%% -*- coding:utf-8 -*-
\chapter{Ring extensions, norms and traces, reduction modulo $p$}
We build a tool for finding elements in Galois groups, learning to use
the reduction modulo $p$. For this, we have to talk a little bit about
integral ring extensions and also about norms and traces.

\section{Integral elements over a ring}

Let $P \in \mathbb{Z}\left[X\right]$. We want to know what is
$Gal\left(P\right)$. Just a reminder that $Gal\left(P\right) =
Gal\left(K/\mathbb{Q}\right)$ where $K$ is a
\mynameref{def:splittingfield} of $P$. We have already done the work for
several types of polynomials: \mynameref{def:cyclotomicpolynomial}s,
\mynameref{sec:kummerextension} and so on.

Sometimes, if our polynomial is a kind of combination of them, then the
explicit information about the roots helps to calculate the Galois
group. For instance if we have polynomial $X^5 - 2$ we know it's
roots: $\sqrt[5]{2}, j^k \sqrt[5]{2}$, where
$j= e^{\frac{2 \pi i}{5}}, 1 \le k \le 4$. Now we have a lot about the
\mynameref{def:galoisgroup}. If $K$ is the splitting field of the
polynomial then we have the following towers:

  \begin{tikzpicture}[descr/.style={fill=white,inner sep=2.5pt}]
    \matrix (m) [matrix of math nodes, row sep=3em,
      column sep=3em]
            { & \mathbb{Q}\left(j\right) & \\
              \mathbb{Q} & & \mathbb{Q}\left(\sqrt[5]{2}, j\right) = K\\
              & \mathbb{Q}\left(\sqrt[5]{2}\right) & \\ };
            \path[->,font=\scriptsize]
            (m-2-1) edge node[descr] {$ 4 $} (m-1-2)
            (m-1-2) edge node[descr] {$ 5 $} (m-2-3)
            (m-2-1) edge node[descr] {$ 5 $} (m-3-2)
            (m-3-2) edge node[descr] {$ 4 $} (m-2-3);
  \end{tikzpicture}

From that we know we can conclude that it follows that our Galois
group, contains a normal cyclic subgroup of a order 5:
$\mathbb{Z}/5\mathbb{Z}$.
\footnote{
  We have the following towers:
  $\mathbb{Q} \subset \mathbb{Q}\left(j\right) \subset
  \mathbb{Q}\left(j, \sqrt[5]{2}\right) = K$, where
  $\mathbb{Q}\left(j\right)$ and $\mathbb{Q}\left(j,
  \sqrt[5]{2}\right)$ are \mynameref{def:galoisextension}s.
  By \mynameref{thm:galoiscorrespondence} one can get that
  \[
  Gal\left(K/\mathbb{Q}\left(j\right)\right) \triangleleft
  Gal\left(K/\mathbb{Q}\right) = Gal\left(P\right).
  \]
  As soon as (see theorem \ref{thm:lec5_4}),
  \[
  \left|Gal\left(K/\mathbb{Q}\left(j\right)\right)\right| =
  \left[K:\mathbb{Q}\left(j\right)\right] = 5,
  \]
  the $Gal(P)$ has a normal subgroup of order 5.
}
And then
the quotient is the Galois group of cyclotomic extension
\footnote{
  The quotient is
  \[
  Gal\left(K/\mathbb{Q}\right)/Gal\left(K/\mathbb{Q}\left(j\right)\right).
  \]
  It has order $20/5=4$ and therefore it is
  $\left(\mathbb{Z}/5\mathbb{Z}\right)^\times$. It is also
  (see claim \ref{claim:galoisquotient})
  $Gal\left(\mathbb{Q}\left(j\right)/\mathbb{Q}\right)$
}
, so this is
$\left(\mathbb{Z}/5\mathbb{Z}\right)^\times$.
So this is a group of
order 20.
\footnote{
  There is a General affine group:
  $Gal(P) \cong GA(1,5) \triangleleft S_5$
  \cite{groupwiki:s5structure, groupwiki:generalaffinegroup}
}
You can show that this
is noncommutative, and from this exact sequence, you have some
information about it. But what will we do if we don't know the
roots. One of the tool that we will use is the reduction of modulo
prime and this will be the subject of the lecture.

\begin{gapexample}[Galois group]
  Let investigate the \mynameref{def:galoisgroup}
  $Gal\left(K/\mathbb{Q}\right)$ via GAP:
  \begin{verbatim}
    gap> x:=Indeterminate(Rationals,"x");;
    gap> n:=5;;
    gap> i:=GaloisType(x^n-2);;
    gap> g:=TransitiveGroup(n,i);;
    gap> g;
    F(5) = 5:4
    gap> Order(g);
    20
  \end{verbatim}
\end{gapexample}

\subsection{Ring extensions}
\begin{definition}[Integral element]
  Let $A$ be an \mynameref{def:integraldomain}, i.e. a ring without zero
  divisors and let $B$ is an extension of $A$. The element $\alpha
  \in B$ is called integral over $A$ if $\alpha$ is a root of a
  \mynameref{def:monicpolynomial} $P \in A\left[X\right]$.

  So one can write the following relation
  \[
  \alpha^n + a_{n-1}\alpha^{n-1} + \dots + a_1 \alpha + a_0 = 0, a_i
  \in A.
  \]
  \label{def:integralelement}
\end{definition}

\begin{example}
  $\frac{1}{2}$ is not integral element over $\mathbb{Z}$ but
  $\sqrt{2}$ is an \mynameref{def:integralelement} over $\mathbb{Z}$.

  This is because the polynomial in the definition
  \ref{def:integralelement} is monic i.e. the leading coefficient is
  1.
\end{example}

\begin{lemma}
  The following conditions are equvivalent
  \begin{enumerate}
  \item $\alpha$ is integral over $A$.
  \item $A\left[\alpha\right]$ is a finitely generated $A$-module
    (see definition \ref{def:fgmodule}).
  \item $A\left[\alpha\right] \subset C \subset B$ where $C$ is a a
    finitely generated
    $A$-module (see definition \ref{def:fgmodule}). I.e.
    $A\left[\alpha\right]$ is contained in a finitely generated $A$-module.
  \end{enumerate}
  \label{lem:lec9_1}
  \begin{proof}
    $1 \to 2 \to 3$ is easy
    \footnote{
      $1 \to 2$ is really easy because the finite set $\{1,\alpha,
      \dots, \alpha^{n-1}\}$ generates
      $A\left[\alpha\right]$ i.e.
      $\forall x \in A\left[\alpha\right], \exists \{x_i\}
      \subset A$ such that $x = \sum_{i=0}^{n-1} x_i \alpha^{i}$.

      $2 \to 3$ is even more easy because $C=A\left[\alpha\right]$ will work.
    } and we will concentrate on $3 \to 1$.

    Let $x_1, \dots, x_r$ generate $C$ as $A$-module then we can write
    \footnote{
      As soon as $\alpha x_i \in C$ is an element of $C$ that can be written
      as a linear combination of
      $x_1, \dots, x_r$ with coefficients from $A$.
    }
    \[
    \alpha x_i = \sum \lambda_{ij}x_j,
    \]
    where $\lambda_{ij} \in A$. Consider the matrix
    $\Lambda = \{\lambda_{ij}\}$ and let $M = \alpha \cdot id -
    \Lambda$. Then
    \[
    M \cdot
    \begin{pmatrix}
      x_1 \\ \vdots \\ x_r
    \end{pmatrix} = 0.
    \]
    Thus (see (\ref{eq:lec4_det}) at theorem
    \ref{thm:structurefinitekalgebra} proof)
    \[
    \det M \cdot
    \begin{pmatrix}
      x_1 \\ \vdots \\ x_r
    \end{pmatrix} = 0.
    \]
    Therefore $\det M \cdot C = 0$ but $1 \in C$ thus $\det M = 0$.
    \footnote{
      There is a question why do we need $\det M \cdot C = 0$, may be
      we can directly write $\det M = 0$? Staff provided a good
      example why it's necessary: Sending generators to zero is not
      sufficient for having zero determinant. For example, consider a
      ring $A$, its ideal $I \subset A$ and $A/I$ as a
      $A$-\mynameref{def:cyclicmodule} i.e. module with one generator.
      If $x \in A$ is the generator then we can just say that $\det M x
      \in I$ and as result $\det M$ not necessary to be zero.
    }
    The equation $\det M = 0$ can be considered as a polynomial with
    $\alpha$ as a root.
    \footnote{
      The polynomial will be a \mynameref{def:monicpolynomial} because
      the matrix element can have $\alpha$ with coefficient equal $1$.
    }
  \end{proof}
\end{lemma}

\section{Integral extensions, integral closure, ring of integers of a
  number field}

\subsection{Integral extensions and integral closure}

\begin{definition}[Integral extension]
  Let $A \subset B$. $B$ is integral over $A$ if $\forall \alpha \in
  B$, $\alpha$ is an \mynameref{def:integralelement} over $A$.
  \label{def:integralextension}
\end{definition}

The following proposition is not a part of the lectures but it is
required for propositions \ref{prop:lec9_1} and \ref{prop:lec9_2}
proof.
\begin{proposition}
  Let $A$ is a sub-ring of $C$ and
  $\alpha_1, \alpha_2, \dots, \alpha_n \in C$,
  $\alpha_1$ is an integral over $A$,
  $\alpha_2$ is an integral over $A\left[\alpha_1\right]$, and so on,
  $\alpha_n$  is an integral over
  $A\left[\alpha_1, \alpha_2, \dots,   \alpha_{n-1}\right]$ then
  $A\left[\alpha_1, \alpha_2, \dots,\alpha_{n}\right]$ is a finitely
  generated $A$ module.
  \begin{proof}
    Lets proof by induction. The case $n=1$ follows directly from
    lemma \ref{lem:lec9_1}.

    Induction hypothesis gives us that
    $B = A\left[\alpha_1, \alpha_2, \dots,\alpha_{n-1}\right]$ is a finitely
    generated $A$ module and we have to prove that
    $S = A\left[\alpha_1, \alpha_2, \dots, \alpha_{n}\right]$ is a finitely
    generated $A$ module.

    For $B$ as a finitely generated $A$ module we have
    \[
    B = \sum_i A b_i,
    \]
    but $S = B\left[\alpha_n\right]$ is a finitely generated $B$
    module (thanks lemma \ref{lem:lec9_1}) and therefore
    \[
    S = \sum_j B s_j
    \]
    and as result
    \[
    S = \sum_{ij} A b_i s_j =
    \sum_{ij} A k_{ij}
    \]
    where $b_i s_j = k_{ij} \in S$ form a finite set of generators for $S$.
  \end{proof}
  \label{prop:lec9_add}
\end{proposition}

\begin{proposition}
  Let $A \subset B \subset C$. $B$ integral over $A$, $C$ integral
  over $B$ then $C$ is an \mynameref{def:integralextension} over $A$.
  \begin{proof}
    Proof is left as an exercise.
    \footnote{
      Let $x \in C$ then
      \[
      x^n + b_{n-1} x^{n-1} + \dots + b_1 x + b_0 = 0,
      \]
      where $b_i \in B$. Thus $x$ is an integral element over
      $A\left[b_0, b_1, \dots, b_{n-1}\right]$. We also have that $B$
      is integral over $A$ therefore $b_i$ is an integral over $A$. As
      result we have $b_0$ is integral over $A$, $b_1$ is integral
      over $A\left[b_0\right]$, and so on $x$ is integral over
       $A\left[b_0, b_1, \dots, b_{n-1}\right]$. As result with
      proposition \ref{prop:lec9_add} we have that
      $A\left[b_0, b_1, \dots, b_{n-1}, x\right]$ is a finitely
      generated $A$ module and as
      $A\left[x\right] \subset A\left[b_0, b_1, \dots, b_{n-1},
        x\right]$, the lemma \ref{lem:lec9_1} gives us that $x$ is an
      integral element over $A$.
    }
  \end{proof}
  \label{prop:lec9_1}
\end{proposition}

\begin{proposition}
  $B$ is a finitely generated over $A$ as a module (see definition
  \ref{def:fgmodule}) if and only if
  $B=A\left[\alpha_1, \dots, \alpha_r\right]$ where each $\alpha_i$ is
  an \mynameref{def:integralelement} over $A$.
  \begin{proof}
    Proof is left as an exercise
    \footnote{
      If $\alpha_i$ is an integral over $A$ then it also an integral
      over $A\left[\alpha_1, \alpha_2, \dots, \alpha_{i-1}\right]$ and
      therefore by proposition \ref{prop:lec9_add},
      $A\left[\alpha_1, \alpha_2, \dots, \alpha_r\right] = B$ is finitely
      generated $A$-module.

