  \begin{pmatrix}
    1 & 2 & 3 & 4 & 5 \\
    5 & 1 & 4 & 3 & 2
  \end{pmatrix} = (1,5,2)(3,4)
  \]

  As result we have a \mynameref{def:group} of permutations.
  \label{ex:permutation}
\end{example}

\begin{definition}[Parity of a permutation]
  When $X$ is a finite set of at least two elements, the permutations of
  X (i.e. the bijective functions from $X$ to $X$) fall into two classes
  of equal size: the even permutations and the odd permutations. If
  any total ordering of $X$ is fixed, the parity (oddness or evenness)
  of a permutation $\sigma$ of $X$ can be defined as the
  parity of the number of inversions for $\sigma$, i.e., of pairs of elements
  $x, y$ of $X$ such that $x < y$ and $\sigma (x) > \sigma (y)$
  \cite{wiki:paritypermutation}.
  \label{def:paritypermutation}
\end{definition}

\begin{example}[Parity of a permutation]
  For the following permutation $(2,5,4,1,3)$ we have the following
  inversions
  \begin{eqnarray}
    (2,5,4,1,3) \to_{(1,2)}
    (1,5,4,2,3) \to_{(5,2)}
    \nonumber \\
    (1,2,4,5,3) \to_{(3,4)}
    (1,2,3,5,4) \to_{(5,4)}
    (1,2,3,4,5)
    \nonumber
  \end{eqnarray}
  We have made 4 inversions and as result the permutation is even.

  The same result can be got if we use the following equation $l - 1$
  where $l$ is the circle length (5 in our case)
  \label{ex:paritypermutation}
\end{example}

\begin{definition}[Alternating group]
  Alternating group \cite{wiki:alteringgroup} is the group of even
  permutations (see definition \ref{def:paritypermutation}) of a finite
  set. The alternating group on a set of $n$ elements is called the
  alternating group of degree n, or the alternating group on n letters
  and denoted by $A_n$.
  \label{def:alternatinggroup}
\end{definition}

\begin{example}[$S_n$ group]
  If we a have a permutation of $n$ elements then it's possible to do
  by means of $n!$ ways.
  \label{ex:sngroup}
\end{example}

\begin{example}[$S_1$ group]
  $S_1$ permutation of 1 element consists of only one element $e$ -
  the simplest possible group
  \label{ex:s1group}
\end{example}

\begin{example}[$S_2$ group]
  $S_2$ permutation consists of 2 elements:
  \begin{enumerate}
  \item identity:
    \(
    e = \begin{pmatrix}
      1 & 2 \\
      1 & 2
    \end{pmatrix}
    \)
  \item transposition:
    \(
    \tau = \begin{pmatrix}
      1 & 2 \\
      2 & 1
    \end{pmatrix}
    \)
  \end{enumerate}
  It's easy to see that the Cayley table has the form \ref{tab:CayleyS2}
    \begin{table}
    \centering
    \caption{Cayley table for $S_2$}
    \label{tab:CayleyS2}
    \begin{tabular}{l|ll}
      \toprule
      $\circ$ & $e$ & $\tau$ \\
      \midrule
      $e$ & $e$ & $\tau$ \\
      $\tau$ & $\tau$ & $e$ \\
      \bottomrule
    \end{tabular}
    \end{table}
  \label{ex:s2group}
\end{example}

\begin{example}[$S_3$ group]
    $S_3$ permutation consists of 6 elements: $e, \tau, \tau_1, \tau_2,
    \sigma, \sigma_1$. The most important are $e, \tau$ and $\sigma$
    and all others can be obtained from this ones (see table
    \ref{tab:CayleyS3}).
  \begin{enumerate}
  \item identity
    \(
    e = \begin{pmatrix}
      1 & 2 & 3\\
      1 & 2 & 3
    \end{pmatrix}
    \)
  \item transposition:
    \(
    \tau = \begin{pmatrix}
      1 & 2 & 3\\
      2 & 1 & 3
    \end{pmatrix}
    \)
  \item circle:
    \(
    \sigma = \begin{pmatrix}
      1 & 2 & 3\\
      2 & 3 & 1
    \end{pmatrix}
    \)
  \end{enumerate}

  Another elements of $S_3$:
  \(
  \tau_1 = \begin{pmatrix}
    1 & 2 & 3\\
    1 & 3 & 2
  \end{pmatrix}
  \),
  \(
  \tau_2 = \begin{pmatrix}
    1 & 2 & 3\\
    2 & 1 & 3
  \end{pmatrix}
  \) and
  \(
  \sigma_1 = \begin{pmatrix}
    1 & 2 & 3\\
    3 & 1 & 2
  \end{pmatrix}
  \).

  \begin{table}
    \centering
    \caption{Cayley table for $S_3$ \cite{wiki:permutationgroups}}
    \label{tab:CayleyS3}
    \begin{tabular}{l|llllll}
      \toprule
      $\circ$ & $e$ & $\sigma$ & $\sigma_1$ & $\tau$ & $\tau_1$ & $\tau_2$\\
      \midrule
      $e$ & $e$ & $\sigma$ & $\sigma_1$ & $\tau$ & $\tau_1$ & $\tau_2$\\
      $\sigma$ & $\sigma$ & $\sigma_1$ & $e$ & $\tau_2$ & $\tau$ & $\tau_1$\\
      $\sigma_1$ & $\sigma_1$ & $e$ & $\sigma$ & $\tau_1$ & $\tau_2$ & $\tau$\\
      $\tau$ & $\tau$ & $\tau_1$ & $\tau_2$ & $e$ & $\sigma_1$ & $\sigma$\\
      $\tau_1$ & $\tau_1$ & $\tau_2$ & $\tau$ & $\sigma$ & $e$ & $\sigma_1$\\
      $\tau_2$ & $\tau_2$ & $\tau$ & $\tau_1$ & $\sigma_1$ & $\sigma$ & $e$\\
      \bottomrule
    \end{tabular}
  \end{table}
  As we can see from the table \ref{tab:CayleyS3} the elements $e,
  \sigma, \sigma_1$ forms a subgroup of $S_3$ moreover all the
  permutation (see definition \ref{def:paritypermutation}). I.e. there
  we will have \mynameref{def:alternatinggroup} $A_3$.
  \label{ex:s3group}
\end{example}

\begin{example}[$S_3/A_3$ quotient group]
  Lets consider the following \mynameref{def:quotientgroup}
  $S_3/A_3$. As we can see all elements of $S_3$ can be divided into 2
  classes each of them with size $3 = \left|A_3\right|$: $E= A_3 =
  \left\{e, \sigma, \sigma_1\right\}$
  and $G = \left\{\tau, \tau_1, \tau_2\right\}$. If we take an element
  $x_1 \in E$ and multiply it on another element of $x_2 \in E$ we
  will get $x_1 x_2 \in E$ (see table \ref{tab:CayleyS3}) i.e. $E
  \cdot E =
  E$. For $G$ we can get $G \cdot G = E$ and $E \cdot G = G \cdot E =
  G$. Therefore $S_3/A_3 = \{E, G\}$ forms a group of order 2. Thus
  \[
  S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}
  \]
  \label{ex:s3a3quotientgroup}
\end{example}

\begin{definition}[Cycle]
  A cyclic permutation (or cycle) is a permutation of the elements of
  some set $X$ which maps the elements of some subset $S$ of $X$ to each
  other in a cyclic fashion, while fixing (that is, mapping to
  themselves) all other elements of $X$. If $S$ has $k$ elements, the cycle
  is called a $k$-cycle \cite{wiki:cyclicpermutation}.
  \label{def:cycle}
\end{definition}

\begin{example}[Cycle]
  The following permutation is a 3-cycle:
  \[
  (1,2,3) = \begin{array}{c}
                1 \to 2 \\
                2 \to 3 \\
                3 \to 1
              \end{array}
  \]
  \label{ex:cycle}
\end{example}

\begin{definition}[Transposition]
  A cycle with only two elements is called a transposition.
  \cite{wiki:cyclicpermutation}
  \label{def:transposition}
\end{definition}

\begin{example}[Transposition]
  The following permutation is a transposition:
  \[
  (1,2) = \begin{array}{c}
                1 \to 2 \\
                2 \to 1
              \end{array}
  \]
  \label{ex:transposition}
  See also example \ref{ex:transposition_product}
\end{example}

\begin{example}[Transposition product]
  \label{ex:transposition_product}
  The example shows a product of 2 \mynameref{def:transposition}s with a
  same element:
  \begin{eqnarray}
    (a,c)(a,b) =
    \begin{array}{c}
      a \to b \\
      b \to a \to c\\
      c \to a
    \end{array} = (a,b,c)
    \nonumber
  \end{eqnarray}
\end{example}

\begin{theorem}
  Every permutation can be represented as a product of
  \mynameref{def:transposition}s
  \begin{proof}
    This is because any cycle $(a_1, a_2, a_3, \dots, a_{n-1}, a_n)$
    can be represented as a product of transpositions as follows
    \[
    (a_1, a_2, a_3, \dots, a_{n-1}, a_n) =
    (a_1, a_n)(a_1, a_{n-1}) \dots (a_1, a_3) (a_1, a_2)
    \]
  \end{proof}
  \label{thm:permutationrepresent}
\end{theorem}

\begin{corollary}
  $S_n$ is generated by any combination of a \mynameref{def:transposition} and
  $n$-cycle if and only if $n$ is prime
  \label{cor:sn}
  \begin{proof}
    See \cite{mathstackexchange:sn}
  \end{proof}
\end{corollary}

\section{Rings and Fields}

\subsection{Rings}

\begin{definition}[Ring]
  Consider a set $R$ with 2 binary operations defined. The first one
  $\oplus$ (addition) and elements of $R$ forms an
  \mynameref{def:abeliangroup}
  under this operation. The second one is $\odot$ (multiplication) and
  the elements of $R$ forms a \mynameref{def:monoid} under
  the operation. The two binary operations are connected each other
  via the following distributive law
  \begin{itemize}
  \item Left distributivity:
    $\forall a,b,c \in R$:
    $a \odot \left(b \oplus c\right) =
    a \odot b \oplus a \odot c$
  \item Right distributivity:
    $\forall a,b,c \in R$:
    $\left( a \oplus b \right) \odot c =
    a \odot c \oplus b \odot c$

  The identity element for $\left(R, \oplus\right)$ is denoted as $0$
  (additive identity).
  The identity element for $\left(R, \odot\right)$ is denoted as $1$
  (multiplicative identity).

