02.tex

\section{19/02/2024}

If $E/K$ is an extension and $x\in E$ is algebraic
over $K$, then the evaluation homomorphism 
$K[X]\to E$, $p\mapsto p(x)$, is not injective. In particular,
its kernel is a non-zero ideal. Hence 
it is generated by a monic polynomial $f$. 

\begin{definition}
\index{Minimal polynomial}
	Let $E/K$ be an extension and $x\in E$ be an algebraic element.  The monic
	polynomial that generates the kernel of $K[X]\to E$, $f\mapsto f(x)$, is
	known as the \emph{minimal polynomial} of $x$ over $K$ and it will be
	denoted by $f(x,K)$. The \emph{degree} of $x$ over $K$ is then $\deg
	f(x,K)$. 
\end{definition}

Some basic properties of the minimal polynomial of an algebraic element:

\begin{proposition}
	Let $E/K$ be an extension and $x\in E$. Assume that $x$ is algebraic over $K$. 
	\begin{enumerate}
		\item If $g\in K[X]\setminus\{0\}$ is such that $g(x)=0$, then $f(x,K)$ divides $g$ and  
		$\deg f(x,K)\leq\deg g$. 
%		\item If $g(x)=0$ and $g\ne 0$, then $\deg g\geq\gr f(x,K)$.
		\item $f(x,K)$ is irreducible in $K[X]$.
		%\item If $g(x)=0$ and $g(X)$ is monic and irreducible, then
		%	$g=f(x,K)$. 
		\item If $F/K$ is a subextension of $E/K$, then $f(x,F)$ divides
			$f(x,K)$. 
	\end{enumerate}
\end{proposition}

\begin{proof}
	Write $f=f(x,K)$ to denote the minimal polynomial of $x$. 
	To prove 1) note that $g(x)=0$ implies that	$g$ belongs to the kernel of
	the evaluation map, so $g$ is a multiple of $f$. To prove 2) 
	note that if $f=pq$ for some $p,q\in K[X]$ such that
	$0<\deg p,\deg q<\deg f$, then $f(x)=0$ implies that 
	either $p(x)=0$ or $q(x)=0$, a
	contradiction. Finally, we prove 3). Since $f\in K[X]\subseteq F[X]$ 
	and $f(x)=0$, it follows from 1) that $f(x,F)$ divides $f$. 
\end{proof}

Some easy examples: $f(i,\Q)=X^2+1$, 
$f(i,\C)=X-i$ and 
$f(\sqrt[3]{2},\Q)=X^3-2$. Here is the code:  
\begin{lstlisting}
> R<x> := PolynomialRing(Rationals());
> Q<i> := QuadraticField(-1);
> MinimalPolynomial(i);
x^2 + 1
\end{lstlisting}
A bit harder: $f(\sqrt[3]{2}+1,\Q)=X^3 - 3X^2 + 3X - 3$. Here is the code: 
\begin{lstlisting}
> R<x> := PolynomialRing(Rationals());
> E<z> := NumberField(x^3-2);
> MinimalPolynomial(z+1);
x^3 - 3*x^2 + 3*x - 3
\end{lstlisting}

% composite field 
% https://magma.maths.usyd.edu.au/magma/handbook/text/410

% Q<i> := QuadraticField(-1);
% f := MinimalPolynomial(i);
% x<x>:=Parent(f);
% f;
\begin{example}
	Let us compute 
	$f(\sqrt{2}+\sqrt{3},\Q)$. Let $\alpha=\sqrt{2}+\sqrt{3}$. 
	Then 
	\begin{align*}
		\alpha-\sqrt{2}=\sqrt{3} & \implies 
		(\alpha-\sqrt{2})^2=3 \implies \alpha^2-2\sqrt{2}\alpha+2=3\\
		&\implies \alpha^2-1=2\sqrt{2}\alpha \implies
		(\alpha^2-1)^2=8\alpha^2\implies
		\alpha^4-10\alpha^2+1=0.
	\end{align*}
	Thus $\alpha$ is a root of $g=X^4-10X^2+1$. To prove that $g=f(\alpha,\Q)$ 
	it is enough to prove that 
	$g$ is irreducible in $\Q[X]$. First note that 
	the roots
	of $g$ are $\sqrt{2}+\sqrt{3}$, $\sqrt{2}-\sqrt{3}$, 
	$-\sqrt{2}+\sqrt{3}$ and $-\sqrt{2}-\sqrt{3}$. This means that
	if $g$ is not irreducible, 
	then $g=hh_1$ for some polynomials $h,h_1\in\Q[X]$ such that
	$\deg h=\deg h_1=2$. This is not possible, as 
	\begin{align*}
   & (\sqrt{2}+\sqrt{3})+(\sqrt{2}-\sqrt{3})=2\sqrt{2}\not\in\Q,\\
   & (\sqrt{2}+\sqrt{3})+(-\sqrt{2}+\sqrt{3})=2\sqrt{3}\not\in\Q,\\ 
   & (\sqrt{2}+\sqrt{3})(-\sqrt{2}-\sqrt{3})=-5-2\sqrt{6}\not\in\Q.
    \end{align*}
\end{example}