      Let $B$ is finitely generated $A$ module i.e.
      $B = \sum_{i=1}^r \alpha_i x_i$, where
      $\alpha_i \in B$ - the module generators and $x_i \in A$.
      If we look at any $\alpha_i$ we can notice that
      $A\left[\alpha_i\right] \subset B$ - finitely generated $A$
      module and therefore by lemma \ref{lem:lec9_1},
      $\alpha_i$ is an integral element. We also have (from definition
      \ref{def:fgmodule}) that $\forall b \in B, \exists \{x_i\} \in
      A$ such that $b = \sum_{i=1}^r \alpha_i x_i$. Therefore
      $B = A\left[\alpha_1, \alpha_2, \dots, \alpha_r\right]$.
    }
  \end{proof}
  \label{prop:lec9_2}
\end{proposition}

\begin{proposition}
  Let $A \subset B$. I.e. $B$ is an arbitrary extension of $A$. The
  elements of $B$ which are integral over $A$ form a subring of $B$
  (one calls it as the integral closure of $A$ in $B$).
  \begin{proof}
    Let $\alpha, \beta$ are integral over $A$ then $A\left[\alpha,
      \beta\right]$ - finitely generated $A$-module (see definition
    \ref{def:fgmodule}). This follows directly from lemma
    \ref{lem:lec9_1}. It contains $\alpha + \beta$ and $\alpha \beta$
    and by lemma \ref{lem:lec9_1} the $\alpha + \beta$ and $\alpha
    \beta$ are integral over $A$. But this is exactly we need to
    proof.
    \footnote{
      $A\left[\alpha + \beta\right] \subset A\left[\alpha,
      \beta\right]$ and as soon as $A\left[\alpha,
        \beta\right]$ - finetely generated $A$ module then
      $\alpha+\beta$ is an integral element by lemma \ref{lem:lec9_1}.

      Same result can be got for $\alpha \beta$ from the following
      inclusion:  $A\left[\alpha \beta\right] \subset A\left[\alpha,
      \beta\right]$
    }
  \end{proof}
  \label{prop:lec9_3}
\end{proposition}

\begin{definition}[Integrally closed]
  Let $A \subset B$. $A$ is integrally closed in $B$ if the integral
  closure of $A$ in $B$ equals to $A$.

  $A$ is integrally closed (without mention of any $B$) if it is
  integrally closed in \mynameref{def:fractionfield} $\mathrm {Frac}(A)$.
  \label{def:integrallyclosed}
\end{definition}

\begin{example}
  $\mathbb{Z}$ is \mynameref{def:integrallyclosed}.
  \footnote{
    $\mathrm {Frac}(\mathbb{Z}) = \mathbb{Q}$. An arbitrary element
    $q \in \mathbb{Q}$ is an integral element over
    $\mathbb{Z}$ if and only if $q \in \mathbb{Z}$. For example
    $5 \in \mathbb{Z} \subset \mathbb{Q}$ is integral over $\mathbb{Z}$ as a root of a
    monic polynomial $P = X - 5$. But $\frac{5}{2} \in \mathbb{Q}$ is
    not integral over $\mathbb{Z}$ because it is a root of a non-monic
    polynomial $P = 2 X - 5$.
  }
\end{example}

\begin{remark}
  More generally any
  \mynameref{def:ufd} is \mynameref{def:integrallyclosed}.
  \begin{proof}
    Let $A$ be a \mynameref{def:ufd} and $x \in \mathrm {Frac}(A)$ such
    that $x \ne 0$. So
    $x = \frac{p}{q}$ such that $p,q \in A, \left(p, q\right) = 1$
    (this means no common prime divisor). If $x$ integral over $A$
    then
    \[
    \left(\frac{p}{q}\right)^n + a_{n-1}\left(\frac{p}{q}\right)^{n-1}
    + \dots + a_1 \frac{p}{q} + a_0 = 0.
    \]
    Thus
    \[
    \frac{p^n + q a_n p^{n-1} + q^2 a_{n-1} p^{n-2} + \dots + q^{n-1}
      a_1 p + q^n a_0}{q^n} = 0
    \]
    therefore $q \mid p^n$
    \footnote{
      This is because we have
      \[
      p^n = q \left(-a_n p^{n-1} - q a_{n-1} p^{n-2} - \dots - q^{n-2}
      a_1 p - q^{n-1} a_0\right)
      \]
      and
      \[
      \left(-a_n p^{n-1} - q a_{n-1} p^{n-2} - \dots - q^{n-2}
      a_1 p - q^{n-1} a_0\right) \in A
      \]
    }
    which is in contradiction with
    $\left(p, q\right) = 1$. As result we have that $q$ is invertible
    and therefore $x \in A$.
    \footnote{
      $q^{-1} \in A$ and $x = \frac{p}{q} = p q^{-1} \in A$.
    }
    \end{proof}
\end{remark}

\subsection{Ring of integers in a number field}

\begin{definition}[Number field]
  Let $K$ is a finite extension of $\mathbb{Q}$ i.e.
  $\left[K:\mathbb{Q}\right] < \infty$. In the case $K$ is a number
  field.
  \label{def:numberfield}
\end{definition}

Let $K$ is a \mynameref{def:numberfield} and $\left[K:\mathbb{Q}\right]
= N$.
\begin{definition}[Ring of integers]
  Let $K$ is a \mynameref{def:numberfield}. The ring of integers
  $O_K \subset K$ is the integral closure of $\mathbb{Z}$
  in $K$.

  Note: We know that integral closure of $\mathbb{Z}$ in $\mathbb{Q}$
  is $\mathbb{Z}$ but now we consider the closure in $K$ but not in
  $\mathbb{Q}$.
  \label{def:ringintegers}
\end{definition}

\begin{property}
  \begin{enumerate}
  \item $\forall \alpha \in K, \exists d \in \mathbb{Z} \setminus
    \{0\}$ such that $d\alpha \in O_K$.
  \item If $\alpha \in O_K$ then $P_{min}\left(\alpha,
    \mathbb{Q}\right) \in \mathbb{Z}\left[X\right]$.
  \end{enumerate}
  \begin{proof}
    For the first part lets
    $P_{min}\left(\alpha,
    \mathbb{Q}\right) = X^m + a_{m-1}X^{m-1} + \dots + a_1 X + a_0
    \in \mathbb{Q}\left[X\right]$.

    $\exists d \in \mathbb{Z}$ (the common denominator) such that
    $\forall i: d a_i \in \mathbb{Z}$. So
    $b_i = d^{m-i}a_i \in \mathbb{Z}$ for any $i$. Therefore
    \[
    \left(d \alpha\right)^m + b_{m-1} \left(d \alpha\right)^{m-1} +
    \dots + b_0 = 0.
    \]
    Thus $d \alpha \in O_K$.

    The second part is also easy. If we have such
    $\alpha \in O_K$, it is a
    root of some \mynameref{def:monicpolynomial}
    $Q \in \mathbb{Z}\left[X\right]$.
    Then the $P_{min} \mid Q$. So $Q = P_{min} R$.
    If we pick $P_{min}$ to be monic, then by an
    argument very similar to that of the \mynameref{lem:gauss},
    we conclude  that both $P_{min}, R \in \mathbb{Z}\left[X\right]$.
    \footnote{
      I.e. we can always write
      \[
      Q = m n P_1 R_1
      \]
      where $P_1, R_1 \in \mathbb{Z}\left[X\right]$. Choose $p$ is a
      prime divisor of $m n$ we can write
      \[
      \bar{P_1}\bar{R_1} = \bar{Q} = 0
      \]
      where $\bar{Q} = Q \mod p$, $\bar{P}_1 = P_1 \mod p$,
      $\bar{R_1} = R_1 \mod p$ thus $\bar{P}_1$ or $\bar{R}_1$ are
      equal to 0. Let $\bar{R}_1 = 0$ thus all coefficients of $R_1$ are
      divided by $p$ i.e. $R_1 = \frac{R_2}{p}$ where
      $R_2 \in \mathbb{Z}\left[X\right]$. Therefore
      \[
      Q = \frac{mn}{p}P_1 R_2
      \]
      Continue this way we can conclude that
      \[
      Q = P_s R_t,
      \]
      where $P_s, R_t \in \mathbb{Z}\left[X\right]$. As soon as $Q$ is
      monic then both $P_s$ and $R_t$ are monic. Using the fact that
      $P_s = z P_{min}$ where $z \in \mathbb{Z}$ we have
      $P_s\left(\alpha\right) = 0$ and therefore we can
      conclude that $P_s$ is the minimal polynomial.
    }
  \end{proof}
  \label{property:lec9_2}
\end{property}

\section{Norm and trace}

\subsection{Norms and traces}
(The material was given inside the proof of theorem \ref{thm:lec9_1}
and can be considered as a recall. The remarks
\ref{rem:lec9_embedding} and \ref{rem:lec9_add} are not parts of the
lectures and were given for better understanding the material)

\begin{remark}[$K$ embedding of $E$ into the algebraic closure
    $\bar{K}$]
  Let $K \subset E \subset \bar{K}$. When we say about $K$ embedding
  of $E$ into the algebraic closure $\bar{K}$ we assume
  $\sigma \in Hom_K\left(E, \bar{K}\right)$ i.e. $\sigma$ is a
  \mynameref{def:homomorphism} of \mynameref{def:kalgebra}s i.e. the map
  that preserves the structure and especially $\sigma(K) = K$ i.e.
  $\forall k \in K: \sigma(k) = k$.

  If $E$ is a normal extension then all such homomorphisms have the
  same image accordingly theorem \ref{thm:lec5_3}. The image is $E$
  and therefore the homomorphisms can be considered as automorphisms
  i.e. $Gal\left(E/K\right) = Hom_K\left(E, \bar{K}\right)$.
  \label{rem:lec9_embedding}
\end{remark}

\begin{definition}[Norm]
  Let $K \hookrightarrow E$ - finite separable field extension. Let $\alpha \in
  E$.   Define the norm of alpha with respect to this extension as
  \[
  \mathrm {N}_{E/K}\left(\alpha\right) =
  \prod_{\sigma_i: E \hookrightarrow \bar{K}} \sigma_i\left(\alpha\right)
  \]
  i.e. we took a product by all $K$ embeddings of $E$ into the
  algebraic closure of $K$ (see remark \ref{rem:lec9_embedding}).
  And we also assume that $E$ is
  finite and as result $i = 1, \dots, r$.
  \label{def:norm}
\end{definition}

\begin{definition}[Trace]
  Let $K \hookrightarrow E$ - finite separable field extension. Let $\alpha \in
  E$.   Define the norm of alpha with respect to this extension as
  \[
  \mathrm {Tr}_{E/K}\left(\alpha\right) =
  \sum_{\sigma_i: E \hookrightarrow \bar{K}} \sigma_i\left(\alpha\right)
  \]
  i.e. we took a sum by all $K$ embeddings of $E$ into the
  algebraic closure of $K$ (see remark \ref{rem:lec9_embedding}). And
  we also assume that $E$ is finite and as result $i = 1, \dots, r$.
  \label{def:trace}
\end{definition}

In the definitions \ref{def:norm} and \ref{def:trace} we assume that
the extension $E$ is \mynameref{def:separableextension}. If the
extension is not separable then you have to take it to the power
equal to the pure inseparable degree of $E/K$,
\footnote{
  \cite{bib:lang} p. 284 gives the following definitions for norm and
  trace in not separable case:
  \begin{eqnarray}
  \mathrm {N}_{E/K}\left(\alpha\right) =
  \left(
  \prod_{\sigma_i: E \hookrightarrow \bar{K}}
  \sigma_i\left(\alpha\right)
  \right)^{\left[E:K\right]_i},
  \nonumber \\
  \mathrm {Tr}_{E/K}\left(\alpha\right) =
  \left[E:K\right]_i
  \sum_{\sigma_i: E \hookrightarrow \bar{K}} \sigma_i\left(\alpha\right),
  \nonumber
  \end{eqnarray}
  where $\left[E:K\right]_i$ is \mynameref{def:inseparabledegree}.
}
but for simplicity,
we shall suppose that everything is separate.

\begin{remark}
  We have $\mathrm {N}_{E/K}\left(\alpha\right) \in K$ and
  $\mathrm {Tr}_{E/K}\left(\alpha\right) \in K$
  \label{rem:lec9_add}
  \begin{proof}
    Let $g \in Gal\left(E/K\right)$. We have to prove that
    $g\left(\mathrm {N}_{E/K}\left(\alpha\right)\right) =
    \mathrm {N}_{E/K}\left(\alpha\right)$. In the case
    $\mathrm {N}_{E/K}\left(\alpha\right) \in K$ because
    $E^{Gal\left(E/K\right)} = K$.