  The inverse element to $a$ in $\left(R, \oplus\right)$ is denoted as $-a$
  \end{itemize}

  In this case $\left(R, \oplus, \odot\right)$ is called as ring.
  \label{def:ring}
\end{definition}

The \mynameref{def:ring} is a generalization of integer numbers conception.
\begin{example}[Ring of integers $\mathbb{Z}$]
  The set of integer numbers $\mathbb{Z}$ forms a \mynameref{def:ring}
  under $+$ and $\cdot$ operations i.e. addition $\oplus$ is
  $+$ and multiplication $\odot$ is $\cdot$. Thus for integer
  numbers we have the following \mynameref{def:ring}:
  $\left(\mathbb{Z}, +, \cdot\right)$
  \label{ex:ring}
\end{example}

\begin{definition}[Multiplicative group]
  If $R$ is a ring then the multiplicative group
  $\left(R\right)^\times$
  is a group of
  invertible elements of $R$ with the defined multiplication operation.
  \label{def:multiplicativegroup}
\end{definition}

\begin{example}[Multiplicative group of integers modulo $n$]
  Lets consider the following group
  $\left(\mathbb{Z}/9\mathbb{Z}\right)^\times =
  \left\{1,2,4,5,7,8\right\}$
  \cite{wiki:multiplicativegroup}

  The group has order 6. The group generator is $2$:
  $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 7, 2^5 = 5, 2^6 = 1$.

  Accordingly theorem
  \ref{thm:subgroupofcyclicgroup}, there are only 2
  \mynameref{def:propersubgroup} not equal to $id$.

  One of them is a cyclic group of order 2: $\left\{1,8\right\} \cong
  \mathbb{Z}/2\mathbb{Z}$.

  Another one is also cyclic and has order 3: $\left\{1, 4, 7\right\}
  \cong
  \mathbb{Z}/3\mathbb{Z}$.

  \label{ex:multiplicativegroup}
\end{example}

\subsection{Ideals}

\begin{definition}[Ideal]
  \label{def:ideal}
  Lets we have the \mynameref{def:ring}
  $\left(R, \oplus, \odot\right)$. Subset $I \subset R$ will be an
  ideal if it satisfied the following conditions
  \begin{enumerate}
  \item $\left(I, \oplus\right)$ is \mynameref{def:subgroup} of
    $\left(R, \oplus\right)$
  \item $\forall i \in I$ and $\forall r \in R$:
    $i \odot r \in I$ and $r \odot i \in I$
  \end{enumerate}
\end{definition}

\begin{example}[Ideal $2 \mathbb{Z}$]
  Consider even numbers. They forms an \mynameref{def:ideal} in
  $\mathbb{Z}$. Because multiplication of any even number to any
  integer is an even. The ideal's symbolic name is $2 \mathbb{Z}$.
  \label{ex:ideal}
\end{example}

\begin{example}[Ring of integers modulo $n$: $\mathbb{Z}/n\mathbb{Z}$]
  Let $n \in \mathbb{Z}$ and $n > 1$. Then $n \mathbb{Z}$ is an
  \mynameref{def:ideal}.

  Two integer $a, b \in \mathbb{Z}$  are said to be congruent modulo
  $n$, written
  \[
  a \equiv b ( \mod n )
  \]
  if their difference $a - b$ is an integer multiple of $n$.

  Thus we have a separation of set $\mathbb{Z}$ into subsets of
  numbers that are congruent. Each subset has the following form
  \[
  \left\{r\right\}_n = r + n \mathbb{Z} =
  \left\{r + n k \mid k \in \mathbb{Z}\right\}
  \],
  thus
  \[
  \mathbb{Z} = \left\{0\right\}_n \cup \left\{1\right\}_n
  \cup \dots \cup \left\{n-1\right\}_n.
  \]

  Very often use the following notation
  \[
  \bar{r} = \left\{r\right\}_n.
  \]

  We can define the following operations
  \begin{eqnarray}
    \bar{k} \oplus \bar{l} = \overline{k + l}
    \nonumber \\
    \bar{k} \odot \bar{l} = \overline{k \cdot l}
    \nonumber
  \end{eqnarray}

  The \mynameref{def:ring} where the objects are defined is called as
  $\mathbb{Z}/n\mathbb{Z}$.
  \label{ex:intmodulo}
\end{example}

\begin{definition}[Ideal generated by a set]
  Let $R$ be a \mynameref{def:ring} and $S$ is a sub set of $R$.
  Consider the following set
  \[
  I = \left\{
  r_1 s_1 + \dots + r_n s_n \vert n \in \mathbb{N}, r_i \in R, s_i \in S
  \right\}
  \]
  $I$ is called by an ideal generated by set $S$ if
  $\forall r \in R, i \in I: r \cdot i \in I$.

  The sum in the definition of the ideal should be finite. The ring is
  assumed commutative in the definition.
  \label{def:idealset}
\end{definition}

\begin{definition}[Principal ideal]
  The ideal that is generated by one element $a$ is called as
  principal ideal and is denoted as $\left(a\right)$ i.e.
  left principal ideal:
  $\left(a\right) = \left\{r a \mid \forall r \in R\right\}$ and
  right principal ideal:
  $\left(a\right) = \left\{a r \mid \forall r \in R\right\}$
  \label{def:prinicipalideal}
\end{definition}

\begin{definition}[Integral domain]
In mathematics, and specifically in abstract algebra, an integral
domain is a nonzero commutative \mynameref{def:ring} in which the
product of any two nonzero elements is nonzero.
\label{def:integraldomain}
\end{definition}

\begin{definition}[Principal ideal domain]
  In abstract algebra, a principal ideal domain, or PID, is an
  \mynameref{def:integraldomain} in which every ideal is principal, i.e., can be
  generated by a single element.
  \label{def:pid}
\end{definition}

\begin{definition}[Maximal ideal]
  A maximal ideal is an ideal that is maximal (with respect to set
  inclusion) amongst all \mynameref{def:properideal}s i.e.
  $I$ is a maximal ideal of a ring $R$ if there are no other ideals
  contained between $I$ and $R$ \cite{wiki:maxideal}.
\label{def:maxideal}
\end{definition}

\begin{example}[Maximal ideal]
  If $F$ is a \mynameref{def:field} then the only maximal ideal is
  $\{0\}$ \cite{wiki:maxideal}.
  \label{ex:maxideal}
\end{example}

\begin{definition}[Prime ideal]
  \label {def:primeideal}
  An ideal $I$ of a commutative ring $R$ is prime if it has the
  following 2 properties \cite{wiki:primeideal}
  \footnote{
    There is a generalization of prime numbers in arithmetic
  }
  \begin{enumerate}
  \item If $a,b \in R$ such that $ab \in I$ then $a \in I$ or $b \in I$
  \item$I$ is not equal the whole ring $R$
  \end{enumerate}
\end{definition}

\begin{lemma}[$n\mathbb{Z}$ prime ideal]
  A positive integer $n$ is a prime if and only if $n\mathbb{Z}$ is a
  \mynameref{def:primeideal} in $\mathbb{Z}$ \cite{wiki:primeideal}
  \label{lem:primeideal_in_Z}
\end{lemma}

\begin{lemma}
  All nonzero \mynameref{def:primeideal}s are maximal in a principal
  ideal domain \cite{wiki:maxideal}
  \label{lem:primeideal_is_maximal}
\end{lemma}

\begin{definition}[Proper ideal]
$I$ is a proper ideal of a ring $R$ if $I \subsetneq R$.
\label{def:properideal}
\end{definition}

\begin{theorem}[About proper ideal]
  An ideal $I$ of ring $R$ is proper if and only if $1_R \notin I$.
  \label{thm:properideal}
\end{theorem}

\begin{definition}[Quotient ring]
  Quotient ring is a construction where one
  starts with a ring $R$ and a two-sided ideal $I$ in $R$, and constructs a
  new ring, the quotient ring $R/I$, whose elements are the
  \mynameref{def:coset}s of $I$
  in $R$ subject to special $+$ and $\cdot$ operations.

