Let $f: G \rightarrow H$ be a homomorphism. The \emph{kernel} of $f$ is the set $K$ of all the elements of $G$ which are carried by $f$ onto the neutral element of $H$. That is,
$$K ={x \in G : f(x)= e}$$
\emph{Kernel in the context of Extension fields: \ref{def:extensionkernel}}
\end{definition}
For every homomorphism, the $e \in G$ maps to $e \in H$, so the \emph{kernel} is never empty, it always contains the identity $e_G$, and if the kernel only contains the identity, then $f$ is one-to-one (injective).
@ -439,7 +443,99 @@ From the last two theorems: every integer $m$ can be factored into primes, and t
If $a(x)$ has degree $n$, it has at most $n$ roots.
\end{theorem}
In finite $F$, polynomial $\neq$ polynomial function. If $F$ is infinite, polynomial $=$ polynomial function.
\framebox{WIP: covered until chapter 26, work in progress.}
For every polynomial with rational coefficients, there is a polynomial with integer coefficients having the same roots. See:
$$a(x)=\frac{k_0}{l_0}+\frac{k_1}{l_1} x +\cdots+\frac{k_n}{l_n} x^n$$
$a(x)$ has rational coefficients, $b(x)$ has integer coefficients. $b(x)$ differs from $a(x)$ only by a constant factor ($\frac{1}{l_0\cdots l_n}$), so $a(x)$ and $b(x)$ have the same roots.
$\Longrightarrow~~\forall~p(x)\in\mathbb{Q}[x]$, there is a $f(x)\in\mathbb{R}$ with the same roots (for every polynomial with rational coefficients, there is a polynomial with integer coefficients having the same roots).
\begin{theorem}
If $s/t$ is a root of $a(x)$, then $s|a_0$ and $t|a_n$.
\end{theorem}
\begin{theorem}
Suppose $a(x)$ can be factured as $a(x)= b(x)c(x)$, where $b(x), c(x)$ have rational coefficients. Then there are polynomials $B(x), C(x)$ with integer coefficients, which are constant multiples of $b(x)$ and $c(x)$ respectively, such that $a(x)= B(x)C(x)$.
Let $a(x)= a_0+ a_1 x +\cdots+ a_n x^n$ be a polynomial with integer coefficients.
If there is prime $p$ such that $p | a_i, ~\forall i\in\{0, n-1\}$, and $p \ndiv a_n$ and $p^2\ndiv a_0$, then $a(x)$ is irreducible over $\mathbb{Q}$.
% Suppose there is a prime number $p$ which divides every coefficient of $a(x)$ except the leading coefficient $a_n$; suppose $p$ does not divide $a_n$ and $p^2$ does not divide $a_0$. Then $a(x)$ is irreducible over $\mathbb{Q}$.
The \emph{kernel} of $\sigma_c$ consists of all polynomials $a(x)\in F[x]$ such that $c$ is a root of $a(x)$.
\emph{Kernel in the context of Homomorphisms: \ref{def:homomorphismkernel}}
\end{definition}
\begin{definition}[Algebraic]
$c \in E$ is called \emph{algebraic over}$F$ if it is the root of some nonzero polynomial $a(x)\in F[x]$.
Otherwise, $c$ is called \emph{transcendental over}$F$.
\end{definition}
$E/K$ denotes the (field) extension of $E$ over $K$.
\begin{theorem}[Basic theorem of field extensions]
Let $F$ be a field and $a(x)\in F[x]$ a nonconstant polynomial. There exists an extension field $E/F$ and an element $c \in E$ such that $c$ is a root of $a(x)$.
\end{theorem}
Let $a(x)\in F[x]$ be a polynomial of degree $n$. There is an extension field $E/F$ which contains all $n$ roots of $a(x)$.
\section{Vector spaces}
\begin{definition}[Vector space]
A \emph{vector space} over a field $F$ is a set $V$, with two operations $+, \cdot$, called \emph{vector addition} and \emph{scalar multiplication}, such that
\begin{itemize}
\item$V$ with vector addition is an abelian group
\item$\forall k \in F$ and $\overrightarrow{a}\in V$, the scalar product $k \overrightarrow{a}$ is an element of $V$, subject to the following conditions:
$\forall k, l \in F,~\overrightarrow{a},\overrightarrow{b}\in V$
is called a \emph{linear combination} of $\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}$.
The set of all the linear combinations of $\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}$ is a \emph{subspace of}$V$.
\end{definition}
\begin{definition}[Linear dependancy]
Let $S =\{$\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}$\}$ be a set of distinct vectors in a vector space $V$. $S$ is said to be \emph{linearly dependent} if there are scalars $k_1, \ldots, k_n$, not all zero, such that $k_1\overrightarrow{a_1}+ k_2\overrightarrow{a_2}+\cdots+ k_n \overrightarrow{a_n}=0$.
Which is equivalent to saying that at least one of the vectors in $S$ is a linear combination of the others.
If $S$ is not linearly dependent, then it is \emph{linearly independent}. $S$ is linearly independent iff $k_1\overrightarrow{a_1}+ k_2\overrightarrow{a_2}+\cdots+ k_n \overrightarrow{a_n}=0$ implies $k_1= k_2=\cdots= k_n =0$.
Which is equivalent to saying thatno vector in $S$ is equal to a linear combination of the other vectors in $S$.
\end{definition}
If $\{\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}\}$ is linearly dependent, then some $a_i$ is a linear combination of the preceding ones.
If $\{\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}\}$ spans $V$, and $a_i$ is a linear combination of the preceding vectors, then $\{\overrightarrow{a_1}, \ldots, \crossover{\overrightarrow{a_i}}, \ldots, \overrightarrow{a_n}\}$ still spans $V$.
\begin{theorem}
Any two bases of a vector space $V$ have the same number of elements.
(This comes from the fact that all bases of $\mathbb{R}^n$ contain exactly $n$ vectors)
\end{theorem}
If the set $(\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n})$ spans $V$, it contains a basis of $V$.
\framebox{WIP: covered until chapter 28, work in progress.}