    The $g$ just permutes the homomorphisms $Hom_K\left(E,
    \bar{K}\right)$ i.e.
    $\left|g Hom_K\left(E, \bar{K}\right)\right| =
    \left|Hom_K\left(E, \bar{K}\right)\right|$. If this is not the
    truth then $\exists \sigma_i, \sigma_j, \sigma_k \in Hom_K\left(E,
    \bar{K}\right)$ such that
    $g\sigma_i = \sigma_k, g \sigma_j = \sigma_k$ and
    $\sigma_i \ne \sigma_j$.
    Therefore
    \[
    \sigma_j = g^{-1} \sigma_k = g^{-1} g \sigma_i = \sigma_i
    \]
    that is contradiction.
    By  the field definition
    (\ref{def:field}) the product in
    \mynameref{def:norm} does not depend on the order and the
    result become the same after the permutation.

    The same result is for \mynameref{def:trace}.
  \end{proof}
\end{remark}

\begin{property}
  \begin{enumerate}
  \item $\mathrm{N}_{E/K}: E^\times \to K^\times$
    \footnote{
      $E^\times = E \setminus \{0\}$ and
      $K^\times = K \setminus \{0\}$
    }
    is multiplicative
    i.e. homomorphism of groups.
    $\mathrm{Tr}_{E/K}: E \to K$ is additive, $K$-linear
    i.e. homomorphism of $K$-vector spaces.
    \footnote{
      The property statement is the truth i.e.
      there really should be $K$ (not $\bar{K})$(see remark
      \ref{rem:lec9_add}).
    }
  \item If $E=K\left(\alpha\right)$, $n = \left[E:K\right]$ and
    $P_{min}\left(\alpha, K\right) = X^n + a_1 X^{n-1} + \dots +
    a_{n-1} X + a_{n}$ then $\mathrm{N}_{E/K}\left(\alpha\right) = (-1)^n
    a_n$ and $\mathrm{Tr}_{E/K}\left(\alpha\right) = -a_1$.
  \item If we have the tower of extensions $K \subset F \subset E$
    then
    \[
    \mathrm{N}_{E/K} =
    \mathrm{N}_{F/K} \circ
    \mathrm{N}_{E/F}
    \]
    and the same for trace
    \[
    \mathrm{Tr}_{E/K} =
    \mathrm{Tr}_{F/K} \circ
    \mathrm{Tr}_{E/F}
    \]
    \footnote{
      Or in other words
      \[
      \mathrm{N}_{E/K}\left(\alpha\right) =
      \mathrm{N}_{F/K} \left(
      \mathrm{N}_{E/F}\left(\alpha\right)
      \right)
      \]
      and
      \[
      \mathrm{Tr}_{E/K}\left(\alpha\right) =
      \mathrm{Tr}_{F/K} \left(
      \mathrm{Tr}_{E/F}\left(\alpha\right)
      \right).
      \]
      Lets also note \cite{bib:KeithConradTraceNorm2} that the
      following expression does not make sense:
      \[
      \mathrm{Tr}_{E/F} \left(
      \mathrm{Tr}_{F/K}\left(\alpha\right)\right)
      \]
      because $\mathrm{Tr}_{E/F}: E \to F$ and
      $\mathrm{Tr}_{F/K}: F \to K$ (see remark
      \ref{rem:lec9_add}).
      The same is valid for \mynameref{def:norm}.
    }
    \item Consider $T: E \times E \xrightarrow[(x, y) \to
      \mathrm{Tr}_{E/K}\left(xy\right)  ]{} K$. This is a
      non-degenerate $K$-bilinear form (see
      definition \ref{def:nondegeneratebf})
    \item $\alpha$ integral over $\mathbb{Z}$. Then
      $\mathrm{N}_{E/\mathbb{Q}}\left(\alpha\right),
      \mathrm{Tr}_{E/\mathbb{Q}}\left(\alpha\right)$ are integers.
      \footnote{
        We have $K = \mathbb{Q}$ in the property.
      }
  \end{enumerate}
  \label{property:lec9_norm_trace}
  \begin{proof}
    The first property is obvious from the definition.
    \footnote{
      Let $\alpha, \beta \in E$ then
      \begin{eqnarray}
        \mathrm{N}_{E/K}\left(\alpha \beta\right) =
          \prod_{\sigma_i: E \hookrightarrow \bar{K}}
          \sigma_i\left(\alpha \beta\right) =
          \nonumber \\
          =
          \prod_{\sigma_i: E \hookrightarrow \bar{K}}
          \sigma_i\left(\alpha\right)
          \sigma_i\left(\beta\right)
          =
          \mathrm{N}_{E/K}\left(\alpha\right)
          \mathrm{N}_{E/K}\left(\beta\right).
        \nonumber
      \end{eqnarray}
      With remark \ref{rem:lec9_add} we can get that
      \[
      \mathrm{N}_{E/K}: E^\times \to K^\times
      \]

      If we take $a,b \in K, \sigma \in Hom_K\left(E, \bar{K}\right)$
      then one can get that
      $\sigma\left(a \alpha\right) = a
      \sigma\left(\alpha\right)$. This is because $a$ is fixed under
      every embedding of $E$ over $K$ (\cite{bib:lang} page 286), see
      also remark \ref{rem:lec9_embedding}.

      Therefore one can get
      \[
      \mathrm{Tr}_{E/K}\left(a \alpha + b \beta\right)=
      a \mathrm{Tr}_{E/K}\left(\alpha\right) +
      b \mathrm{Tr}_{E/K}\left(\beta\right)
      \]
      I.e. there is a $K$-linear map of $E$ to $K$
      (see remark \ref{rem:lec9_add}).
    }

    The second one uses the following fact:
    $\sigma_i\left(\alpha\right)$ are roots of $P_{min}\left(\alpha,
    K\right)$. The \mynameref{def:norm} is a product and it's assigned
    to its constant term ($a_n$) and the sum is the first coefficient
    term ($a_1$) (see also example \ref{ex:lec8_generalequation}).

    The third property is somewhat less trivial, so this follows from,
    the fact that if $\tau_1, \dots, \tau_k$ are $K$ embeddings of $F$
    into $\bar{K}$ and, $\mu_1, \dots, \mu_s$ are $F$ embeddings of
    $E$ into $\bar{K}$ then the embedings of $E$ into $\bar{K}$ are
    just the compositions $\{\tau_j \mu_i\}$.
    \footnote{
      We have $K \subset F \subset \bar{K}$, and (see remark
      \ref{rem:lec9_embedding}):
      \[
      \begin{cases}
        \tau_j: F \to \bar{K} \text{ such that }
        \forall k \in K: \tau_j(k) = k, \\
        \mu_i: E \to \bar{K}  \text{ such that }
        \forall f \in F: \mu_i(f) = f.
      \end{cases}
      \]
      Lets extend homomorphism $\tau_j: F
      \to \bar{K}$ to the automorphism
      $\bar{K} \to \bar{K}$ and denote the result with the same
      $\tau_j$ (\cite{bib:lang} p. 285).
      As result we have $\tau_j \mu_i: E \to \bar{K}$. We also have
      $\forall k \in K \subset F$
      \(
      \mu_i(k) = k
      \) (as soon as $k \in F$) and therefore
      $\tau_j(\mu_i(k)) = \tau_j(k) = k$. Thus
      $\tau_j \mu_i \in Hom_K\left(E, \bar{K}\right)$.
    }

    For the 4th property. Indeed if $x \in \ker T$,
    \footnote{
      $T: E \times E \to K$
    }
    that means
    $\mathrm{Tr}_{E/K}\left(xy\right) = 0, \forall y \in E$ (see
    definition \ref{def:nondegeneratebf}), but this
    can't be a case when $xy \in K \setminus \{0\}$ by definition
    \ref{def:trace}
    \footnote{
      and by taking into consideration the following fact
      (see remark \ref{rem:lec9_embedding}):
      if $xy \in K$
      then $\sigma_i(xy) = xy$
    }
    $\mathrm{Tr}_{E/K}\left(xy\right) =
    \left[E:K\right] xy$.
    \footnote{
      we proved that $T$ has a trivial kernel i.e. only $x = 0$ is in
      the $\ker T$. There is one of definitions of the non-degenerate
      $K$-bilinear form (see definition \ref{def:nondegeneratebf})
    }

    For the 5th property we know that
    \begin{eqnarray}
    \mathrm{Tr}_{E/\mathbb{Q}}\left(\alpha\right) =
    \mathrm{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\left(
    \mathrm{Tr}_{E/\mathbb{Q}(\alpha)}\left(\alpha\right)
    \right) =
    \nonumber \\
    =
    \mathrm{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\left(
    \left[E:\mathbb{Q}(\alpha)\right] \alpha
    \right) =
    \left[E:\mathbb{Q}(\alpha)\right]
    \mathrm{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\left(
     \alpha
    \right)
    \nonumber
    \end{eqnarray}
    but
    \(
    \mathrm{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\left(
    \alpha
    \right) \in \mathbb{Z}
    \)
    because
    \(
    \mathrm{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\left(
    \alpha
    \right)
    \) is a coefficient of $P_{min}\left(\alpha, \mathbb{Q}\right) \in
    \mathbb{Z}\left[X\right]$.
    \footnote{
      Property \ref{property:lec9_2} says that $P_{min}\left(\alpha,
      \mathbb{Q}\right) \in \mathbb{Z}\left[X\right]$. But 2d item of
      the property \ref{property:lec9_norm_trace} says that the
      \mynameref{def:trace} is a coefficient of the polynomial i.e.
      \(
      \mathrm{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\left(
      \alpha
      \right) \in \mathbb{Z}
      \)
    }
  \end{proof}
\end{property}

Why such names are used? Consider the following map
(multiplication by $a$)
\[
f_a: E \xrightarrow[x \to a x]{} E
\]
then the $\mathrm{Tr}_{E/K}\left(a\right)$ is exactly the trace of the
linear map (i.e. sum of diagonal elements of the linear map matrix in
a basis) and the $\mathrm{N}_{E/K}\left(a\right)$ is the
determinant
\footnote{
  Consider an easy case when $E = K\left(a\right)$. In the case
  $a$ is a root of $P_{min}\left(a, K\right) = a_n + a_{n-1} X + \dots +
  a_1 X^{n-1} + X^{n}$.
  We have the following basis $1, a, \dots, a^{n-1}$. The basis is
  transformed by multiplication via the following rules
  \[
  \begin{cases}
    1 \to a, \\
    a \to a^2, \\
    \vdots \\
    a^{n-1} \to a^n = -a_n - a_{n-1} a - \dots -
    a_1 a^{n-1}.
  \end{cases}
  \]
  Therefore the \mynameref{def:endomorphism} matrix can be written as
  follows:
  \[
  M =
  \begin{bmatrix}
    0 & 1 & 0 & \cdots & 0 & 0 \\
    0 & 0 & 1 & \cdots & 0 & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
    0 & 0 & 0 & \cdots & 0 & 1 \\
    -a_n & -a_{n-1} & -a_{n-2} & \cdots & -a_2 & -a_1
  \end{bmatrix}
  \]
  It can be easy seen (with 2d item of
  the property \ref{property:lec9_norm_trace}) that
  \[
  \mathrm{Tr}\left(M\right) = -a_1 =
  \mathrm{Tr}_{E/K}\left(a\right)
  \]
  and
  \[
  \det\left(M\right) =
  (-1)^n a_n =
  \mathrm{N}_{E/K}\left(a\right).
  \]
}
. Now this $f_a$ is a a $K$-linear map. It's an
\mynameref{def:endomorphism} of a vector space $E/K$, and
the $\mathrm{Tr}_{E/K}\left(a\right)$ is
the trace of this endomorphism, and the
$\mathrm{N}_{E/K}\left(a\right)$ is the determinant of this endomorphism.