  Given a ring $R$ and a two-sided ideal $I \subset R$, we may define
  an equivalence relation $\sim$ on $R$ as follows:
  $a \sim b$ if and only if $a - b \in I$.
  The equivalence class of the element $a$ in $R$ is given by
  \[
  \bar{a} = \left\{a\right\} = a + I := \left\{ a + r : r \in I \right\}.
  \]
  This equivalence class is also sometimes written as a mod $I$ and
  called the "residue class of a modulo I" (see also example
  \ref{ex:intmodulo}).

  The special $+$ and $\cdot$ operations are defined as follows
  \[
  \forall \bar{x},\bar{y} \in R/I:
  \bar{x} + \bar{y} = \left(x + I\right) + \left(y + I\right) =
  \left(x+y\right) + I = \overline{x+y}.
  \]
  \[
  \forall \bar{x},\bar{y} \in R/I:
  \bar{x} \cdot \bar{y} = \left(x + I\right) \cdot \left(y + I\right) =
  \left(x \cdot y\right) + I = \overline{x \cdot y}.
  \]

  As result we will get the following ring $\left(R/I, +,
  \cdot\right)$ is called the quotient ring of $R$ by $I$.

  See also \mynameref{def:quotientgroup}
  \label{def:quotientring}
\end{definition}

\subsection{Polynomial ring $K\left[X\right]$}

Let we have a commutative \mynameref{def:ring} $K$. Lets create a new
\mynameref{def:ring} $B$ with the following infinite sets as elements:
\begin{equation}
  f = \left(f_0, f_1, \dots \right), \, f_i \in K,
  \label{eq:polynomial}
\end{equation}
such that only finite number of elements of the sets are non zero.

We can define addition and multiplication on $B$ as follows
\begin{eqnarray}
  f + g = \left(f_0 + g_0, f_1 + g_1, \dots \right),
  \nonumber \\
  f \cdot g = h = \left(h_0, h_1, \dots \right),
  \label{eq:polynomialops}
\end{eqnarray}
where
\[
h_k = \sum_{i + j =k} f_i g_j.
\]

The sequences (\ref{eq:polynomial}) forms a \mynameref{def:ring} with
the following identities:
\begin{itemize}
\item Additive identity: $\left(0, 0, \dots \right)$
\item Multiplicative identity: $\left(1, 0, \dots \right)$
\end{itemize}

The
sequences $k = \left(k, 0, \dots \right)$ added and multiplied as
elements of $K$ this allows say that such elements are elements of
original \mynameref{def:ring} $K$. Thus $K$ is sub-ring of the new ring
$B$.

Let
\begin{eqnarray}
  X = \left(0, 1, 0, \dots \right),
  \nonumber \\
  X^2 = \left(0, 0, 1, \dots \right)
  \nonumber
\end{eqnarray}
thus if we have
\[
f = \left(f_0, f_1, f_2, \dots, f_n, 0, \dots \right),
\]
where $f_n$ is the last non-zero element of (\ref{eq:polynomial}),
when one can get
\[
f = f_0 + f_1 X + f_2 X^2 + \dots + f_n X^n.
\]

\begin{definition}[Polynomial ring]
  The \mynameref{def:ring} of sequences (\ref{eq:polynomial}) with
  operations defined by (\ref{eq:polynomialops}) is called as
  polynomial ring $K\left[X\right]$.
  \label{def:polynomial}
\end{definition}

\begin{lemma}[Bézout]
  Let $a$ and $b$ be nonzero integers and let $d$ be their greatest common
  divisor. Then there exist integers $x$ and $y$ such that
  \[
  a x + by = d.
  \]
  \label{lem:bezout}
\end{lemma}

\begin{definition}[Monic polynomial]
  Monic polynomial is a univariate polynomial in which the leading
  coefficient (the nonzero coefficient of highest degree) is equal to
  1. Therefore, a monic polynomial has the form
  \[
  x^n + a_{n-1}x^{n-1}+ \dots + a_1 x + a_0
  \]
  \label{def:monicpolynomial}
\end{definition}

\begin{definition}[Irreducible polynomial]
  An irreducible polynomial is, roughly speaking, a non-constant
  polynomial that cannot be factored into the product of two
  non-constant polynomials.
  \label{def:irreducible}
\end{definition}

\begin{example}[Irreducible polynomial]
  The following polynomial is irreducible in
  $\mathbb{R}\left[X\right]$: $X^2 + 1$. The following one is also
  irreducible despite it has a root: $X+1$.
  \label{ex:irreducible}
\end{example}

\begin{theorem}[About irreducible polynomials]
  Let $\pi(X)$ is an \mynameref{def:irreducible} in $K\left[X\right]$ and let
  $\alpha$ be a root of $\pi(X)$ in a some larger field.
  $\forall h(x) \in K\left(X\right)$ if have the following statement:
  $h\left(\alpha\right) = 0$ if and only if $\pi(X) \mid h(X)$ in
  $K\left[X\right]$.
  \begin{proof}
    If $h(X) = \pi(X) g(X)$ then $h(\alpha) = 0$

    From other side let $\pi \nmid h$ in $K\left[X\right]$ this means
    that they are relatively prime in $K\left[X\right]$ and by
    \mynameref{lem:bezout} we can get $Q,R \in K\left[X\right]$ such
    that
    \[
    \pi(X) R(X) + h(X) Q(X) = 1,
    \]
    and especially for $X = \alpha$ we will get that $0 = 1$ that is
    impossible.
  \end{proof}
  \label{thm:irreduciblediv}
\end{theorem}

\begin{theorem}[About ideal generated by irreducible polynomial]
  Let $P \in K\left[X\right]$ is a polynomial and $I = \left(P\right)$ is
  an \mynameref{def:ideal} generated by the polynomial. The $I$ is
  \mynameref{def:maxideal} if and only if $P$ is irreducible in
  $K\left[X\right]$
  \begin{proof}
    Let $P$ is reducible i.e. $P = G F$. In the case $(P) \subset (G)$
    and $(P) \subset (F)$ i.e. by definition it is not a maximal
    ideal.

    If $P$ is irreducible then $K\left[X\right]/(P)$ is a field (see
     claim \ref{claim:lec1_sec14})
    and by theorem \ref{thm:maxideal} $(P)$
    is a maximal ideal.
  \end{proof}
  \label{thm:irreducibleideal}
\end{theorem}

\subsection{Fields}

\begin{definition}[Field]
  The ring $\left(R, \oplus, \odot\right)$ is called as a field if
  $\left(R \setminus \{0\}, \odot\right)$ is an \mynameref{def:abeliangroup}.

  The inverse element to $a$ in
  $\left(R \setminus\{0\}, \odot\right)$ is denoted as $a^{-1}$
  \label{def:field}
\end{definition}

\begin{example}[Field $\mathbb{Q}$]
  Note that $\mathbb{Z}$ is not a field because not for every integer
  number an inverse exists. But if we consider a set of fractions
  $\mathbb{Q} = \left\{a/b \mid a \in \mathbb{Z}, b \in
  \mathbb{Z}\setminus\{0\}\right\}$ when it will be a field.

  The
  inverse element to $a/b$  in
  $\left(\mathbb{Q}\setminus\{0\}, \cdot\right)$  will be $b/a$.
  \label{ex:field}
\end{example}

\begin{definition}[Unique factorization domain]
   Unique factorization domain (UFD) is a commutative ring, which is
   an \mynameref{def:integraldomain}, and in which every non-zero non-unit element
   can be written as a product of prime elements (or irreducible
   elements), uniquely up to order and units, analogous to the
   fundamental theorem of arithmetic for the integers.
  \label{def:ufd}
\end{definition}

\begin{theorem}[About Quotient Ring and Maximal Ideal]
  Let $\left(R, +, \cdot\right)$ is a commutative \mynameref{def:ring}
  with additive identity $0_R$ and multiplicative identity $1_R$. Let
  $I$ be an \mynameref{def:ideal} of $R$ then $I$ is
  \mynameref{def:maxideal} if and only if \mynameref{def:quotientring}
  $R/I$ is a \mynameref{def:field}
  \label{thm:maxideal}
  \begin{proof}
    See the end of section \ref{sec:lec2_ideals}.
  \end{proof}
\end{theorem}

\begin{definition}[Fraction field]
  The field of fractions of an integral domain is the smallest field
  in which it can be embedded. The elements of the field of fractions
  of the integral domain $R$ are equivalence classes
  (see the construction below) written as $\frac{a}{b}$ with $a, b \in
  R$ and $b\neq 0$. The field of
  fractions of $R$ is sometimes denoted by
  $\mathrm{Quot} (R)$ or $\mathrm {Frac}(R)$ \cite{wiki:fractionfield}.
  \label{def:fractionfield}
\end{definition}

\subsection{Characters}

\begin{definition}[Character]
  For an abelian group $G$, finite or infinite, a character of $G$ is
  a group homomorphism $\phi: G \to F^\times$ where $F$ is a field.

  The definition was taken from \cite{bib:KeithConradLinearchar}
  \label{def:character}
\end{definition}

\begin{example}[Character]
  A field homomorphism $K \to F$ is a character by restricting it to the
  non- zero elements of $K$ (that is, using $G = K^\times$) and ignoring the
  additive aspect of a field homomorphism.