% > R<x> := PolynomialRing(Rationals());
% > f := x^2-2;
% > g := x^2-3;
% > E := CompositeFields(NumberField(f), NumberField(g));

%\begin{lstlisting}
%julia> E, x = quadratic_field(2);
%julia> minpoly(x)
%x^2 - 2
%\end{lstlisting}

\begin{proposition}
\label{pro:multiplicativity of degree}
	Let $F/K$ be a subextension of $E/K$. Then
	\[
	[E:K]=[E:F][F:K].
	\]
\end{proposition}

\begin{proof}
	Let $\{e_i:i\in I\}$ be a basis of $E$ over $F$
	and $\{f_j:j\in J\}$ be a basis of $F$ over $K$. If $x\in E$,
	then $x=\sum_i \lambda_ie_i$ (finite sum) 
	for some $\lambda_i\in F$. For each $i$, 
	$\lambda_i=\sum_j a_{ij}f_j$ (finite sum)
	for some $a_{ij}\in K$. Then 
	$x=\sum_i\sum_j a_{ij}(f_je_i)$. This means
	that $\{f_je_i:i\in I,j\in J\}$ generates
	$E$ as a $K$-vector space. Let us prove that 
	$\{f_je_i:i\in I,j\in J\}$
	is linearly independent. If $\sum_i\sum_j a_{ij}(f_je_i)=0$ (finite sum)
	for some $a_{ij}\in K$, 
	then
	\begin{align*}
		0=\sum_i\left(\sum_j a_{ij}f_j\right)e_i&\implies
		\sum_j a_{ij}f_j=0\text{ for all $i\in I$}\\
		&\implies 
		a_{ij}=0\text{ for all $i\in I$ and $j\in J$}.\qedhere
	\end{align*}
\end{proof}

We state a lemma:

\begin{lemma}
If $A$ is a finite-dimensional commutative algebra over $K$ 
and $A$ is an integral domain, then $A$ is a field. 
\end{lemma}

\begin{proof}
	Let $a\in A\setminus\{0\}$. We need to prove that there exists $b\in A$
	such that $ab=1$. Let $\theta\colon A\to A$, $x\mapsto ax$. Note that 
	$\theta$ is $K$-linear transformation, as 
    \[
    \theta(x+y)=a(x+y)=ax+ay=\theta(x)+\theta(y),\quad
    \theta(\lambda x)=a(\lambda x)=\lambda (ax)=\lambda\theta(x),
    \]
    for all $x,y\in A$ and $\lambda\in K$. 
 It is injective since $A$ is an
	integral domain.  Since $\dim_KA<\infty$, it follows that $\theta$ is an
	isomorphism. In particular, $\theta(A)=A$, which implies that there exists
	$b\in A$ such that $1=ab$. 
\end{proof}

Let $E/K$ be an extension and $x\in E$. 
Then 
\[
K[x]=\{f(x): f\in K[X]\}
\]
is a subring of $E$ that contains $K$. Note that 
$K[x]$ is a $K$-vector space. 