\subsection{Theorem about rings of integers}
\begin{theorem}
  $O_k$ is a finitely generated (see definition \ref{def:fgmodule})
  $\mathbb{Z}$-module that is a \mynameref{def:freemodule} of
  rank (see definition \ref{def:rankfreemodule}) $n$, where $n=
  \left[K:\mathbb{Q}\right]$.
  \label{thm:lec9_1}
  \begin{proof}
    If $e_1, \dots, e_n$ is a $\mathbb{Q}$-basis of $K$ then
    $\forall i \exists d_i \in \mathbb{Z} \setminus
    \{0\}$ such that $d_i e_i \in O_K$ (see property
    \ref{property:lec9_2}). Therefore $O_K$ contains a free
    $\mathbb{Z}$-submodule of rank $n$
    \footnote{
      This is because $d_1 e_1, \dots, d_n e_n$ are linearly
      independent and form a basis of a free $\mathbb{Z}$-module. The
      number $n$ is the cardinality of the basis.

      It can be proved by contradiction i.e. let there exists a set
      $\{c_i\}$ such that $\exists j: c_j \ne 0$ and
      \[
      \sum_{i=1}^n c_i d_i e_i = 0,
      \]
      as soon as $d_j \ne 0$ then $k_j = c_j d_j \ne 0$ and
      \[
      \sum_{i=1}^n k_i e_i = 0,
      \]
      i.e. $\{e_i\}$ are not linearly independent. This is in
      contradiction with the initial conditions.
    }.

    What is the $\mathbb{Z}$-module this is a finitely generated
    \mynameref{def:fgagroup} and we know a lot of things about such
    groups. The \mynameref{def:fgagroup} is the same as
    finitely generated $\mathbb{Z}$-module. Any such group is isomorphic
    to (see theorem \ref{thm:fgagroup})
    \[
    \mathbb{Z}^n \oplus A,
    \]
    where $A$ is a finite group (torsion part). A subgroup
    $B \subset \mathbb{Z}^n$ is itself a free module ($B \cong \mathbb{Z}^m$)
    of rank $m \le n$.

    We have to show that $O_K \subset A$ where $A$ is a free
    $\mathbb{Z}$-submodule of rank $n = \left[K:\mathbb{Q}\right]$.
    Let $e_1, \dots, e_n$ is a $\mathbb{Q}$-basis of $K$ (as above)
    contained in $O_K$. Consider the following map
    ($T: K \times K \to \mathbb{Q}$):
    \[
    (x,y) \to \mathrm{Tr}_{K/\mathbb{Q}}\left(xy\right)
    \]
    this is
    \mynameref{def:nondegeneratebf} (see 4th property
    \ref{property:lec9_norm_trace}) therefore $\exists v_1, \dots,
    v_n$ - \mynameref{def:dualbasis} ($\mathbb{Q}$-basis of $K$)
    and we have the property that
    $\mathrm{Tr}_{K/\mathbb{Q}}\left(e_i v_j\right) = \delta_{ij}$.
    \footnote{
      Lets consider the following map $T: K \times K \to
      \mathbb{Q}$. We can use it to construct a linear map by the
      following rule: $\forall x \in K$ we have $f_x(y) =
      T(x,y): K \to \mathbb{Q}$.
      We have $f_{x+y} = f_x + f_y$ and $f_{ax} = a f_x$
      i.e. the set $\{f_x\} = K^\ast$ is the linear space.
      The map $K \xrightarrow[x \to f_x]{} K^\ast$ is
      \mynameref{def:surjection} by the $f_x$ construction. The map is
      also \mynameref{def:injection} as soon as the map $T: K \times K \to \mathbb{Q}$ is
      \mynameref{def:nondegeneratebf} and as result the $f_x(y) =
      T(x,y)$ has a trivial kernel (if $x \ne 0$): $\ker f =
      \{0\}$. Therefore we can conclude that $K^\ast$ is
      \mynameref{def:dualspace}. I.e. exists a set of elements
      $\{f_{v_j}\} \subset K^\ast$  which
      form the \mynameref{def:dualbasis} to $\{e_i\}$ i.e. $f_{v_j}(e_i) =
      \delta_{ij}$. Each element $f_{v_j}$ of dual space corresponds
      to $v_j \in K$. For such $v_j$ we have:
      \[
      \delta_{ij} = f_{v_j}(e_i) = T(v_j, e_i) =
      \mathrm{Tr}_{K/\mathbb{Q}}\left(v_j e_i\right).
      \]
    }

    We claim that $\mathbb{Z}$ submodule generated by $v_1, \dots,
    v_n$ contains $O_K$. Indeed let $\alpha \in O_K$ and write
    \(
    \alpha = \sum \alpha_i v_i, \alpha_i \in \mathbb{Q}
    \). We can do it because $\{v_i\}$ is a $\mathbb{Q}$ basis of
    $K$. But one can see that $\alpha_i \in \mathbb{Z}$ because
    $\alpha_i = \mathrm{Tr}_{K/\mathbb{Q}}\left(\alpha e_i\right)$ (by
    definition of $v_j$).
    \footnote{
      \begin{eqnarray}
        \mathrm{Tr}_{K/\mathbb{Q}}\left(\alpha e_i\right) =
        \mathrm{Tr}_{K/\mathbb{Q}}\left(\sum_{j=1}^n\alpha_j v_j
        e_i\right) =
        \nonumber \\
        = \sum_{j=1}^n\alpha_j
        \mathrm{Tr}_{K/\mathbb{Q}}\left( v_j
        e_i\right) =
        \sum_{j=1}^n\alpha_j \delta_{ij} = \alpha_i
        \nonumber
      \end{eqnarray}
    }
    Since $\alpha$ and $e_i$ are elements of
    $O_K$ then $\alpha e_i \in O_K$ too. Therefore
    $\mathrm{Tr}_{K/\mathbb{Q}}\left(\alpha e_i\right) \in
    \mathbb{Z}$. So $\alpha_i \in \mathbb{Z}$ and this one is what we
    want to proof. We have expressed any element of $O_K$ as a
    combination of $v_i$ with integral coefficients. So $O_K$ is contained
    in a $\mathbb{Z}$ submodule, generated by $\{v_i\}$.
  \end{proof}
\end{theorem}

\section{Reduction modulo a prime}

Let $P \in \mathbb{Z}\left[X\right]$ is an irreducible polynomial with integer
coefficients. $K$ is a \mynameref{def:splittingfield} of $P$ over
$\mathbb{Q}$ and $n = \left[K:\mathbb{Q}\right]$. Let
$G = Gal\left(P\right) \eqdef Gal\left(K/\mathbb{Q}\right)$. We denote
roots of $P$ as $\alpha_1, \dots, \alpha_n$ and they are elements of
$O_K$. $G$ acts on the set of roots, and on $O_K$. We will denote
$O_K$ as $A$. Let $p$ is a prime number and we will consider $A/pA$.
As we have seen
\footnote{
  see proposition \ref{prop:lec4_prop2} where $M = A, A=\mathbb{Z}$
  and $I = p\mathbb{Z}$.
}
\[
A/pA \cong A \otimes_{\mathbb{Z}} \mathbb{Z}/p\mathbb{Z} =
A \otimes_{\mathbb{Z}} \mathbb{F}_p
\]
there $A \otimes_{\mathbb{Z}} \mathbb{F}_p$ is a
$n$-dimension vector space over $\mathbb{F}_p$
\footnote{
  As soon as $A=O_K$ is a free
  $\mathbb{Z}$-module of rank $n$ (see theorem \ref{thm:lec9_1}) then
  proposition
  \ref{prop:lec4_Addon} gives us that $A \otimes_{\mathbb{Z}}
  \mathbb{F}_p$ is a free $\mathbb{F}_p$-module with rank equal
  to $n$.
}.
Maximal ideals
of $A/pA$ are in one-to-one correspondence with maximal ideals of $A$
containing $p$. As we know (see theorem
\ref{thm:structurefinitekalgebra}) there are only finitely many
maximal ideal in a finite algebra over a field. Therefore $A$ also has
finitly many
maximal ideals $J_1, \dots, J_r$ containing $p$. Our group $G$ acting
on $A$ must permute these maximal ideals in some way.
\footnote{
  Let $J$ is an \mynameref{def:ideal} of $A$, then $A J = J$
  (i.e. $\forall a \in A, j \in J: aj \in J$). Let also $g \in G$ then
  $g\left(A\right) = A$ (see \cite{bib:lang} p. 341) and
  \[
  A g\left(J\right) = g\left(A\right) g\left(J\right) =
  g\left(AJ\right) = g\left(J\right)
  \]
  i.e. $g\left(J\right)$ is also an ideal of $A$.
  \label{note:lec9_permute_ideal}
}

Lets consider a subgroup $D_i \subset G$ which stabilizes $J_i$ (see
definition \ref{def:stabilizersubgroup}) i.e.
\[
D_i = \{g \in G \mid g J_i = J_i\}.
\]
Let also $k_i = A/J_i$ - this is a field
\footnote{
  see theorem \ref{thm:maxideal}
}
and there is a finite
extension of $\mathbb{F}_p$.
\footnote{
  We have that $A$ is a $\mathbb{F}_p$-algebra. $J_i$ is a
  \mynameref{def:maxideal} of $A$ then by remark
  \ref{rem:lec5_quotientkalgebra} we have that $k_i = A/J_i$ is also
  $\mathbb{F}_p$-algebra i.e. $k_i$ is a field extension of
  $\mathbb{F}_p$ (see definition \ref{def:fextension2}). The extension
  is finite as soon as $A$ is finite.
}
Then there exists a natural homomorphism
\(
D_i \to Gal\left(k_i/\mathbb{F}_p\right)
\).
Since $D_i$ stabilizes $J_i$ and it acts on the residual classes of
modulo $J_i$ so there is a homomorphism of $D_i$ into the
\mynameref{def:galoisgroup}.
\footnote{
  We can notice that both $k_i$ and $D_i$
  are related to $J_i$ because any $g^{(k)} \in
  Gal\left(k_i/\mathbb{F}_p\right)$ is an automorphism of $k_i$
  i.e. it should preserve $O_{k_i} = J_i$ (as soon as it is an
  injection). Or in other words any element of
  $Gal\left(k_i/\mathbb{F}_p\right)$ acts in the same way as elements
  of $D_i$ (preserves $J_i$). See also theorem \ref{thm:lec9_2} about
  the map structure (it's surjection and in several situations is also
  bijection).
}