  The example was taken from \cite{bib:KeithConradLinearchar}
  \label{ex:character}
\end{example}

\begin{theorem}[Dedekind]
  If we have $n$ distinct \mynameref{def:character}s $\phi_1, \dots,
  \phi_n: G \to F^\times$ then they are linearly independent i.e. if
  $c_1, \dots, c_n \in F$ satisfy
  \[
  c_1 \phi_1(g) + \dots + c_n \phi_n(g) = 0
  \]
  for all $g \in G$ then $c_1 = \dots = c_n = 0$
  (see also \cite{bib:KeithConradLinearchar}).
  \label{thm:dedekind}
\end{theorem}

\section{Modules and Vector spaces}

\subsection{Modules}

A module over a ring is a generalization of the notion of vector space
over a field, wherein the corresponding scalars are the elements of an
arbitrary given ring (with identity) and a multiplication (on the left
and/or on the right) is defined between elements of the ring and
elements of the module.

\begin{definition}[Module]
  \label{def:module}
  Let $R$ is a \mynameref{def:ring} and $1_R$ is it's multiplicative
  identity. A left R-module $M$ consists of an \mynameref{def:abeliangroup}
  $\left(M, +\right)$ and an operation $\cdot: R \times M \to M$ such
  that $\forall r,s \in R$ and $\forall x,y \in M$ the following
  relations are hold:
  \begin{enumerate}
  \item $r \cdot \left(x+y\right) = r \cdot x + r \cdot y$
  \item $\left(r + s \right) \cdot x = r \cdot x + s \cdot x$
  \item $\left(rs\right) \cdot x = r \cdot \left(s \cdot x\right)$
  \item $1_R \cdot x = x$
  \end{enumerate}
\end{definition}

\begin{example}[Module]
  If $K$ is a \mynameref{def:field} then concepts of
  K-\mynameref{def:vectorspace} and $K$-module are the same
\end{example}

\begin{definition}[Generating set of a module]
  A generating set $G$ of a module $M$ over a ring $R$ is a subset of
  $M$ such that the smallest submodule of $M$ containing $G$ is $M$
  itself \cite{wiki:gsmodule}
  \label{def:gsmodule}
\end{definition}

\begin{definition}[Free module]
  \label{def:freemodule}
  The \mynameref{def:module} that has a basis (i.e. linearly independent
  generating set)
  is called as free module \cite{wiki:freemodule}.

  For a $R$-module $M$ the set $E \subseteq M$ is a basic for $M$ if
  \begin{enumerate}
  \item $E$ is a generating set (see definition \ref{def:gsmodule})
    for $M$ i.e. $\forall m \in M$
    $\exists n < \infty$: $\exists e_i \in E, r_i \in R$:
    $m = \sum_{i = 1}^n r_i e_i$
  \item $E$ is linearly independent, i.e. if $r_1 e_1 + \dots + r_n
    e_n = 0_M$ for distinct elements $e_1, \dots, e_n \in E$ then
    $r_1 = \dots = r_n = 0_R$.
  \end{enumerate}
\end{definition}

\begin{definition}[Rank of free module]
  The cardinality of any (and therefore every) basis is called the
  rank of the free module $M$ \cite{wiki:freemodule}.
  \label{def:rankfreemodule}
\end{definition}

\begin{definition}[Direct sum of modules]
  In abstract algebra, the direct sum is a construction which combines
  several modules into a new, larger module. The direct sum of modules
  is the smallest module which contains the given modules as
  submodules with no "unnecessary" constraints, making it an example
  of a coproduct. Contrast with the direct product, which is the dual
  notion \cite{wiki:directsummodules}.
  \label{def:directsummodules}
\end{definition}

\begin{example}[Direct sum of modules]
  If we have 2 \mynameref{def:freemodule}s $M$ and $N$ with bases $m_1,
  m_2, \dots, m_m$ and $n_1, n_2, \dots, n_n$. Then the
  \mynameref{def:directsummodules} $A = M \oplus N$ will also be a free
  module with composite basis: $m_1, m_2, \dots, m_m, n_1, n_2, \dots, n_n$
  \label{ex:directsummodules}
\end{example}

\begin{definition}[Finitely generated module]
  Finitely generated module is a module that has a finite generating
  set (see also definition \ref{def:gsmodule}) \cite{wiki:fgmodule}.
  \label{def:fgmodule}
\end{definition}

\begin{definition}[Cyclic module]
  A (left) $R$-\mynameref{def:module} $M$ is called cyclic if $M$ can be
  generated by
  a single element i.e.
  \[
  M = \left<x\right>=R x = \left\{rx \mid r \in R\right\}
  \]
  for some $x \in M$ \cite{wiki:cyclicmodule}.
  \label{def:cyclicmodule}
\end{definition}

\begin{claim}
  $M$ is a cyclic $R$ module if and only if exists left ideal $I
  \subset R$ such that $M \cong R/I$
  \begin{proof}
    See \cite{mathstackexchange:cyclicmodule}. Note that
    \cite{bib:lang} p. 149 has
    the claim as a definition of the \mynameref{def:cyclicmodule}.
  \end{proof}
  \label{claim:cyclicmodule}
\end{claim}

\subsection{Linear algebra}

\begin{definition}[Vector space]
  \label{def:vectorspace}
  Let $F$ is a \mynameref{def:field}. The set $V$ is called as vector
  space under $F$ if the following conditions are satisfied
  \begin{enumerate}
  \item We have a binary operation $V \times V \rightarrow V$
    (addition): $(x,y) \rightarrow x + y$ with the following
    properties:
    \begin{enumerate}
    \item $x + y = y + x$
    \item $(x + y) + z = x + ( y + z )$
    \item $\exists 0 \in V$ such that $\forall x \in V: x + 0 = x$
    \item $\forall x \in V \exists -x \in V$ such that $x + (-x) = x -
      x = 0$
    \end{enumerate}
  \item We have a binary operation $F \times V \rightarrow V$ (scalar
    multiplication) with the following properties
    \begin{enumerate}
    \item $1_F \cdot x = x$
    \item $\forall a,b \in F, x \in V$: $a\cdot\left(b \cdot x\right)
      = \left(a b\right) \cdot x$.
    \item $\forall a,b \in F, x \in V$:
      $(a+b)\cdot x = a \cdot x + b \cdot x$
    \item $\forall a \in F, x, y \in V$:
      $a\cdot(x+y) = a\cdot x + a \cdot y$
    \end{enumerate}
  \end{enumerate}
\end{definition}

\begin{lemma}[About vector space isomorphism]
  2 vector spaces $L$ and $M$ with same dimension $dim L = dim M$ then
  there exists an \mynameref{def:isomorphism} between them
  \label{lem:vsisomorphism}
\end{lemma}

\begin{definition}[Image]
  The image or range of a linear map $f: V \to W$ is the following
  set \cite{wiki:linearmap}:
  \[
  \Ima f = \left\{
  w \in W: w = f(v), v \in V
  \right\}
  \]
  \label{def:image}
\end{definition}

\begin{definition}[Kernel]
  The kernel of a linear map $f: V \to W$ is the following
  set \cite{wiki:kernel}:
  \[
  \ker f = \left\{
  v \in V: f(v) = 0
  \right\}
  \]
  \label{def:kernel}
\end{definition}

\begin{definition}[Rank]
  The rank of a linear map $f: V \to W$ is dimension of
  \mynameref{def:image}: $rg f = \dim \Ima f$ \cite{wiki:rank}:
  \label{def:rank}
\end{definition}

\begin{definition}[Nullity]
  The nullity of a linear map $f: V \to W$ is dimension of
  \mynameref{def:kernel}: $nul f = \dim \ker f$ \cite{wiki:kernel}:
  \label{def:nullity}
\end{definition}

\begin{theorem}[Rank–nullity theorem]
  Let $V$ and $W$ be vector spaces over some field and let $T: V \to
  W$ be a linear map. Then the \mynameref{def:rank} of $T$ is the
  dimension of the image of $T$ and the \mynameref{def:nullity} of $T$
  is the dimension of the kernel of $T$, so we have
  \[
  \dim\left(\Ima T\right) + \dim\left(\ker T\right) = \dim V
  \]
  or, equivalently
  \[
  rg(T) + nul(T) = \dim(V)
  \]
  \label{thm:ranknullity}
  \begin{proof}
    See \cite{wiki:ranknullity}
  \end{proof}
\end{theorem}

\begin{definition}[General linear group of a vector space]
  If $V$ is a \mynameref{def:vectorspace} over field $K$  the general
  linear group of $V$, written $GL\left(V\right)$ or
  $Aut\left(V\right)$, is the group of all automorphisms of $V$,
  i.e. the set of all bijective linear transformations $V \to V$,
  together with functional composition as group operation \cite{wiki:glg}.
  \label{def:glv}
\end{definition}

\begin{definition}[Dual space]
  Given any vector space $V$ over a field $F$, the dual space $V^\ast$
  is defined as the set of all linear maps $\phi: V \to F$ (linear
  functionals). The dual space $V^\ast$ itself becomes a vector space
  over $F$ when equipped with an addition and scalar multiplication
  satisfying:
  \begin{eqnarray}
    (\varphi +\psi )(x)=\varphi (x)+\psi (x)
    \nonumber \\
    (a\varphi )(x)=a\left(\varphi (x)\right)
    \nonumber
  \end{eqnarray}
  for all $\phi, \psi \in V^\ast$, $x \in V$, and $a \in F$.