More generally,
if $x_1,\dots,x_n\in E$, then
\[
K[x_1,\dots,x_n]=\{f(x_1,\dots,x_n):f\in K[X_1,\dots,X_n]\}
\]
is a subring of $E$. 
Note that $K[x_1,\dots,x_n]$ is a $K$-vector space. 
Clearly, $K[x_1,\dots,x_n]$ is a domain
and 
\[
K(x_1,\dots,x_n)=\left\{\frac{f(x_1,\dots,x_n)}{g(x_1,\dots,x_n)}:f,g\in K[X_1,\dots,X_n]\text{ with $g(x_1,\dots,x_n)\ne 0$}\right\}
\]
is the extension of $K$ generated by $x_1,\dots,x_n$. 
Note that 
\[
K(x_1,\dots,x_n)=(K(x_1,\dots,x_{n-1}))(x_n).
\]
The previous construction
can be generalized. Let $I$ be a non-empty set. 
For each $i\in I$, let $X_i$ be a variable. Consider
the polynomial ring $K[\{X_i:i\in I\}]$ and let 
$S=\{x_i:i\in I\}$ be a subset of $E$. There exists a unique 
algebra homomorphism 
\[
K[\{X_i:i\in I\}]\to E
\]
such that $X_i\mapsto x_i$ for all $i\in I$. The image 
is denoted by $K[S]$. In particular, an element $z\in K[S]$ 
is of the form 
\[
z=h(x_1,\dots,x_n)
\]
for a polynomial $h\in K[X_1,\dots,X_n]$ 
in finitely many variables $X_1,\dots,X_n$ and 
$x_1,\dots,x_n\in S$. 

\begin{exercise}
\label{xca: Q[sqrt2]=Q(sqrt2)}
    Prove that $\Q[\sqrt{2}]=\Q(\sqrt{2})$. 
\end{exercise}

The exercise is not an accident. 

\begin{theorem}
\label{thm:simple extesnions}
	Let $E/K$ be an extension and $x\in E\setminus K$.
	The following statements are equivalent:
	\begin{enumerate}
		\item $x$ is algebraic over $K$.
		\item $\dim_KK[x]<\infty$.
		\item $K[x]$ is a field.
		\item $K[x]=K(x)$. 
	\end{enumerate}
\end{theorem}

\begin{proof}
	We first prove $1)\implies 2)$. Let $z\in K[x]$, say $z=h(x)$ for some $h\in K[X]$. There exists
	$g\in K[X]$ such that $g\ne 0$ and $g(x)=0$. Divide $h$ by $g$ to obtain 
	polynomials $q,r\in K[X]$ such that $h=gq+r$, where $r=0$ or $\deg r<\deg g$. This implies that
	\[
		z=h(x)=g(x)q(x)+r(x)=r(x).
	\]
	If $\deg g=m$, then $r=\sum_{i=0}^{m-1}a_iX^i$ for some $a_0,\dots,a_{m-1}\in K$. Thus
	\[
 z=\sum_{i=0}^{m-1}a_ix^i 
 \]
 and hence $K[x]\subseteq\langle 1,x,\dots,x^{m-1}\rangle$. 

	The previous lemma proves that $2)\implies 3)$. 

	It is trivial that $3)\implies 4)$. 

	It remains to prove that $4)\implies 1)$. 
	Since $x\ne 0$, $1/x\in K(x)=K[x]$. There exists $a_0,\dots,a_n\in K$ such that
	$1/x=a_0+a_1x+\cdots+a_nx^n$. Thus
	\[
		a_nx^{n+1}+\cdots+a_1x^2+a_0x-1=0, 
	\]
	and hence $x$ is a root of $a_nX^{n+1}+\cdots+a_0X-1\in K[X]\setminus\{0\}$. 
\end{proof}

Note that if $x$ is algebraic over $K$, then
$K[x]\simeq K[X]/(f(x,K))$. 

\begin{exercise}
\label{xca:degree_of_x}
    Let $E/K$ be an extension and $x\in E$ be an algebraic element over $K$.
    Prove that the degree of $x$ over $K$ is equal to $[K(x):K]$. 
\end{exercise}

\begin{corollary}
\label{cor:finite=>algebraic}
	If $E/K$ is finite, then $E/K$ is algebraic. 
\end{corollary}

\begin{proof}
	Let $n=[E:K]$ and $x\in E\setminus K$. The set $\{1,x,\dots,x^n\}$ has $n+1$ elements, so it is linearly dependent. 
	There exist $a_0,\dots,a_n\in K$, not all zero, such that
	\[
        a_0+a_1x+\cdots+a_nx^n=0.
        \]
        Thus $x$ is a root of the non-zero
	polynomial $a_0+a_1X+\cdots+a_nX^n\in K[X]$. 
\end{proof}

In Example~\ref{exa:two_algebraics} we 
proved that $\sqrt{2}+\sqrt[3]{3}$ and $\sqrt{2}\sqrt[3]{3}$ 
are algebraic over $\Q$. This can be easily proved
now with Corollary~\ref{cor:finite=>algebraic}. 

\begin{exercise}
\label{xca:algebraic}
    Let $E/K$ be an extension and 
    $a$ and $b$ be algebraic over $K$. Prove 
    that $a+b$ and $ab$ are algebraic over $K$. 
\end{exercise}