\begin{theorem}
  \begin{enumerate}
  \item $G$ acts transitively (see definition \ref{def:transitive}) on
    $\{J_1, \dots, J_r\}$ and the map
    $D_i \to Gal\left(k_i/\mathbb{F}_p\right)$ is a
    \mynameref{def:surjection} i. e.
    $D_i \twoheadrightarrow Gal\left(k_i/\mathbb{F}_p\right)$
    \item If the reduction $\bar{P} = P \mod p$ has no multiple roots
      then the map $D_i \to Gal\left(k_i/\mathbb{F}_p\right)$ is
      bijection and $k_i$ is a splitting field of the reduction
      $\bar{P}$.
  \end{enumerate}
  \label{thm:lec9_2}
  \begin{proof}
    For the first part. Suppose that for some $i$ and $\forall g \in G,
    g\left(J_1\right) \ne J_i$ i.e. suppose that there is not a
    \mynameref{def:transitive}. By \mynameref{thm:chineseremainder}
    $\exists x \in A$ such that
    $x \equiv 0 (\mod J_i),  x \equiv 1 (\mod g\left(J_1\right))
    \forall g \in G$.
    \footnote{
      We have that all $J_i$ are relatively prime (see note
      \ref{note:lec9_primeideal} below) and as result
      the map $\pi\left(a\right)$ from (\ref{eq:lec4_crt_map}) is
      \mynameref{def:surjection}
      i.e. for any residuals and therefore for the chosen ones
      ($\equiv 0 \mod J_i$ and $\equiv 1 \mod J_1$) exists $x$ that
      produces the residuals.
      \label{note:lec9_crt}
    }
    Consider a product of all such things:
    \[
    a = \prod_{g} g x
    \]
    it's an integer $a \in \mathbb{Z}$.
    \footnote{
      This is because $\prod_{g} g x =
      \mathrm{N}_{A/\mathbb{Q}}\left(x\right)$, but by 5th item of
      property \ref{property:lec9_norm_trace} one can get that such
      \mynameref{def:norm}
      (in the definition of norm we have homomorphism but not
      automorphism but accordingly remark \ref{rem:lec9_embedding} it
      is not important as soon as $A$ is a normal extension of
      $\mathbb{Q}$)
      is an integer.
    }
    But since $x \in J_i$
    (remember that $\equiv 0 \mod J_i$)
    then $a$
    is also in $J_i$:
    \footnote{
      $a = \prod_{g} g x = x \prod_{g \ne id} g x$ and by ideal
      definition \ref{def:ideal} we can conclude that $a \in J_i$.
    }
    $a \in \mathbb{Z} \cap J_i = (p)$
    \footnote{
      We have $J_i$ is a \mynameref{def:maxideal} and $A$ is a
      \mynameref{def:pid}. Therefore by lemma \ref{lem:primeideal_is_maximal},
      $J_i$ is a \mynameref{def:primeideal}
      that contains $p
      \in \mathbb{Z}$, but the prime ideal in $\mathbb{Z}$ that
      contains $p$ is $(p)$ (see lemma \ref{lem:primeideal_in_Z}).
      \label{note:lec9_primeideal}
    }
    - the ideal generated by the prime
    number $p$. So one has $a \in J_1$ since all $J_i$, and
    especially $J_1$, contains $p$. But this is impossible as soon as
    $J_1$ is a \mynameref{def:primeideal} (see note
    \ref{note:lec9_primeideal}). That is because if we have
    $\prod_k x_k \in J_1$ ($x_k = g_k(x)$)
    then $\exists i$ such that $x_i \in J_1$ but
    there is not a case in our construction.
    \footnote{
      We have $x
      \in J_i$ and $x_i \in J_1$ but $x_i$ is obtained from $x$ by
      applying an element from $G$ i.e. $\exists g_i \in G$ such that
      $x_i = g_i(x)$. But if we consider a reverse element: $g =
      g^{-1}_i$ then $g(x_i) = x \in
      J_i$ that is in contradiction with $g(J_1) \ne J_i$.
    }

    We still need to proof that $D_i \twoheadrightarrow
    Gal\left(k_i/\mathbb{F}_p\right)$.
    \footnote{
      i.e. that there is a
      \mynameref{def:surjection} or in other words that $\forall \bar{g}_i
      \in Gal\left(k_i/\mathbb{F}_p\right), \exists d_i \in D_i$ such that
      $d_i \to \bar{g}_i$.
    }
    We may assume that $i=1$. By the
    \mynameref{thm:primitiveelement} $\exists z \in
    \mathbb{F}_p$ such that $k_1 = \mathbb{F}_p\left(z\right)$
    i.e. $z$ generates $k_i/\mathbb{F}_p$.
    By \mynameref{thm:chineseremainder} $\exists y \in A$ such that
    $y \in J_i, i \ne 1,  y \equiv z (\mod J_1)$.
    \footnote{
      I.e. using the same argument as in note \ref{note:lec9_crt} one
      can conclude that $\exists y \in A$ such that its residual are
      the following $\forall i \ne 1$: $\equiv 0 \mod J_i$ and $\equiv z
      \mod J_1$.
    }
    Consider polynomial
    $Q = \prod_{g \in G} \left(X - g\left(y\right)\right)$. There is a
    polynomial with integral coefficients i.e.
    $Q \in \mathbb{Z}\left[X\right]$. This is because we know that
    coefficients are $G$ invariant
    i.e. $Q \in
    \mathbb{Q}\left[X\right]$
    \footnote{
      $Q = P_{min}\left(y, \mathbb{Q}\right)$.
    }
    moreover the roots
    are \mynameref{def:integralelement}s over $\mathbb{Z}$
    \footnote{
      as soon as $y \in A = O_K$.
    }
    and therefore the polynomial arguments are in $\mathbb{Z}$.
    \footnote{
      Ekaterina also said that ???
      as soon as
      $\mathbb{Z}$ is \mynameref{def:integrallyclosed}.
    }

    Lets study $\bar{Q} = Q \mod J_1$. If $g \notin D_1$ then $\exists
    i$ such that $g\left(J_i\right) = J_1$
    \footnote{
      If $g \notin D_1$ then $g^{-1} \notin D_1$ i.e.
      $g^{-1}\left(J_1\right) \ne J_1$ i.e. as soon as $g \in G$
      permutes the ideals (see note \ref{note:lec9_permute_ideal})
      then $\exists i$ such that
      $g^{-1}\left(J_1\right) = J_i$ i.e. $g\left(J_i\right) = J_1$.
    }
    and particularly $g(y) \in
    J_1$. Therefore for such $g$ we have
    \[
    \overline{X -g\left(y\right)} = X -g\left(y\right) \mod J_1 = X.
    \]
    So we have for $\bar{Q} \in \mathbb{F}_p\left[X\right]$
    \[
    \bar{Q} = \prod_{g \in G \setminus D_1} X \prod_{g \in D_1}
    \left(X - \overline{g\left(y\right)}\right),
    \]
    but
    \(
    \prod_{g \in D_1}
    \left(X - \overline{g\left(y\right)}\right)
    \) has $z$ as a root
    \footnote{
      Let $g = id \in D_1$ then
      $\overline{g\left(y\right)} = y \mod J_1 = z$
    }
    and $D_1$ acts transitively on its roots.

    Now recall that $z$ generates $k_1$ (i.e. $k_1 =
    \mathbb{F}_p\left(z\right)$). Thus an element of
    $Gal\left(k_1/\mathbb{F}_p\right)$ is determined by the image of
    $z$.
    \footnote{
      Because we have bijection $g \to g(z)$.
    }
    And we have an element of $D_1$ which sends $z$ to any
    possible image of it. But this means that
    \(
    D_1 \twoheadrightarrow
    Gal\left(k_1/\mathbb{F}_p\right)
    \)

    For the second part of the theorem we assume that $\bar{P}$ has no
    multiple roots. So $\alpha_1, \dots, \alpha_n$ -roots of $P$ and
    $\bar{\alpha}_1, \dots, \bar{\alpha}_n$ -roots of $\bar{P}$ where
    $\bar{\alpha}_i = \alpha_i \mod J_1$.

    Lets $g \in D_1$ acts as $id$ on $k_1$. Then, of course,
    $g\left(\bar{\alpha}_i\right) = \bar{\alpha}_i$.
    But $g\left(\alpha_i\right) \in \{\alpha_1, \dots, \alpha_n\}$ and
    it can not be different from $\alpha_i$ since they will have
    diffeent reduction $\mod J_1$. So
    $\forall i,  g\left(\alpha_i\right) = \alpha_i$ and therefore $g =
    id$.
    \footnote{
      This proves injection of the map
      $D_1 \to
      Gal\left(k_1/\mathbb{F}_p\right)$
      (see lecture 6 note \ref{note:lec6_staff_comment}).
    }
    Thus conclusion that $D_1 \cong
    Gal\left(k_1/\mathbb{F}_p\right)$.

    By the same argument
    \footnote{
      i.e.
      $\forall g \in Gal\left(k_1/\mathbb{F}_p\left[\bar{\alpha}_1,
        \dots, \bar{\alpha}_n\right]\right)$ and
      $\forall x \in \mathbb{F}_p\left[\bar{\alpha}_1,
        \dots, \bar{\alpha}_n\right]$ we have $g(x) = x$ i.e.
      $g = id$.
    }
    \[
    Gal\left(k_1/\mathbb{F}_p\left[\bar{\alpha}_1,
      \dots, \bar{\alpha}_n\right]\right) = id
    \]
    therefore
    $k_1 = \mathbb{F}_p\left[\bar{\alpha}_1,
      \dots, \bar{\alpha}_n\right]$ i.e. $k_1$ is a splitting field.
  \end{proof}
\end{theorem}

\section{Finding elements in Galois groups}
How can we apply the above material to study Galois groups? One uses
this theorem to construct elements of a certain type in the
Galois group to show that the Galois group is large.

So let $P \in \mathbb{Z}\left[X\right]$ be an
irreducible polynomial and suppose that there is a prime
$p \in \mathbb{Z}$ such that $\bar{P} = P \mod p$ is also irreducible.
Then $Gal\left(P\right)$ contains a subgroup that is isomorphic to
$Gal\left(\bar{P}\right)$.
\footnote{
  See theorem \ref{thm:lec9_2} there
  $D_i \cong Gal\left(k_i/\mathbb{F}_p\right)$
}
But we know Galois group of finite fields
and we conclude that this Galois group contains an $n$ cycle.
This is because $Gal\left(\bar{P}\right)$ is cyclic generated by $n$
cycle (see corollary \ref{cor:lec3_2}).

Sometimes, there is no such prime, but of course, a variant of this
argument exists also in other cases. Suppose, for instance that $P$ is
irreducible of degree 5 and that
$\bar{P} = R_2 R_3$ where $R_i$ is irreducible of degree $i$.
Then the same argument, gives that
$Gal\left(P\right)$ contains the permutation $(1,2)(3,4,5)$.
\footnote{
  The $Gal\left(P\right)$ contains a subgroup of 2-cycle (associated
  with the first 2 roots) and a
  subgroup of 3-cycle (associated with the last 3 roots).
}

And in this way one can construct elements of particular type in the
Galois group and use this to show that those groups are very large.