  This is also named as algebraic dual space at \cite{wiki:dualspace}.
  \label{def:dualspace}
\end{definition}

\begin{definition}[Dual basis]
  If $\{e_i\}$ is a basis of $V$ then exists a basis $\{v_j\}$ of
  \mynameref{def:dualspace} such that $v_j(e_i) = \delta_{ij}$ and is
  called as dual basis.
  \cite{wiki:dualbasis}
  \label{def:dualbasis}
\end{definition}

\begin{definition}[Degenerate bilinear form]
  A degenerate bilinear form $f(x, y)$ on a vector space $V$ is a
  bilinear form such that the map from $V$ to $V^\ast$ (the
  \mynameref{def:dualspace}
  of $V$) given by $v \to (x \to f(x, v) )$ is not an isomorphism
  \cite{wiki:dbf}.

  An equivalent definition when $V$ is finite-dimensional is that it
  has a non-trivial kernel: there exist some non-zero $x \in V$ such
  that
  \(
  \forall y \in V f(x,y) = 0
  \)
  \label{def:degeneratebf}
\end{definition}

\begin{definition}[Non-degenerate bilinear form]
  A nondegenerate or nonsingular form is one that is not degenerate,
  meaning that the map from $V$ to $V^\ast$ (the \mynameref{def:dualspace}
  of $V$) given by $v \to (x \to f(x, v) )$ is an isomorphism
  \cite{wiki:dbf} or equivalently when $V$ is finite-dimensional if
  and only if
  \(
  \forall y \in V f(x,y) = 0
  \)
  implies $x = 0$, i.e. the map $f$ has a trivial kernel.
  \label{def:nondegeneratebf}
\end{definition}

\begin{definition}[Eigenspace]
  Consider a linear map $T: V \to V$.
  The set of all eigenvectors of $T$ corresponding to the same
  eigenvalue, together with the zero vector, is called an eigenspace
  or characteristic space of $T$ \cite{wiki:eigenvalues}
  \label{def:eigenspace}
\end{definition}

\begin{definition}[Diagonalizable map]
  If $V$ is a finite-dimensional vector space, then a linear map
  $T: V \to V$ is called diagonalizable if there exists an ordered
  basis of $V$ with respect to which $T$ is represented by a diagonal
  matrix.

  A linear map $T: V \to V$ is diagonalizable if and only if the sum
  of the dimensions of its
  eigenspaces is equal to $n = dim(V)$, which is the case if and only if
  there exists a basis of $V$ consisting of eigenvectors of $T$. With
  respect to such a basis, $T$ will be represented by a diagonal
  matrix. The diagonal entries of this matrix are the eigenvalues of
  $T$ \cite{wiki:diagonalizable_matrix}.
  \label{def:diagonalizable_map}
\end{definition}

\begin{theorem}[About eigenvalues of a diagonalizable linear map]
  A linear map $T: V \to V$ with $n = dim(V)$ is diagonalizable if it
  has $n$ distinct eigenvalues \cite{wiki:diagonalizable_matrix}.
  \label{thm:diagonalizable_map_eigenvalues}
\end{theorem}

\begin{theorem}[About invertible matrix]
  Let $A$ be a matrix over $F$. If $A$ is diagonalizable, then so is
  any power of it. Conversely, if $A$ is invertible, $F$ is
  algebraically closed, and $A^n$ is diagonalizable for some $n$ that
  is not an integer multiple of the characteristic of $F$, then $A$ is
  diagonalizable.
  \begin{proof}
    See \cite{wiki:diagonalizable_matrix}
  \end{proof}
  \label{thm:diagonalizable_matrix}
\end{theorem}

\section{Functions aka maps}

\subsection{Functions}

\begin{definition}[Surjection]
  The function $f: X \rightarrow Y$ is surjective (or onto) if
  $\forall y \in Y$, $\exists x \in X$ such that
  $f\left(x\right) = y$.
  \label{def:surjection}
\end{definition}

\begin{definition}[Injection]
  The function $f: X \rightarrow Y$ is injective (or one-to-one function) if
  $\forall x_1, x_2 \in X$, such that $x_1 \ne x_2$ then
  $f\left(x_1\right) \ne f\left(x_2\right)$.
  \label{def:injection}
\end{definition}

\begin{definition}[Bijection]
  The function $f: X \rightarrow Y$ is bijective (or one-to-one
  correspondence) if it is an \mynameref{def:injection} and a
  \mynameref{def:surjection}.
  \label{def:bijection}
\end{definition}

\begin{definition}[Homomorphism]
  The homomorphism is a function (map) between two sets that preserves
  its algebraic structure. For the case of groups
  $\left(X, \circ\right)$ and $\left(Y, \odot\right)$ the function
  $\phi: X \rightarrow Y$ is called homomorphism if
  $\forall x_1, x_2 \in X$ it holds
  $\phi\left(x_1 \circ x_2\right) = \phi\left(x_1 \right) \odot
  \phi\left( x_2\right)$.
  \label{def:homomorphism}
\end{definition}

\begin{definition}[Field homomorphism]
  Let \cite{proofwiki:fieldhomomorphism} $\left(F, +, \times\right)$
  and $\left(K, \oplus, \otimes\right)$ are
  \mynameref{def:field}s. Let $\phi: F \rightarrow K$ is a mapping
  such that both
  $\left(F, +\right) \rightarrow \left(K, \oplus\right)$
  and
  $\left(F, \times\right) \rightarrow \left(K, \otimes\right)$
  are \mynameref{def:homomorphism}s under $\phi$.

  That is $\forall a, b \in F$:
  \begin{enumerate}
  \item $\phi\left(a + b\right) = \phi\left(a\right) \oplus \phi\left(b\right)$
  \item $\phi\left(a \times b\right) = \phi\left(a\right) \otimes \phi\left(b\right)$
  \end{enumerate}

  Then $\phi: \left(F, +, \times\right) \rightarrow \left(K, \oplus,
  \otimes\right)$ is a field homomorphism.
  \label{def:fieldhomomorphism}

  Note that in the lectures we often refer the Field homomorphism as
  just \mynameref{def:homomorphism}.
\end{definition}

\begin{definition}[Isomorphism]
  If a map is \mynameref{def:bijection} as well as
  \mynameref{def:homomorphism} when it is called as isomorphism.

  We use the following symbolic notation for isomorphism between $X$
  and $Y$: $X \cong Y$.
  \label{def:isomorphism}
\end{definition}

\begin{definition}[Endomorphism]
   An endomorphism is a morphism (or homomorphism) from a mathematical
   object to itself \cite{wiki:endomorphism}
   \label{def:endomorphism}
\end{definition}

\begin{definition}[Automorphism]
  Automorphism is an isomorphism from a mathematical object to itself.
  \label{def:automorphism}
\end{definition}

\begin{definition}[Embedding]
  When some object X is said to be embedded in another object $Y$, the
  embedding is given by some injective and structure-preserving map
  $f : X \to Y$. The precise meaning of "structure-preserving" depends on
  the kind of mathematical structure of which $X$ and $Y$ are
  instances.

  The fact that a map $f : X \to Y$ is an embedding is often indicated
  by the use of a "hooked arrow", thus: $f:X\hookrightarrow Y$. On the
  other hand, this notation is sometimes reserved for inclusion maps.
  \label{def:embedding}
\end{definition}

\begin{theorem}[First isomorphism]
  Let $G$ is a group and $\phi: G \to H$ is a
  surjective \mynameref{def:homomorphism}. Then if $N = \ker \phi$ we
  have
  \[
  H \cong G/N
  \]
  \label{thm:firstisomorphism}
\end{theorem}

\begin{theorem}[Isomorphism extension theorem]
  Let $F$ is a \mynameref{def:field} and $E$ is an
  \mynameref{def:algebraicextension} of $F$.
  $F'$ is another \mynameref{def:field} and $E'$ the
  \mynameref{def:algebraicextension} of $F'$.

  If there exists an \mynameref{def:isomorphism} $\phi: F \to F'$ then
  it can be extended to an isomorphism $\tau: E \to E'$.

  \begin{proof}
    The proof of the isomorphism extension theorem depends on
    \mynameref{lem:zorn}.

    ??? The theorem seems to be very close to the theorem
    \ref{thm:lec2_3}.
  \end{proof}

  \label{thm:isomorphismextension}
\end{theorem}

\begin{theorem}[About group homomorphism]
  A group homomorphism is \mynameref{def:injection} iff the kernel is
  trivial \cite{wiki:grouphomomorphism}
  \label{thm:grouphomomorphsim}
\end{theorem}

\subsection{Category theory}

\begin{definition}[Commutative diagram]
  A commutative diagram is a diagram of objects (also known as
  vertices) and morphisms (also known as arrows or edges) such that
  all directed paths in the diagram with the same start and endpoints
  lead to the same result by composition
  \label{def:commutativediagram}

  The following diagram commutes if $f_{AB} = f_{CB} f_{AC}$ or
  $f_{AB}\left(x\right) = f_{CB} \left(f_{AC}\left(x\right)\right)$.
  \begin{tikzpicture}[description/.style={fill=white,inner sep=2pt}]
    \matrix (m) [matrix of math nodes, row sep=3em,
      column sep=2.5em, text height=1.5ex, text depth=0.25ex]
            { A& & B \\
              & C & \\ };
            %\draw[double,double distance=5pt] (m-1-1) – (m-1-3);
            \path[->]
            (m-1-1) edge node[description] {$ f_{AB} $} (m-1-3)
            edge node[description] {$  f_{AC} $} (m-2-2)
            (m-2-2) edge node[description] {$  f_{CB} $} (m-1-3);
  \end{tikzpicture}

\end{definition}

\section{Number theory}

\begin{definition}[Euler's totient function]
  In number theory, Euler's totient function counts the positive
  integers up to a given integer $n$ that are relatively prime to $n$. It
  is written using the Greek letter phi as $\phi\left(n\right)$, and
  may also be called Euler's phi function. It can be defined more formally as
  the number of integers $k$ in the range $1 \le k \le n$ for which the
  greatest common divisor $gcd\left(n, k\right)$ = 1. The integers $k$ of this
  form are sometimes referred to as totatives of $n$.