We note that the converse of Corollary~\ref{cor:finite=>algebraic} result does not hold. 

\begin{corollary}
\label{cor:finite type finite}
	If $E/K$ is an extension and $x_1,\dots,x_n\in E$ 
	are algebraic over $K$, then 
	$K(x_1,\dots,x_n)/K$ is finite and
	$K(x_1,\dots,x_n)=K[x_1,\dots,x_n]$. 
\end{corollary}

\begin{proof}
	We proceed by induction on $n$. The case $n=1$ follows immediately from 
	the theorem. So assume the result holds for some $n\geq1$. Since the extensions 
	\[
    K(x_1,\dots,x_n)/K(x_1,\dots,x_{n-1})\quad\text{and}\quad 
    K(x_1,\dots,x_{n-1})/K
    \]
    are
	both finite, it follows that $K(x_1,\dots,x_n)/K$ is finite. Moreover, 
	\begin{align*}
	K(x_1,\dots,x_n)&=K(x_1,\dots,x_{n-1})(x_n)\\
	&=K(x_1,\dots,x_{n-1})[x_n]=K[x_1,\dots,x_{n-1}][x_n]=K[x_1,\dots,x_n].\qedhere
    \end{align*}
\end{proof}

\begin{corollary}
\label{cor:finite type algebraic}
	Let $E=K(S)$ for some set $S$. Then $E/K$ is algebraic if and only if
	$x$ is algebraic over $K$ for all $x\in S$. 
\end{corollary}

\begin{proof}
	Let us prove the non-trivial implication. Let $z\in K(S)$. In particular, 
	there exists a finite subset $T\subseteq S$ such that 
	$z\in K(T)$. The previous result implies that $K(T)/K$ is algebraic, and
	hence $z$ is algebraic. 
\end{proof}

If $E/K$ is an extension, let 
\[
	\overline{K}_E=\{x\in E:x\text{ is algebraic over }K\}. 
\]
%is the \emph{algebraic closure} of $K$ in $E$. 

\begin{corollary}
	If $E/K$ is  an extension, then $\overline{K}_E$ 
	is a subfield of $E$ that contains $K$. Moreover, 
	$K(\overline{K}_E)=\overline{K}_E$ and 
        $K(\overline{K}_E)/K$ is algebraic. 
\end{corollary}	

\begin{proof}
    By definition, $K(\overline{K}_E)/K$ is algebraic. 
    Thus $K(\overline{K}_E)\subseteq\overline{K}_E$. From this, it follows that
    $K(\overline{K}_E)=\overline{K}_E$. 
\end{proof}

The following exercise is now almost trivial:

\begin{exercise}
\label{xca:finite type}
\index{Extension!of finite type}
    Let $E/K$ be an extension of finite type; this means 
    that  
    $E=K(S)$ for some finite
    set $S$.  
    Prove that $E/K$ is algebraic if and only if $E/K$ 
    is finite. 
\end{exercise}

Let $\overline{\Q}=\{\alpha\in\C:\alpha\text{ is algebraic over }\Q\}$. 
Then $\overline{\Q}$ is the field of algebraic numbers. 
Can you compute $[\overline{\Q}:\Q]$?

\begin{exercise}
\label{xca:degree of sqrt[3]2}
    Prove that $[\Q[\sqrt[3]{2}]:\Q]=3$. 
\end{exercise}

For the previous exercise, you may use Eisenstein's criterion. 

\begin{exercise}
\label{xca:Q(i,sqrt2)}
    Let $E=\Q[i,\sqrt{2}]=\Q[\sqrt{2}][i]$. Prove that $[E:\Q]=4$.  
\end{exercise}

\begin{exercise}
\label{xca:Q(sqrt2,sqrt[3]5)}
    Let $E=\Q[\sqrt{2},\sqrt[3]{5}]$. 
    \begin{enumerate}
        \item Compute $[E:\Q]$.
        \item Prove that $E=\Q[\sqrt{2}+\sqrt[3]{5}]$.
        \item Find the minimal polynomial of $\sqrt{2}+\sqrt[3]{5}$ over $\Q$. 
    \end{enumerate}
\end{exercise}

\begin{exercise}
\label{xca:isqrt[4]3}
    Find the minimal polynomials of $\sqrt[4]{3}i$ over $\Q[i]$ and over $\Q[\sqrt{3}]$. 
\end{exercise}

\begin{exercise}
\label{xca:sqrt{2}+sqrt[3]{5}i}
    Find the minimal polynomial of $\sqrt{2}+\sqrt[3]{5}i$ over $\Q[i]$.
\end{exercise}