\section{Matplotlib} % (fold)
\label{sec:section_name}

\begin{frame}
\frametitle{What is Matplotlib?}

\begin{itemize}
    \item Plotting library for Python
    \item Works well with Numpy
    \item Syntax similar to Matlab
\end{itemize}

\end{frame}

\begin{frame}
\frametitle{Scatter Plot}
\lstset{basicstyle=\scriptsize}
\codeblock{code/mpl_line_plot.py}
\begin{figure}[h]
\includegraphics[width=.6\textwidth]{img/line_plot.png}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{Seaborn makes plot pretty}
\lstset{basicstyle=\scriptsize}
\codeblock{code/mpl_line_plot_sns.py}
\begin{figure}[h]
\includegraphics[width=.6\textwidth]{img/lineplot_sns.png}
\end{figure}
\end{frame}

\begin{frame}
Adding titles and labels
\frametitle{Scatter Plot}
\codeblock{code/mpl_line_plot_plus.py}
\end{frame}

\begin{frame}
\frametitle{Scatter Plot}
\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/line_plot_plus.pdf}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{Scatter Plot}
Adding multiple lines and a legend
\lstset{basicstyle=\small}
\codeblock{code/mpl_line_plot_plus2.py}
\end{frame}

\begin{frame}
\frametitle{Scatter Plot}
\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/line_plot_plus2.pdf}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{Histogram}
\lstset{basicstyle=\small}
\codeblock{code/mpl_histogram.py}
\end{frame}

\begin{frame}
\frametitle{Histogram}
\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/histogram.pdf}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{Box Plot}
\codeblock{code/mpl_boxplot.py}
\end{frame}

\begin{frame}
\frametitle{Box Plot}
\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/boxplot.pdf}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{Image Plot}
\codeblock{code/mpl_imageplot.py}
\end{frame}

\begin{frame}
\frametitle{Image Plot}
\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/imageplot.pdf}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{Wire Plot}
matplotlib toolkits extend funtionality for other kinds of visualization
\codeblock{code/mpl_wire.py}
\end{frame}

\begin{frame}
\frametitle{Wire Plot}
\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/wire.pdf}
\end{figure}
\end{frame}

\begin{frame}\frametitle{Possibilities}

A lot is possible, but not always easy to figure out how...

\begin{figure}[h]
\centering
\includegraphics[width=.9\textwidth]{img/sf_vectors.png}
\end{figure}

\end{frame}

% section section_name (end)


\documentclass[12pt]{article}
\usepackage{fullpage}
\usepackage{amsthm}
\usepackage{amsfonts,amsmath, amssymb,latexsym,mathrsfs}
\usepackage[margin=1.15in]{geometry}
\usepackage{enumitem}
\setlength{\parindent}{0pt}
\usepackage{tikz-cd}
\usepackage{fancyhdr}

\theoremstyle{definition}
\newtheorem{thm}{Theorem}[section]

\newtheorem{prop}{Proposition}[section]

\theoremstyle{definition}
\newtheorem{definition}{Definition}[section]

\theoremstyle{remark}
\newtheorem*{remark}{Remark}

\theoremstyle{definition}
\newtheorem{example}{Example}[section]

\theoremstyle{definition}
\newtheorem{lem}{Lemma}[section]

\theoremstyle{definition}
\newtheorem{cor}{Corollary}[section]

\date{}
\title{Exam 1 summary}

\begin{document}
\maketitle

\section{Finding Antiderivative, introduction}

\begin{definition}
	A definite integral of $f$ from $a$ to $b$ is definted as
\[\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{i=1}^n f(x_i)\Delta x\text{\bf (Limit of Right-hand sum)}\]
or
\[\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{i=0}^{n-1} f(x_i)\Delta x\text{\bf (Limit of Left-hand sum)}\]

Here, Left-hand sum and Right-hand sum are equal after taking limits and it the the so-called Riemann Sum.
\end{definition}

There are two more type of Riemann sum I would like to discuss in the future, which is the Mid sum and the Trapezoidal sum.

I will only give definition here.
\[\int_a^bf(x)dx\approx  \sum_{i=0}^{n-1} f(\frac{x_i+x_{i+1}}{2})\Delta x\text{\bf (Mid sum)}\]
and
\[\int_a^bf(x)dx\approx \sum_{i=0}^{n-1} \frac{f(x_i)+f(x_{i+1})}{2}\Delta x\text{\bf (Trapezoid sum)}\]

In all these Riemann sum we discussed, we are assuming $\Delta(x)=\dfrac{b-a}{n}$, thus as $n\to \infty$, $\Delta x \to 0$.

Note that \[\frac{LEFT(n)+RIGHT(n)}{2}=TRAP(n)\], \[MID(n)\neq TRAP(n)\].

\newpage

\begin{remark}
	Properties of definite integral:
	\begin{enumerate}
		\item \[\int^a_b f(x) dx = -\int^b_a f(x) dx\]
		\item \[\int^a_b f(x) dx+\int^b_c f(x) dx=\int^a_c f(x) dx\]
		\item \[\int^a_b (f(x)\pm g(x)) dx=\int^a_b f(x) dx \pm \int^a_b g(x) dx\]
		\item \[\int^a_b cf(x) dx = c \int^a_b f(x) dx\]
		\item Symmetry due to the oddity of the function.
	\end{enumerate}
\end{remark}

\hrulefill

\begin{remark}
	Interpretation of Define Integral as Area under graph of $f$ between $x=a$ and $x=b$, counting positivity.
\end{remark}

Important cases discussed on Friday's course:
\begin{enumerate}
	\item Compute \[\int^{1}_{-1}\sqrt{1-x^2}dx\]
	\item How about \[\int^{1}_{-1}(\sqrt{1-x^2}-1)dx?\]
	\item Maybe try\[\int^{0.5}_{-0.5}\tan(x)dx\]
\end{enumerate}

Find out the answer yourself only geometrically, even you know more techniques!

More Importantly there are two major topics I want to mention here and maybe discuss:

\begin{itemize}
	\item When is the estimation done by Riemann sum a underestimate/overestimate?

\textbf{It is also covered in 7.5, check it out and try problem 2.}

	\item Error estimation

	Think about the case where you know that $f(x)$ lies between any pair of $LEFT(n)$ and $RIGHT(n)$, then we see that $|LEFT(n) - f(x)|<|LEFT(n)-RIGHT(n)|=(f(b)-f(a))\Delta x$. This usually gives a bound for $n$.
\end{itemize}

\hrulefill
\begin{thm}
\textbf{The Fundamental Theorem of Calculus} is basically the theorem defined below.

If $f$ is continuous on interval $[a,b]$ and $f(t)=F'(t)$, then \[\int^b_a f(t) dt = F(b)-F(a).\]
\end{thm}
\hrulefill

\textbf{Application of Definite Integral}

\begin{enumerate}
	\item Average value of function $f(x)$ in $[a,b]$ is \[\frac{1}{b-a} \int_{a}^{b}f(x)dx\].

	Here, think about the integration is analogue to summation in the discrete world, then this average value is the analogue of \[x_{\text{average}}=\frac{x_1+x_2+\ldots+x_n}{1+1+1+\dots}=\frac{x_1+x_2+\ldots+x_n}{n}.\] where in the integration case is actually \[x_{\text{average}}=\frac{\lim_{n\rightarrow \infty} \sum_{i=0}^{n-1} f(x_i)\Delta x}{\lim_{n\rightarrow \infty} \sum_{i=0}^{n-1} 1 \times \Delta x}=\frac{\int_{a}^{b}f(x)dx}{\int_{a}^{b}1 dx}=\frac{1}{b-a} \int_{a}^{b}f(x)dx\]
\end{enumerate}

\textbf{Anti-derivative of usual functions}

\begin{enumerate}
	\item Try to find the antiderivatives by the graphs
	\item Compute an antiderivative using definite integrals.
\end{enumerate}

Suggested Problems: $\mathsection$ 6.1 3,7,9,13, 17,29,31,33

\hrulefill

\textbf{Construct antiderivative analytically}

\begin{definition}
	We define the general antiderivative family as indefinite integral.
\end{definition}

\begin{remark}
	\[\int C dx = 0\]
	\[\int kdx=kx+C\]
	\[\int x^ndx=\frac{x^{n+1}}{n+1}+C, (n \neq -1)\]
	\[\int \frac{1}{x}dx = \ln|x|+C\]
	\[\int e^xdx=e^x+C \]
	\[\int \cos xdx=\sin x + C \]
	\[\int \sin xdx=-\cos x + C \]

\end{remark}

\hrulefill

Properties of antiderivatives:

\begin{enumerate}
\item \[\int (f(x) \pm g(x))dx=\int f(x) dx \pm \int g(x) dx\]
\item \[\int cf(x) dx = c \int f(x) dx\]
\end{enumerate}

Suggested Problems: $\mathsection$  6.2 51-59, 65,71,75

\hrulefill

\textbf{Second FTC (Construction theorem for Antiderivatives)}

\begin{thm}
If $f$ is a continuous function on an interval, and if $a$ is any number in that interval then the function $F$ defined on the interval as follows is an antiderivative of $f$:

\[F(x)=\int^x_a f(t) dt\]
\end{thm}

Suggested Problems: $\mathsection$ 6.4  5,7,9,11,17,27,31-34

\section{Integration Techniques}

In this section, we will focus more on how to find the family of antiderivatives that is determined by the function we provided, since if we know such family, then given an initial value of $F(x)$, we can recover the specific antiderivative that we need.

\subsection{Guess and Check method}

Frankly speaking, this is gambling based on experience.\\
Although it is one of the fundamental method we use to solve problems, for example, it appears in also cryptography with the name Bruce force attack, and it also appears in computer science with the name Brute force search. However, as you probably know, this is usually not the most systematic or the most efficient way to the solution, and it usually requires either experience on the work or machine that do not rest.

\subsection{Integration using Substitution}
Exercise as motivation: \textsection 6.4 35,37.

Similarly, the chain rule of derivative is also extremely useful in finding integral.

Theoretically, integration using substitution is using the philosophy that given an integration as $\int f(g(x)) g'(x) dx$, we will have $F(g(x))+C$ as its derivative, which is due to $\frac{d}{dx}(F(g(x))) = f(g(x))g'(x)$, and doing integration on both side will give us the ''substitution rule'' we want.

Technically, you want to recognize the ''$g(x)$'' part in your function, then substitude $g(x)$ as $w$, with the rule $dw=w'(x)dx =\frac{dw}{dx} dx$ (naively, we do have $\frac{dw}{dx} dx =dw$! However, this does not make much sense here which if you are particular interesting, you can ask me in person.)

Now let us consider an example, to see how is it actually applies.
Personally, my favorite almost trivial example is \[\int^{\infty}_0 e^{-x}dx=-\int_{0}^{-\infty}e^w -dw=\int^{0}_{-\infty}e^w dw=1.\text{ (Here I let }w=-x)\]

I will explain why is this one interesting in the next subsection, but here there is a warning you may see, \textbf{Check the up and low bound for your integration!} Let us see another example to see this clearer.

\[\int_{0}^{\pi/4}\frac{\tan^3 \theta}{\cos^2 \theta}d \theta = \int_{0}^{1}w^3dw=\frac{1}{4},\text{ ( Here I take } w = \cos \theta.)\]

Fast question: Can I do the following integrals? Why?$$(1) \int_{0}^{\pi/2}\frac{\tan^3 \theta}{\cos^2 \theta}d \theta  \text{ or } (2)\int_{-\pi/4}^{\pi/4}\frac{\tan^3 \theta}{\cos^2 \theta}d \theta$$

Suggested Problems: $\mathsection$7.1 some ex, 81,89,90-96,97,99,109,111,114,125,141,145,147

\subsection{Integration using partial fractions}

Small break after the introduction to substitution, we have this partial fractions.

Using the method of the substitution, we can already compute many weird integration like $\int \frac{2}{x-10} dx$.

Now, how about $\int \frac{1}{x^2-1} dx$? We are out of knowledge about how to do it right now, but there is an important method we ususally can apply, which is the partial fractions.

The idea here is to break $\frac{1}{x^2-1}$ as a sum: $$\frac{1}{x^2-1}=\frac{1}{2(x-1) }+(-\frac{1}{2(x+1)})$$

But then now we can compute the integration in the question, since we can definitely compute $\int \frac{1}{2(x-1) } dx$ and $\int \frac{1}{2(x+1) }dx$.

If you run into repeated factors? The most general case (example): $$\frac{p(x)}{(x+c_1)^2(x+c_2)(x^2+c_3)}=\frac{A}{x+c_1}+\frac{B}{(x+c_1)^2}+\frac{C}{x+c_2}+ \frac{Dx+E}{x^2+c_3}$$

\subsection{Integration by parts}
If we call the integration using substitution the masterpiece established on the Chain Rule, then I have to say integration by parts is the same thing established on product rule.

Similar to the substitution, we reverse our thoughts here.