  The definition was taken from \cite{wiki:eulerfunction}
  \label{def:eulerfuction}
\end{definition}

\begin{example}[Euler's totient function]
  For example \cite{wiki:eulerfunction}, the totatives of $n = 9$ are
  the six numbers 1, 2, 4, 5,
  7 and 8. They are all relatively prime to 9, but the other three
  numbers in this range, 3, 6, and 9 are not, because
  $gcd\left(9, 3\right)$ =
  $gcd\left(9, 6\right) = 3$ and $gcd\left(9, 9\right) = 9$.
  Therefore, $\phi\left(9\right) = 6$. As another
  example, $\phi\left(1\right) = 1$ since for $n = 1$ the only integer in the range from
  1 to n is 1 itself, and $gcd\left(1, 1\right) = 1$.
\end{example}

\end{appendices}


\section{Topology of Level Sets}

Let $P$ be a probability measure on a product space $\mathcal{X} \times \mathcal{Y}$,
where we assume $\mathcal{X}$ and $\mathcal{Y}$ are Euclidean vector spaces for simplicity.
Let $\{ (x_i, y_i)\}_i$ be an iid sample of size $L$ drawn from $P$ defining the training set.
We consider the classic empirical risk minimization of the form
\begin{equation}
\label{emp_risk_min}
\Fem(\theta) = \frac{1}{L} \sum_{l=1}^L \| \Phi(x_i;\theta) - y_i \|^2 + \kappa \mathcal{R}(\theta)~,
\end{equation}
where $\Phi(x ; \theta)$ encapsulates the feature representation
that uses parameters $\theta \in \R^S$ and $\mathcal{R}(\theta)$ is a regularization term.
 In a deep neural network, $\theta$
contains the weights and biases used in all layers.
For convenience, in our analysis we will also use the oracle risk minimization:
\begin{equation}
\label{risk_min}
\Forr(\theta) = \E_{(X,Y) \sim P} \| \Phi(X;\theta) - Y \|^2 + \kappa \mathcal{R}(\theta)~.
\end{equation}
Our setup considers the case where $\mathcal{R}$
consists on either $\ell_1$ or $\ell_2$ norms, as we shall describe below.
They correspond to well-known sparse and ridge regularization respectively.
%
% on all layers: $\mathcal{R}(\theta) = \| \theta \|_1$,
%which provides a sparse regularization prior, and the Ridge regression case
%$\mathcal{R}(\theta) = \| \theta\|_2^2$ (also referred as weight decay).

\subsection{Poor local minima characterization from topological connectedness}

We define the level set of $F(\theta)$ as
\begin{equation}
\Omega_F(\lambda) = \{ \theta \in \R^S~;~F(\theta) \leq \lambda \}~.
\end{equation}

The first question we study is the structure of critical points of $\Fem(\theta)$ and $\Forr(\theta)$
when $\Phi$ is a multilayer neural network.
For simplicity, we consider first a strict notion of local minima:
$\theta \in \R^S $ is a strict local minima of $F$ if there is $\epsilon>0$ with $F(\theta') > F(\theta)$ for all $\theta' \in B(\theta,\epsilon)$ and $\theta'\neq \theta$.
%$\theta \in \R^N $ is a local minima of $F \in C^\infty(\R^N,\R)$ if for any differentiable curve $\gamma: (-\epsilon, \epsilon) \to \R^N$ passing through $\theta$ at $t=0$, there exists $K>1$ even such that the function $g(t) = F(\gamma(t))$ satisfies $g^{(k)}(0) = 0 $ for all $k < K$ and
%$g^{(K)}(0) > 0$.
In particular, we are interested to know whether
$\Fem$ has local minima which are not global minima.
 %\footnote{this implies $\nabla F(\theta) =0$ and $H F(\theta) \succeq 0$ but avoids degenerate cases where $F$ is constant along low
This question is answered by
knowing whether $\Omega_F(\lambda)$ is connected at each energy level $\lambda$:

\begin{proposition}
\label{connectedminima}
If $\Omega_F(\lambda)$ is connected for all $\lambda$ then every local minima of $F(\theta)$ is a global minima.
\end{proposition}

Strict local minima implies that $\nabla F(\theta) =0$ and $H F(\theta) \succeq 0$, but avoids degenerate cases where $F$ is constant along a manifold intersecting $\theta$. In that scenario, if $\mathcal{U}_\theta$ denotes that manifold, our reasoning immediately
implies that if $\Omega_F(\lambda)$ are connected, then for all $\epsilon > 0$ there exists $\theta'$ with $\text{dist}(\theta',\mathcal{U}_\theta) \leq \epsilon$ and $F(\theta') < F(\theta)$. In other words, some element at the boundary of $\mathcal{U}_\theta$ must be a saddle point.
A stronger property that eliminates the risk of gradient descent getting stuck at $\mathcal{U}_\theta$ is
  that \emph{all} elements at the boundary of $\mathcal{U}_\theta$ are saddle points. This can be guaranteed if one can show that
  there exists a path connecting any $\theta$ to the lowest energy level such that $F$ is strictly decreasing along it.

Such degenerate cases arise in deep linear networks in absence of regularization. If $\theta = (W_1, \dots, W_K)$ denotes any parameter value, with $N_1, \dots N_K$ denoting the hidden layer sizes, and $F_k \in \mathbf{GL}_{N_k}^{+}(\R)$ are arbitrary elements of the general linear group of invertible $N_k \times N_k$ matrices with positive determinant, then
$$\mathcal{U}_\theta = \{ W_1 F_1^{-1}, F_1 W_2 F_2^{-1}, \dots, F_K W_K  ~;~ F_k \in \mathbf{GL}_{N_k}^{+}(\R) \}~. $$
In particular, $\mathcal{U}_\theta$ has a Lie Group structure. In the half-rectified nonlinear case, the general linear group is replaced by the Lie group of homogeneous invertible matrices $F_k = \text{diag}(\alpha_1, \dots, \alpha_{N_k})$ with $\alpha_j > 0$.

This proposition shows that a sufficient condition to prevent the existence of poor local minima is having connected level sets, but this condition is not necessary: one can have isolated local minima lying
at the same energy level. This can be the case in systems that are defined up to
a discrete symmetry group, such as multilayer neural networks. However, as we shall see next, this case puts the system in a brittle position, since one needs to be able to account for all the local minima (and there can be exponentially many of them as the parameter dimensionality increases) and verify that their energy is indeed equal.

\subsection{The Linear Case}

We first consider the
particularly simple case where
$F$ is a multilayer network defined by
\begin{equation}
\label{linearcase}
\Phi(x;\theta) = W_K \dots W_1 x~,~\theta = (W_1, \dots, W_K)~.
\end{equation}
and the ridge regression $\mathcal{R}(\theta) =\| \theta \|^2$. This model defines a non-convex (and non-concave) loss $\Fem(\theta)$.
When $\kappa = 0$, it has been shown in \cite{saxe2013exact} and \cite{kawaguchi2016deep} that in this case,
every local minima is a global minima.
We provide here an alternative proof of that result that uses
a somewhat simpler argument and allows for $\kappa > 0$ in the case $K=2$.

%For that purpose, let $W_1, W_2, \dots, W_K$ be weight matrices of sizes
%$n_k \times n_{k+1}$, $k < K$. Assume first that $n_j \geq \min(n_1, n_K)$ for $j=2 \dots K-1$.
%and let us define the following multilinear regression problem:
%\begin{equation}
%\label{multilinloss}
%L_0(W_1, \dots, W_K) = \sum_i \| W_K, \dots W_1 x_i - y_i \|^2~,
%\end{equation}
%where $\{ (x_i, y_i)\,; x_i \in \mathbb{R}^{n_1}, y_i \in \mathbb{R}^{n_K} \}_i$ is a given
%training set.

\begin{proposition}
\label{proplinear}
Let $W_1, W_2, \dots, W_K$ be weight matrices of sizes
$n_k \times n_{k+1}$, $k < K$, and let $\Fem(\theta)$, $\Forr(\theta)$
denote the risk minimizations using $\Phi$ as in (\ref{linearcase}). %and ridge regression.
Assume that $n_j > \min(n_1, n_K)$ for $j=2 \dots K-1$.
Then $\Omega_{\Fem}(\lambda)$ (and $\Omega_{\Forr}$) is connected for all $\lambda$ and all $K$ when $\kappa=0$, and for $\kappa>0$ when $K=2$; and therefore there are no poor local minima in these cases.
Moreover, any $\theta$ can be connected to the lowest energy level with a strictly decreasing path. %when $\kappa=0$.
\end{proposition}

%Let us highlight that this result is weaker than that obtained by \cite{kawaguchi2016deep}, since it does not
%control the behavior of non-strict local minima and does not inform about the strict saddle condition.
Let us highlight that this result is slightly complementary than that of \cite{kawaguchi2016deep}, Theorem 2.3.
Whereas we require $n_j > \min(n_1, n_K)$ for $j=2 \dots K-1$ and our analysis does not inform about the order of the saddle points,
we do not need full rank assumptions on $\Sigma_X$ nor the weights $W_k$. %We leave the consolidation of these results for future work

This result does also highlight a certain mismatch between the picture of having no poor local minima
and generalization error. Incorporating regularization drastically changes the topology, and the
fact that we are able to show connectedness only in the two-layer case with ridge regression is profound; we conjecture that extending it to deeper models requires a different regularization, perhaps using more general atomic norms \cite{bach2013convex}. But we now move our interest to the nonlinear case, which is more relevant to our purposes.