%\begin{theorem}[Galois]
%	\index{Galois' theorem}
%	For every prime number $p$ and every $m\geq1$
%	there exists a field of size $p^m$. 
%\end{theorem}
%
%\begin{proof}
%\end{proof}
%
%
Algebraic field extensions form a nice class of extensions. The same happens
with finite field extensions. 

\begin{proposition}
	Let $F/K$ be a subextension of $E/K$. Then $E/K$ is algebraic 
	if and only if $E/F$ and $F/K$ are algebraic. 
\end{proposition}

\begin{proof}
    If $E/K$ is algebraic, then $E/F$ and $F/K$ are both algebraic, 
    as $K\subseteq F\subseteq E$. 
    Let us assume that $E/F$ and $F/K$ are both algebraic. Let $x\in E$ and 
    let $L$ be the subextension over $K$ generated by the coefficients of $f(x,F)$, 
    the minimal polynomial of $x$ over $F$. Then $L/K$ is finite, since it is generated
    by finitely many algebraic elements. Moreover, $x$ is algebraic over $L$. Since 
    \[
    [L(x):K]=[L(x):L][L:K]<\infty,
    \]
    $L(x)/K$ is algebraic. In particular, $x$ is algebraic over $K$. 
\end{proof}

\begin{exercise}
\label{xca:tower of finite extensions}
	Let $F/K$ be a subextension of $E/K$. Prove that $E/K$ is finite 
	if and only if $E/F$ and $F/K$ are finite. 
\end{exercise}

Let $K$ be a field and $K\subseteq 
F\subseteq L$ and $K\subseteq F\subseteq L$ be fields. The \emph{composite}  
of $E$ and $F$ is defined~as 
\[
EF=K(E\cup F)=F(E)=E(F)
\]
and it is equal to the 
smallest field that contains $E$ and $F$. Here is the
picture:
\[\begin{tikzcd}
	& L \\
	& EF \\
	E && F \\
	& K
	\arrow[no head, from=1-2, to=2-2]
	\arrow[no head, from=2-2, to=3-1]
	\arrow[no head, from=2-2, to=3-3]
	\arrow[no head, from=3-1, to=4-2]
	\arrow[no head, from=4-2, to=3-3]
\end{tikzcd}\]


\begin{exercise}
\label{xca:composite generated by products}
    Let $E/K$ and $F/K$ be algebraic field extensions. 
    Prove that 
    \[
    EF=\left\{\sum_{i=1}^me_if_i:m\in\Z_{>0},e_i\in E,f_i\in F\text{ for all $i\in\{1,\dots,m\}$}\right\}.
    \]
\end{exercise}

\begin{exercise}
\label{xca:sqrt(2),sqrt(3)}
    If $E=\Q(\sqrt{2})$ and $F=\Q(\sqrt{3})$, then $EF=\Q(\sqrt{2},\sqrt{3})$. 
    Compute $[\Q(\sqrt{2},\sqrt{3}):\Q]$ and 
    $\Q(\sqrt{2})\cap\Q(\sqrt{3})$. 
\end{exercise}

\begin{exercise}
\label{xca:sqrt[3]2,3rd root of 1}
    Let $\xi\in\C$ be a primitive cubic root of one. 
    If $E=\Q(\sqrt[3]{2})$ and $F=\Q(\xi)$, then $EF=\Q(\sqrt[3]{2},\xi)$. 
    Compute $[\Q(\sqrt[3]{2},\xi):\Q]$ and 
    $\Q(\sqrt[3]{2})\cap\Q(\xi)$. 
\end{exercise}

\begin{exercise}
\label{xca: shift for algebraic}
	Let $E/K$ and $F/K$ be extensions, where both $E$ and $F$ are subfields of 
	a field $L$. If $F/K$ is algebraic, then $EF/E$ is algebraic.
\end{exercise}

% \begin{proof}
%     If $F/K$ is algebraic, then $EF/E=E(F)/E$ is algebraic, as it is generated by 
%     algebraic elements over $E$.  
% \end{proof}

\begin{exercise}
\label{xca: shift for finite}
	Let $E/K$ and $F/K$ be extensions, where both $E$ and $F$ are subfields of 
	a field $L$. If $F/K$ is finite, then $EF/E$ is finite.
\end{exercise}

The solution to the previous exercise shows, in particular, that $[EF:E]\leq [F:K]$.