What is exactly happen in the integration by parts is that we are just undoing the product rules. Therefore, the key part is recognize your product rule carefully,

Consider the product rule:
\[(uv)'=u'v+uv'=u'v+uv' \]
By some twisted, we will have the following formula for $uv'$ which we have not seen the advantages yet.
\[uv'=(uv)'−u'v\]
But now, integrate both sides will give us:
\[\int udv=\int(uv)′−\int vdu= uv - \int v du\], now if you remember what we talked about in the substitution chapter, you will immediately recognize that this is what we want.

Traditional Example:\[\int \ln(x) dx\] \[\int \cos^2 x\]

Suggested Problems: $\mathsection$ 7.2 49,53,55,65,69,73

\newpage

\section{Integration Approximation}

Estimate the integral is based on the fact that after taking limits of Left/Right/Mid/Trapezoidal Riemann Sum, they are gonna equals and will give the actual integration value, thus if we just compute the Riemann sums in finite many block, we will have a value that is pretty much close to the actual integration, and we can refine our calculation by taking more blocks.

Reminder: we have then four ways of estimating an integral using a Riemann Sum:
\begin{enumerate}
\item LEFT(n)
\item RIGHT(n)
\item MID(n)
\item TRAP(n)
\end{enumerate}
Think about what they are?
\textbf{Think about Pictures!}

Is your estimation over/under?\\

\textbf{Again, Pictures!}(It probably will never make sense without picture.)
\begin{enumerate}
\item If the graph of $f$ is increasing on $[a,b]$, then $$LEFT(n)\leq \int^b_a f(x) dx \leq RIGHT(n)$$
\item If the graph of $f$ is decreasing on $[a,b]$, then $$RIGHT(n)\leq \int^b_a f(x) dx \leq LEFT(n)$$
\item If the graph of $f$ is concave up on $[a,b]$, then $$MID(n)\leq \int^b_a f(x) dx \leq TRAP(n)$$
\item If the graph of $f$ is concave down on $[a,b]$, then $$TRAP(n)\leq \int^b_a f(x) dx \leq MID(n)$$

\end{enumerate}

Suggested Problems: $\mathsection$7.5	1,3,5,16,17,25

\section{Find Area/Volumes by slicing}
\begin{itemize}
	\item Compute the area of triangle;
	\item Compute the area of (semi)circle;
	\item Compute the volume of a sphere;
	\item Compute the volume of a cone with base radius $5$ and height $5$;
	\item Volume of revolution: $y = e^{-x}$ from $0$ to $1$ around $x$-axis.
\end{itemize}

Suggested Problems: $\mathsection$8.1	1-4,10,12,14,16,18,19,21,29,31,37

\section{Volumes of Solids of Revolution}
There are two methods to find such a volume of solids of revolution(here revolution is important as it brings in the symmetries).
\subsection{Disk Method}

 One way is using the slicing which is very similar to the 2D case when we try to find the area. This is officially called the Disk Method.
Check above.

\subsection{Shell Method}

This is not so intuitive that why will we want to consider other method to integrating any function since we already have this disk method. However, consider the following example:

Determine the volume of the solid obtained by rotating the region bounded by $y=(x-1)(x-3)^2$ and the $x$-axis about the $y$-axis.

Try to graph it and see why it is bad.

Thus, the shell method arises naturally as a different approach, as it integrates along an axis perpendicular to the axis of revolution.

\section{Arc length}

\[\text{Arc Length}=\int^b_a\sqrt{1+(f'(x))^2}dx\]

The function we are integrating might not see as intuitive, but when we think about the picture of ''small increment'' as in class, you will see how it is making sense.

Suggested Problems: $\mathsection$8.2 25,27,35,47,49,57,63,65

\section{Relation of Integration and Physics}

Think about what integration is doing. Integral is defined as a limit of the Riemann sum, thus what it is doing is exactly the same as what the Riemann sum is doing, furthermore, you can think definite integral as just infinite block Riemann sum.\\
What we had done in the last section is to find Volumes by integrating, which is the first time we figure out what this mysterious ''f(x)'' in the integration really is, i.e. if it is an area, then $f(x)\Delta x$ is actually a volume, and we are summing many slices/shells of volumes to get the exact amount of volume of the object.\\
Here, the most essential formula we are using is actually $V=Ah$, where $V$ is the volume of prism/cylinder, $A$ is the area, and $h$ is the height. In the most settings above, we will see $S$ as function of $h$(which is exactly disk method), or sometimes $h$ is a function of $A$( what we only deal are the good cases, where $S$ is a function of $r$ and $h$ is a function of  $r$, since $dS$ is not making sense yet).

\subsection{Mass}

Here, the basic formula we are doing is:\begin{enumerate}
\item One dimensional: $M=\delta l$ where $M$ is the total mass, $\delta$ is the density, $l$ is line.
\item
Two dimensional: $M=\delta A$ where $M$ is the total mass, $\delta$ is the density, $A$ is Area.
\item Three dimensional (real world): $M=\delta V$ where $M$ is the total mass, $\delta$ is the density, $V$ is Volume.
\end{enumerate}

It is confusing that when should one use which, but if you think about what the UNIT of the density is, it will make sense of itself.

(Alternative way to think of integration: actually, if you think area as the density of Volume, i.e. think of unit of it as $m^2=m^3/m$, then you will see integration is basically integrating the density to get the total, this is actually the essential picture you should have. In the end of day, they are all just sum.)

\subsection{Work}

Key formula we are using:\\\[
\text{Work done = Force} \cdot \text{Distance}\] or
\[W=F\cdot s\]

Integration version:
\[W = \int^b_a F(x) dx\]
(Question: why usually not $x = x(F)$?)

\end{document}


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\usepackage{makeidx}
\usepackage{url}
\usepackage{graphicx}
\usepackage{color}

\title{Image Processing Projects}
\author{}
\date{}
\begin{document}
\maketitle

\section{Initial downloads and installations}
\begin{itemize}
\item git : \url{http://git-scm.com/downloads}
\item cmake : \url{http://www.cmake.org}
\item qt 4.8.5: \url{http://qt-project.org/downloads}
\item boost: \url{http://www.boost.org/}
\item lapack: \url{http://www.netlib.org/lapack/}
\end{itemize}
A C++ compiling environment\\
\\
Setup an account on github \url{http://www.github.com} \\
Create (if you do not have one already) an ssh key and to deposit it on this site.\\
If you succesfully installed git and setup the git account, you can
checkout the project:\\

{\tt git clone --recursive git@github.com:CBBProject/CBBProject.git}

\section{Processing of 2D gray-scale images}

\subsection{Basic Processing}

%The image class should contain:\\
%\begin{itemize}
%\item image size (N lines, M columns)
%\item image min and max values
%\end{itemize}

\subsubsection{Load an image (already provided)}
\begin{itemize}
\item[Input:] image path
\item[Output:] boolean flag
\end{itemize}
\subsubsection{Save an image (already provided)}
\begin{itemize}
\item[Inputs:] image path, image
\item[Output:] boolean flag
\end{itemize}

\subsubsection{Threshold an image}
\begin{itemize}
\item[Inputs:] image I, scalar S
\item[Output:]thresholded image $\tt J$ such that $\tt J(i,j)=I(i,j)$ if ${\tt I(i,j)\geq S}$, otherwise ${\tt J(i,j)=0}$
\end{itemize}

\subsubsection{Display an image}
\begin{itemize}
\item[Inputs:] image I, \\
  image range (intensities corresponding to min (black) and max (white))
\item[Output:] display of image (thresholded to fit the range)
\end{itemize}

\subsubsection{Multiply 2 images pixel-wise}
\begin{itemize}
\item[Inputs:] images I and J
\item[Output:] image K such that K(i,j)=I(i,j)*J(i,j)
\end{itemize}

\subsubsection{Add two images pixel-wise (already provided)}
\begin{itemize}
\item[Inputs:] images I and J
\item[Output:] image K such that K(i,j)=I(i,j)+J(i,j)
\end{itemize}

\subsubsection{Modify the pixel intensities of an image (e.g. by a look-up table)}
\begin{itemize}
\item[Inputs:] image I\\
 name of function (look-up table) f
\item[Output:] image J such that J(i,j) = f(I(i,j))
\end{itemize}

\subsubsection{Create an image defined by a function f(x,y) }
\begin{itemize}
\item[Inputs:] domain boundaries xmin,xmax,ymin,ymax\\
  image size N,M\\
  name of function f
\item[Output:] image J such that J(i,j) = f(i,j)
\end{itemize}

\subsubsection{Calculate the histogram of an image}
\begin{itemize}
\item[Inputs:] image I\\
  number of bins B
\item[Output:] a list of B bin-centers\\
  a list of B pixel counts belonging to each bin
\end{itemize}

\subsubsection{Subsample an image}
\begin{itemize}     \item[Input:] image I
\item[Output:] image J such that J(i,j) = I(2*i,2*j)
\end{itemize}

\subsection{Intermediate}

\subsubsection{Convolve an image with a mask}
\url{http://en.wikipedia.org/wiki/Kernel_%28image_processing%29}

  \subsubsection{Compute 1D FFT of an image (in x or in y direction)}
  use fftw package

  \subsubsection{Compute 2D FFT of an image}
  \url{https://ia700307.us.archive.org/7/items/Lectures_on_Image_Processing/EECE253_06_FourierTransform.pdf}

  \subsubsection{Convolve an image with a Gaussian, by FFT}
  \url{https://ia700307.us.archive.org/7/items/Lectures_on_Image_Processing/EECE253_08_FrequencyFiltering.pdf}

  \subsubsection{Compute the gradient of an image (in x or in y direction)}
  \url{http://en.wikipedia.org/wiki/Image_gradient}

  \subsubsection{Compute the Laplacian of an image}
  \url{http://en.wikipedia.org/wiki/Discrete_Laplace_operator}

  \subsubsection{Filter the image with a Gaussian filter}
  \url{http://en.wikipedia.org/wiki/Gaussian_blur}

  \subsection{Advanced processing}

  \subsubsection{Compute the contours of an image}
  by using a Gaussian filter + a Laplacian filter

  \subsubsection{Subsample an image while avoiding aliasing}
  by preprocessing with a Gaussian filter, see   \url{https://ia600307.us.archive.org/7/items/Lectures_on_Image_Processing/EECE253_11_SamplingAliasing.pdf}

  \subsubsection{Compute the median filter of an image}
  \url{https://archive.org/download/Lectures_on_Image_Processing/EECE253_16_MedianFilters.pdf}

\end{document}


\subsection{Temperatures on a day}
The member function \texttt{tempOnDay()} (corresponding to the \texttt{class tempTrender}) creates a histogram with the temperatures recorded for a specific day given as an input by the user. It tries to fit the histogram with a Gaussian distribution as well. An example for this is shown in Figure \ref{fig:LundTeml}. In the figures, one can see the change in mean temperature and spread through the different months. This indicates how the temperature distribution changes with the seasons. During summer, around June-September the temperature is much higher compared to winter, around December-March.
\begin{figure}[H]
    \centering
    \subfloat[March 6th]{\includegraphics[width=0.49\linewidth]{Images/Lund3-3.pdf}}
    \subfloat[June 6th]{\includegraphics[width=0.49\linewidth]{Images/Lund6-6.pdf}}
    \quad
    \subfloat[September 6th]{\includegraphics[width=0.49\linewidth]{Images/Lund9-6.pdf}}
    \subfloat[December 6th]{\includegraphics[width=0.49\linewidth]{Images/Lund12-6.pdf}}
    \caption{Temperatures in Lund}
    \label{fig:LundTeml}
\end{figure}

\noindent Since the distribution seemed normal, it was possible to use the ROOT fitting functions to do a gaussian fit and know the average temperatures as well as their standard deviations. The temperatures change over the year as expected and they have standard deviations from the average by 3-4 degrees. This is different to the case in cities which are located in the northernmost part of Sweden such as Luleå where the distribution is not quite normal. In Figure \ref{LuleåTemp}, we can see an almost exponential distribution and therefore the Gaussian fit does not provide any meanful information in this case. Although, the number of data points of temperatures measured might not have been enough to show the actual distribution which might be a Gaussian (as one can see the entries for Luleå were less than half than for Lund).