\subsection{Half-Rectified Nonlinear Case}

We now study the setting given by
\begin{equation}
\label{relucase}
\Phi(x;\theta) = W_K \rho W_{K-1} \rho \dots \rho W_1 x~,~\theta = (W_1, \dots, W_K)~,
\end{equation}
where $\rho(z) = \max(0 ,z)$.
The biases can be implemented by replacing the input vector $x$
with $\overline{x}=(x, 1)$ and by rebranding each parameter matrix as
$$\overline{W}_i = \left(
\begin{array}{c|c}
W_i & b_i \\
\hline
0 & 1
\end{array}
\right)~,$$
where $b_i$ contains the biases for each layer.
For simplicity, we continue to use $W_i$ and $x$ in the following.

\subsubsection{Nonlinear models are generally disconnected}
\label{disconnect}

One may wonder whether the same phenomena of global connectedness also holds
in the half-rectified case. A simple motivating counterexample shows that this is not the case in
general. Consider a simple setup with $X \in \R^2$ drawn from a mixture of two Gaussians $\mathcal{N}_{-1}$
and $\mathcal{N}_{1}$, and let $Y = (X-\mu_Z) \cdot Z $ , where $Z$ is the (hidden) mixture component taking $\{1,-1\}$ values.  Let
$\hat{Y} = \Phi(X; \{ W_1, W_2\} )$ be a single-hidden layer ReLU network, with two hidden units.
%Since the model is homogeneous, one can think about $W_1$ as encoding two unitary vectors without loss of generality.
Let $\theta^A$ be a configuration that bisects the two mixture components,
and let $\theta^B$ the same configuration, but swapping the bisectrices.
One can verify that they can both achieve arbitrarily small risk by letting the covariance of the mixture components go to $0$.
However, any path that connects $\theta^A$ to $\theta^B$
must necessarily pass through a point in which $W_1$ has rank $1$, which leads to an estimator with risk at least $1/2$.

In fact, it is easy to see that this counter-example can be extended to any generic half-rectified architecture, if one is
allowed to adversarially design a data distribution. For any given $\Phi(X; \theta)$ with arbitrary architecture and current parameters
$\theta = (W_i)$, let $\mathcal{P}_\theta=\{ \mathcal{A}_1, \dots, \mathcal{A}_S\}$ be the underlying tessellation of the input space given by our current choice of parameters; that is, $\Phi(X; \theta)$ is piece-wise linear and $\mathcal{P}_\theta$ contains those pieces. Now let
$X$ be any arbitrary distribution with density $p(x) > 0$ for all $x \in \R^n$, for example a Gaussian, and let %$Y ~|~ \{X \in \mathcal{A}_s\} = s$.
%It is the indicator function corresponding to the tesselation $\mathcal{P}$.
$Y ~|~X ~\stackrel{d}{=} \Phi(X ; \theta)$~. Since $\Phi$ is invariant under a subgroup of permutations $\theta_\sigma$ of its hidden layers, it is easy to see that one can find two parameter values $\theta_A = \theta$ and $\theta_B = \theta_\sigma$ such that $\Forr(\theta_A) = \Forr(\theta_B) = 0$, but any continuous path $\gamma(t)$ from $\theta_A$ to $\theta_B$ will have a different tessellation and therefore won't satisfy $\Forr( \gamma(t) ) = 0$.
Moreover, one can build on this counter-example to show that not only the level sets are disconnected, but also that there exist poor local minima. Let $\theta'$ be a different set of parameters, and $Y' ~|~X \stackrel{d}{=} \Phi(X; \theta')$ be a different target distribution. Now consider the data distribution given by the mixture
$$X ~|~p(x) ~~,~z \sim \text{Bernoulli}(\pi)~,~Y ~|~X,z \stackrel{d}{=} z \Phi(X;\theta) + (1-z) \Phi(X; \theta')~.$$
By adjusting the mixture component $\pi$ we can clearly change the risk at $\theta$ and $\theta'$ and make them different, but we conjecture that this preserves the status of local minima of $\theta$ and $\theta'$. Appendix \ref{sec:disconnect} constructs a counter-example numerically.

This illustrates an intrinsic difficulty in the optimization landscape if one is after \emph{universal}
guarantees that do not depend upon the data distribution. This difficulty is non-existent in the linear case
and not easy to exploit in mean-field approaches such as \cite{choromanska2015loss},
and shows that in general
we should not expect to obtain connected level sets. However,
connectedness can be recovered if one is willing to accept a small increase
of energy and make some assumptions on the complexity of the regression task.
 Our main result shows that the amount by which the energy is
allowed to increase is upper bounded by a quantity that trades-off model overparametrization
and smoothness in the data distribution.

For that purpose, we start with a characterization of the oracle loss, and for simplicity let us assume
$Y \in \R$ and let us first consider the case with a single hidden layer and $\ell_1$ regularization:
$\mathcal{R}(\theta) = \| \theta\|_1$.

\subsubsection{Preliminaries}
 Before proving our main result, we need to introduce  preliminary notation and results.
We first describe the case with a single hidden layer of size $m$.

We define
\begin{equation}
\label{bla2}
e(m) = \min_{W_1 \in \R^{m \times n}, \|W_1(i) \|_2 \leq 1, W_2 \in \R^m} \E\{ | \Phi(X; \theta) - Y|^2 \} + \kappa  \| W_2 \|_1~.
\end{equation}
to be the oracle risk using $m$ hidden units with norm $\leq 1$ and using sparse regression.
It is a well known result by Hornik and Cybenko that a single hidden layer
is a universal approximator under very mild assumptions, i.e. $\lim_{m \to \infty} e(m) = 0$.
This result merely states that our statistical setup is consistent, and it should not be
surprising to the reader familiar with classic approximation theory.
 A more interesting question is the rate at which $e(m)$ decays, which depends
on the smoothness of the joint density $(X, Y) \sim P$ relative to the nonlinear activation
family we have chosen.

For convenience, we redefine $W = W_1$ and $\beta = W_2$ and
 $Z(W) = \max(0, W X)$. We also write $z(w) = \max(0, \langle w, X \rangle)$ where $(X, Y) \sim P$ and $w \in \R^N$ is any deterministic vector.
Let $\Sigma_X = \E_{P} XX^T \in \R^{N \times N}$ be the covariance operator of the random input $X$. We assume $\| \Sigma_X \| < \infty$.

 A fundamental property that will be essential to our analysis is that, despite
the fact that $Z$ is nonlinear, the quantity $[ w_1, w_2 ]_Z := \E_P \{ z(w_1) z(w_2) \} $
is locally equivalent to the linear metric $\langle w_1, w_2 \rangle_X = \E_P \{ w_1^T X X^T w_2 \} = \langle w_1, \Sigma_X w_2 \rangle$, and that the linearization error decreases with the angle between $w_1$ and $w_2$. Without loss of generality, we assume here that $\|w_1 \| = \| w_2 \| = 1$, and we write $\| w \|_Z^2 = \E \{ | z(w) |^2 \} $.
\begin{proposition}
\label{localdistprop}
Let $\alpha = \cos^{-1}( \langle w_1, w_2 \rangle )$ be the angle between unitary vectors $w_1$ and $w_2$ and let $w_m =  \frac{w_1 + w_2}{\| w_1 + w_2 \|}$ be their unitary bisector.
Then
\begin{equation}
\label{localdisteq}
 \frac{1 + \cos \alpha}{2}  \| w_m  \|_Z^2 - 2 \| \Sigma_X \| \left( \frac{1-\cos \alpha}{2} + \sin^2 \alpha \right) \leq [ w_1, w_2 ]_Z \leq \frac{1+\cos \alpha}{2}  \| w_m  \|_Z^2 ~.
\end{equation}
\end{proposition}
The term $\| \Sigma_X \| $ is overly pessimistic: we can replace it by the energy of $X$ projected into the subspace spanned by $w_1$ and $w_2$ (which is bounded by $2 \| \Sigma_X \|$).
When $\alpha$ is small, a Taylor expansion of the trigonometric terms reveals that
\begin{eqnarray*}
\frac{2}{3 \| \Sigma_X \|} \langle w_1, w_2 \rangle &=& \frac{2}{3 \| \Sigma_X \|} \cos \alpha = \frac{2}{3\| \Sigma_X \|}(1 - \frac{\alpha^2}{2} + O(\alpha^4)) \\
&\leq& ( 1 - \alpha^2/4)\| w_m \|_Z^2 - \| \Sigma_X \|( \alpha^2/4 + \alpha^2) + O(\alpha^4) \\
&\leq & [ w_1, w_2 ]_Z + O(\alpha^4) ~,
\end{eqnarray*}
and similarly
$$[ w_1, w_2 ]_Z \leq \langle w_1, w_2 \rangle \| w_m \|_Z^2 \leq \| \Sigma_X\| \langle w_1, w_2 \rangle~.$$
The local behavior of parameters $w_1, w_2$ on our regression problem is thus equivalent to that of having a linear layer, provided $w_1$ and $w_2$ are sufficiently close to each other.
This result can be seen as a \emph{spoiler} of what is coming: increasing the hidden layer dimensionality $m$ will increase the chances to encounter pairs of vectors $w_1, w_2$ with small angle; and with it some hope of approximating the previous linear behavior thanks to the small linearization error.

In order to control the connectedness, we need a last definition. Given a
hidden layer of size $m$ with current parameters $W \in \R^{n \times m}$, we define a
``robust compressibility" factor as
\begin{equation}
\label{compress}
\delta_W(l, \alpha; m) = \min_{ \|\gamma \|_0 \leq l, \sup_i |\angle(\tilde{w}_i, w_i)| \leq \alpha} \E \{| Y - \gamma Z(\tilde{W}) |^2 + \kappa \| \gamma \|_1  \}~,~(l \leq m)~.
\end{equation}
This quantity thus measures how easily one can compress the current hidden layer representation,
by keeping only a subset of $l$ its units, but allowing these units to move by a small amount controlled by $\alpha$. It is a form
of $n$-width similar to Kolmogorov width \cite{donoho2006compressed} and is also related to robust sparse coding from \cite{tang2013compressed, ekanadham2011recovery}.

\subsubsection{Main result}

Our main result considers now a non-asymptotic scenario given by some fixed
size $m$ of the hidden layer. Given two parameter values $\theta^A = (W_1^A, W_2^A) \in \mathcal{W}$
and $\theta^B= (W_1^B, W_2^B)$ with $\Forr(\theta^{\{A,B\} } ) \leq \lambda$,
we show that there exists a continuous path
$\gamma: [0,1] \to \mathcal{W}$ connecting $\theta^A$ and $\theta^B$
such that its oracle risk is uniformly bounded by $\max(\lambda, \epsilon)$, where $\epsilon$
decreases with model overparametrization.
\begin{theorem}
\label{maintheo}
For any $\theta^A, \theta^B \in \mathcal{W}$ and $\lambda \in \R$ satisfying $\Forr(\theta^{\{A,B\}}) \leq \lambda$, there exists a continuous path $\gamma: [0,1] \to \mathcal{W}$ such that
$\gamma(0) = \theta^A$, $\gamma(1) = \theta^B$ and
\begin{equation}
\Forr( \gamma(t) )  \leq \max( \lambda, \epsilon)~,\text{ with}
\end{equation}

\begin{align}
\epsilon = \inf_{l, \alpha} \Bigl(\max \Bigl\{ e(l), &\delta_{W_1^A}(m, 0; m ) , \delta_{W_1^A}(m-l, \alpha; m ) ,   \\
&\delta_{W_1^B}(m, 0; m ) ,\delta_{W_1^B} (m-l, \alpha; m ) \Bigr\} + C_1 \alpha  + O(\alpha^2) \Bigr)~,
\end{align}
%with
%$$f(\alpha, M) = C_1 \alpha  + C_2 M \alpha^2 + O(\alpha^2)~,$$
%$$f(\alpha) = ~,$$
where $C_1$ is an absolute constant depending only on $\kappa$ and $P$.
\end{theorem}
Some remarks are in order. First, our regularization term is currently a mix between $\ell_2$ norm constraints on the first layer and $\ell_1$ norm constraints on the second layer. We believe this is an artifact of our proof technique, and we conjecture that more general regularizations yield similar results. Next, this result uses the data distribution through the oracle bound $e(m)$ and the covariance term. The
extension to empirical risk is accomplished by replacing the probability measure $P$ by the empirical measure $\hat{P} = \frac{1}{L} \sum_l \delta\left( (x,y) - (x_l, y_l)\right) $. However, our asymptotic analysis has to be carefully reexamined to take into account and avoid the trivial regime when $M$ outgrows $L$.
%\item Finally, another essential question is the extension of this result to several layers
%\end{itemize}
A consequence of Theorem \ref{maintheo} is that as $m$ increases, the model becomes asymptotically connected, as proven in the following corollary.
\begin{corollary}
\label{maincoro}
As $m$ increases, the energy gap $\epsilon$ satisfies $\epsilon = O( m^{-\frac{1}{n}})$ and therefore the level sets become connected at all energy levels.
%Here we use the fact that $\delta(\lambda m; \epsilon(m); m) \to 0$ as $m \to \infty$. (find the rate).
\end{corollary}
This is consistent with the overparametrization results from \cite{safran2015quality,shamir2} and the general common knowledge amongst deep learning practitioners. Our next sections explore this question, and refine it by considering not only topological properties but also some rough geometrical measure of the level sets.


\section{Natural Boundary; A Gluing Construction; Roots}
\subsection{Natural Boundary}
Sometimes, it is impossible to do analytic continuation.
Write $\mathbb D=D(0,1)$ and $\mathbb T=\partial\mathbb D=S^1$.
Consider a power series $f(z)=\sum_na_nz^n$ with radius of convergence $1$.\\
We say $z_0\in\mathbb T$ is regular if there is a neighbourhood $U$ of $z_0$ such that there exists a function element $(g,U)$ such that $f|_{U\cap\mathbb D}=g|_{U\cap\mathbb D}$.
A point that is not regular is singular.
Easily, the set of regular points is open in $\mathbb T$, so the set of singular points has to be closed.
There are something to beware of which are illustrated in the example below.
\begin{example}
    Consider $f(z)=\frac{1}{1-z}=\sum_{n\ge 0}z^n$.
    We know that the set of singular point is just $\{1\}$.
    However, the power series evaluated at $-1$ does not converge.
    So a regular point needs not guarantee that the power series converge there.
\end{example}
\begin{example}
    Consider the power series
    $$g(z)=\sum_{n\ge 2}\frac{z^n}{n(n-1)}$$
    Then the series converges at $z=1$, but $1$ is not regular for $g$ because it is singular for $f=g^{\prime\prime}$.
\end{example}
The moral od the examples is whether or not a point is regular does not relate directly to whether or not the power series converges there.
\begin{proposition}
    If a power series $f(z)=\sum_{n\ge 0}a_nz^n$ has radius of convergence $1$, then some point of $\mathbb T$ is singular.
\end{proposition}
\begin{proof}
    Just use the compactness of $\mathbb T$.
\end{proof}
\begin{definition}
    If every point of $\mathbb T$ is singular, we say $\mathbb T$ is the natural boundary of $f$.
\end{definition}
\begin{remark}
    We can extend this definition to other simple curves in $\mathbb C_\infty$.
\end{remark}
\begin{example}
    Consider the function $f(z)=\sum_{n\ge 0}z^{n!}$.
    We shall show that $\mathbb T$ is its natural boundary.
    It suffices to show any point in the form $\omega=e^{2\pi ip/q}$ is singular, where $p,q\in\mathbb Z,q\neq 0$.
    Indeed, fix such an $\omega$ with corresponding $p,q$, then for $r\in(0,1)$,
    $$f(r\omega)=\sum_{n=0}^{q-1}r^{n!}\omega^{n!}+\sum_{n\ge q}r^{n!}$$
    But the second term definitely converge.
    To see this, we observe that for any $M$,
    $$\lim_{r\to 1}\sum_{n=q}^{M+q}r^{n!}=M+1\implies \sum_{n=q}^{M+q}r^{n!}>M$$
    for $r$ sufficiently close to $1$.
    So the second term converges.
    Yet the first term is bounded, therefore $f(r\omega)\to\infty$ as $r\to1$, so $\omega$ has to be singular at $\omega$.
\end{example}


\subsection{Lerner index}

Marginal profit. If high it suggests existing power prevents it from raising output.

\(L=\dfrac{P-MC}{P}\)

From \(0\) to \(1\).


\section{Separable Differential Equations}
\noindent
The most basic approach for solving a 1st-order differential equation is simply integrating both sides. You're probably already familiar with this technique from taking indefinite integrals. This approach only works when the independent and dependent variables can be arranged on different sides of the equation. We'll formalize this idea with separability.

\begin{definition}
	A 1st order ODE is separable if it can be written in the form
	\begin{equation*}
		\frac{\mathrm{d} y}{\mathrm{d} x} = f(x)g(y)
	\end{equation*}
\end{definition}

\noindent
Separable equations provide a special way of solving them that can be useful. If we treat the derivative like a fraction (which is not formally allowed but OK here),
\begin{equation*}
	\frac{\mathrm{d} y}{\mathrm{d} x} = f(x)g(y) \implies \frac{\mathrm{d} y}{g(y)} = f(x) \mathrm{d}x \implies \int{\frac{\mathrm{d} y}{g(y)}} = \int{f(x) \mathrm{d}x}
\end{equation*}
We then have a function in $y$ on the left and a function in $x$ on the right, meaning we only have to solve for $y$ to get the solution.

\ifodd\includeFirstOrderLinearODEExamples\input{./1stOrderLinearODE/separability/separability_example.tex}\fi