\begin{figure}[H]
    \centering
    \includegraphics[scale=0.6]{Images/lulea_histogram.pdf}
    \caption{Histogram of the Temperature in Luleå on December 15}
    \label{LuleåTemp}
\end{figure}
One could also compare the temperatures for different cities for the same day as in Fig \ref{fig:Cities}. Although, the mean temperature is not that different for the different cities, it is noticed that as one goes more North the temperature drops slightly.

\begin{figure}[H]
    \centering
    \subfloat[Falsterbo]{\includegraphics[width=0.49\linewidth]{Images/Task1Images/Falsterbo1215.pdf}}
    \subfloat[Borås]{\includegraphics[width=0.49\linewidth]{Images/Task1Images/Boras.pdf}}
    \quad
    \subfloat[Karlstad]{\includegraphics[width=0.49\linewidth]{Images/Task1Images/Karlstad1215.pdf}}
    \subfloat[Falun]{\includegraphics[width=0.49\linewidth]{Images/Task1Images/Falun1215.pdf}}
    \caption{Temperatures in different cities on December 15.}
    \label{fig:Cities}
\end{figure}

\newpage
\subsection{Yearly max and min temperatures}
In this part, the maximum and minimum temperatures for a given city were found and plotted over time. In figure \ref{fig:MinMaxTemp}, the maximum and minimum temperatures are given for 4 cities, Falsterbo, Lund, Borås and Umeå.
\begin{figure}[H]
    \centering
    \subfloat[Falsterbo]{\includegraphics[width=0.5\linewidth]{Images/minmax_falsterbo.pdf}}
    \subfloat[Lund]{\includegraphics[width=0.5\linewidth]{Images/minmax_lund.pdf}}
    \quad
     \subfloat[Borås]{\includegraphics[width=0.5\linewidth]{Images/minmax_Boras.pdf}}
     \subfloat[Umeå]{\includegraphics[width=0.5\linewidth]{Images/minmax_Umea.csv.pdf}}
    \caption{Plots illustrating how the maximum and minimum temperatures change throughout time in (a) Falsterbo, (b) Lund, (c) Borås and (d) Umeå.}
    \label{fig:MinMaxTemp}
\end{figure}
As can be seen in the figures above, both the maximum and minimum temperatures fluctuate over time. Furthermore, between adjacent years, the minimum temperature oscillates more. In general, for all the cities, the hottest days reach a temperature of 20-30$^{\circ}$C, whereas the coldest temperature drops as we go more in the North of Sweden; Falsterbo and Lund reach a minimum of -20$^{\circ}$C, while Umeås temperature drops to -40$^{\circ}$C.
\newline
\newline
Also, at the plot for Lund, one can notice a spike in the Max temperature. This is because in that particular year the temperature data recorded were less than usual, which gave us less information.

\subsection{Average daily temperature and season length}
The following figures in this section all correspond to weather data gathered from Lund and Umeå, these two cities are here chosen to showcase the results as they represent different parts of Sweden, where Lund is located in southern Sweden and Umeå is located in northern Sweden. By then plotting the average daily temperature in Lund and Umeå through time, the following plots were obtained.
\begin{figure}[H]
    \centering
    \subfloat[Lund]{\includegraphics[width=\linewidth]{Images/Lund-daily2.png}}
    \quad
    \subfloat[Umeå]{\includegraphics[width=\linewidth]{Images/Umeå-daily.png}}
    \caption{Plots illustrating how the average daily temperature changes throughout time in (a) Lund, and (b) Umeå.}
    \label{fig:Daily_temp}
\end{figure}
From the plots in Figure~\ref{fig:Daily_temp} it is observed that there seems to be a rather continuous oscillation around a certain temperature for both cities. However, to observe the behavior closer, consider the zoomed-in versions below.
\begin{figure}[H]
    \centering
    \subfloat[Lund]{\includegraphics[width=\linewidth]{Images/Lund-daily-zoom.jpg}}
    \quad
    \subfloat[Umeå]{\includegraphics[width=\linewidth]{Images/Umeå-daily-zoom.jpg}}
    \caption{These plots show how the average daily temperature changes throughout the first 5000 days of recorded temperature measurements in (a) Lund, where "?", marks the season length (i.e. time from warmest day of a year to another), and (b) Umeå.}
    \label{fig:Daily_temp_zoomed}
\end{figure}
By considering the results seen in Figure~\ref{fig:Daily_temp_zoomed}, it is apparent that the average daily temperature is changing with a sinusoidal behavior in both of the considered cities.

To answer the question whether the length of the seasons are changing or not, the distance (i.e. number of days) between each maximum in the plots from Figure~\ref{fig:Daily_temp} are considered. The results are seen below.
\begin{figure}[H]
    \centering
    \subfloat[Lund]{\includegraphics[width=.9\linewidth]{Images/Lund-season-length.png}}
    \quad
    \subfloat[Umeå]{\includegraphics[width=.9\linewidth]{Images/Umeå-season-length.png}}
    \caption{The plots displays how many days that passes between each yearly maximum temperature in (a) Lund, and (b) Umeå.}
    \label{fig:season_length}
\end{figure}

Note that Figures \ref{fig:Daily_temp}-\ref{fig:season_length} all correspond to weather data collected in Lund and Umeå, however, similar behavior as seen in these figures was also observed in all of the investigated data sets.


\documentclass[colorlinks=true,pdfstartview=FitV,linkcolor=blue,
            citecolor=red,urlcolor=magenta]{ligodoc}

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\ligodccnumber{T}{11}{XXXXX}{}{vX}% \ligodistribution{AIC, ISC}

\title{Online Detector Characterization using Neural Networks}

\author{Roxana Popescu}

\begin{document}

\section{Introduction}
\par The data obained from LIGO has noise that comes from many sources. In order to be be able to better distinguish a signal from the noise is it is important to characterize the type of noise observed. Neural networks can be used to characterize the noise.

\par Neural networks can be used to find relationships between the inputed data using hidden layers of connections within the data. A diagram of a neural network is shown in Figure \ref{fig:image1}.

\par  Recurrent neural networks are neural networks that use  loops within them so that previous information can be retained. Long Short Term Memory (LTSM) networks are a type of recurrent neural network that can remeber information for longer than a typical recurrent neural network. A diagram of a LTSM network is shown in Figure \ref{fig:image2}.

\begin{figure}[htbp]
\begin{center}
\includegraphics[width=5in]{neuralnetwork.png}
\caption{Diagram of Neural Network (from \url{http://neuralnetworksanddeeplearning.com/chap1.html})}
\label{fig:image1}
\end{center}
\end{figure}

\begin{figure}[htbp]
\begin{center}
\includegraphics[width=5in]{ltsm.png}
\caption{Diagram of LTSM (from \url{http://colah.github.io/posts/2015-08-Understanding-LSTMs)}}
\label{fig:image2}
\end{center}
\end{figure}

\section{Classification vs. Clustering}

In order to sort data two types of approaches are used: classification and clustering. Classification algorithms search the data and sort the data into already defined categories. Clustering algorithms look for relationships within the data to create categories into which the data is sorted. Classification algorithms are part of supervised learning since the computer determines the structure of the data from data that is already provided. Clustering algorithms are part of unsupervised learning since the computer determines the structure of the data without any previous information.  Clustering algorithms can be used to characterize the noise by identifying common characteristics within the noise depending on its sources. Clustering algortithms can further help with classification. A summary of the differences between classification and clusteirng is shown in the table below.

\begin{center}
 \begin{tabular}{|c | c|}
 \hline
 Classification & Clustering\\ [0.5ex]
 \hline\hline
 Supervised Learning & Unsupervised Learning\\
 \hline
 Classes are known & Classes are unknown\\
 \hline
 Used to classify data & Used to find relationships within data\\
 \hline
\end{tabular}
\end{center}

\par An example of  a clustering visualization code can be found at \url{https://www.mathworks.com/help/stats/visualize-high-dimensional-data-using-t-sne.html}.

\section{Objectives}
The interferometer's gravitational wave signal output is a time series. We will see whether to analyze the raw time series data or the
intermediate data products such as spectograms of the strain data or the band limited RMS of the channels. We will look into whether the data should be clustered according to it's power spectral density or the phase of the signal. We will also look at the various environmental channels to recognize noise states.

\section{Approach}
We will use use the keras python library to use neural networks and clustering algorithms.

\section{Project Schedule}

\begin{itemize}
\item Weeks 1-2: I will familiarize myself with the project
\item Weeks 3-4: I will work on comparing the raw data to intermediate data products
\item Weeks 5-6: I will cluster the data according to power spectral density or phase
\item Weeks 7-8: I will look at the environmental channels
\item Weeks 9-10: I will wrap up project and prepare the final report
\end{itemize}

\end{document}


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\title{Matroids And their Graphs}
\author{o.mcdonnell4@nuigalway.ie }
\date{April 2018}

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\subfile{sections/intro3}
\subfile{sections/intro4}

\pagebreak

\section{Cryptomorphisms}
\subfile{sections/charintro}
\subfile{sections/char1}
\subfile{sections/char12}
\pagebreak
\subfile{sections/char2}
\subfile{sections/char21}

\pagebreak

\section{Graph Theory}
\subfile{sections/g1}

\pagebreak

\section{Greedy Algorithm}
\subfile{sections/algo1}

\pagebreak

\section{Optimization Problems}
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\pagebreak

\section{Other Optimizable set-systems}
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\subfile{sections/conclusion}

\section{Implementation}
\subfile{sections/appenintro}
\subfile{sections/algo3}
\subfile{sections/appen1}

\pagebreak

\begin{thebibliography}{}
\bibitem{ox_book}
    Oxley, James (1992), Matroid Theory, Oxford: Oxford University Press, ISBN 0-19-853563-5, MR 1207587, Zbl 0784.05002.
\bibitem{ox_paper}
    James Oxley : What is a matroid? https://www.math.lsu.edu/~oxley/survey4.pdf
\bibitem{jungnickel}
	Jungnickel, Dieter (2013), Graphs,Networks and Algorithms, Springer-Verlag Berlin Heidelberg, ISBN 978-3-642-32277-8
\bibitem{whitney}
	Whitney, Hassler. “On the Abstract Properties of Linear Dependence.” American Journal of Mathematics, vol. 57, no. 3, 1935, pp. 		509–533. JSTOR, JSTOR, www.jstor.org/stable/2371182.
\bibitem{kruskal}
	Kruskal, Joseph B., Jr. On the shortest spanning subtree of a graph and the traveling salesman problem. Proc. Amer. Math. Soc. 7 	(1956), 48–50.
\bibitem{boruvka}
	Otakar Borůvka, On a minimal problem, Práce Moravské Pridovedecké Spolecnosti, vol. 3, 1926.
\bibitem{welsh}
	Dominic J.A. Welsh, A bound for the number of matroids, Journal of Combinatorial Theory, Volume 6, Issue 3, 1969, Pages 313-316, 	ISSN 0021-9800, https://doi.org/10.1016/S0021-9800(69)80094-3. (http://www.sciencedirect.com/science/article/pii/				S0021980069800943)
\bibitem{greedoids}
	Bjørner, A., Ziegler, G.M.: Introduction to greedoids. In: White, N.(ed) Matroid Applications, pp. 284–357. Cambridge University 	Press, Cambridge (1992)
\bibitem{kortez__lovasz}
	Korte and Lovasz (1984) Korte, B., Lovasz, L.: Greedoids and linear objective functions. SIAM J. Algebr. Discr. Math.5, 229– 			238 (1984)
\bibitem{bryant_brooksbank}
	Bryant,  V.,  Brooksbank,  P.:  Greedy  algorithm  compatibility  and heavy-set structures. Europ. J. Comb. 13 , 81–86 (1992)
\bibitem{github}
	https://github.com/emcd123/Matroids
\bibitem{wiki}
	https://www.wikipedia.org/
\end{thebibliography}
\end{document}