Field Theory - Theorem 2

 

Algebraic:

There exists a polynomial $f$ with coefficients in field $K$ such that $f(u)=0$. In this case, we say $u$ is algebraic over $K$.

Transcendental:

There exists no polynoimal $f$ with coefficients in field $K$ such that $f(u)=0$ with the exception of zero polynomial. In this case, we say $u$ is trascendental over $K$.

A field $L$ containing field $K$ is considered algebraic over $K$ if every element of $L$ is algebraic over $K$.
 
Theorem 2:
 
Suppose $K$ is a field, $u$ an element of larger field, and suppose that $u$ is algebraic over $K$. Let $f$ be a monic polynomial with coefficients in $K$ of the minimal degree $n$ such that $f(u)=0$. Then
(a) $f$ is unique.
(b) $f$ is irreducible over $K$.
(c) $1,u,u^2,\cdots,u^{n-1}$ form a vector space basis for $K(u)$ over $K$.
(d) $[K[u]:K]=n$.
(e) A polynomial $g$ with coefficients in $K$ satisfies $g(u)=0$ iff $g$ is a multiple of $f$.

Proof:

Assume that there is another polynomial $f'$, and let $f_0=f-f'$. Since, both $f$ and $f'$ are monic, the leading terms gets eliminated and $f_0$ becomes a polynomial of degree less than $n$. This means, we have $f_0$ a polynomial with degree less than $n$ going to zero whenever $f,f'$ go to zero. However, this is a contradiction $f$ is supposed to be of minial degree.

$f$ is irreducible. Assume otherwise and write $f=hg$ where $h,g$ are polynomials of degree (obviously less than n). When $f$ goes to zero, one or both of $g,h$ will go to zero, thus contradicting the fact that $f$b is of minimal degree.

Linear relation between $1,u,u^2,\cdots,u^{k-1}$ indicates that there exists a polynomial $g(u)=0$ of degree less than $n$, thus proving that $1,u,u^2,\cdots,u^{n-1}$ form a linearly independent set for $K$.

All we need to show is that $T=K[u]$ is a field. Given all multiplicatons and additions we have been performing so far, not difficult to see that $T$ is a ring.

Next we need to show that $u^k \in T$. Clearly, $1,u,u^2,\cdots,n^{n-1} \in T$.
Now set an element $u^{k-1} = 1+u+u^2+\cdots+u^{n-1}$. Multiply both sidees by $u$. Then, $u^k = u+u^2+u^3+\cdots+u^n$. Since $u^n$ can be expressed in terms of $1,u,u^2,\cdots,u^{n-1}$, we have $u^k$ as a linear combination of these basis elements. Hence $u^k \in T$.

To show that $T$ is a field, we also need to show that every element has an inverse. Indeed if $g$ is another polynomial of degree less than $n$, given that $f$ is irreducible, we will have $gcd(f,g)=1$ which means there exists polynomials $r,s$ of degree less than $n$ such that $fr+gs = 1$. Clear that when $f$ gets sent to zero, the expression $gs=1$ indicating that $s$ is inverse of the $g$.

If $g$ is another polynomial with coefficients in $K$ such that $g(u)=0$ then $g$ is a multiple of $f$. If $g$ is not a multiple of $f$ (which is irreducible) then $gcd(f,g)=0$. This means that there are two polynomials $r,s$ such that $fr+gs=1$. Since $f(u)=0$, then $fr=0$ and $g(u)=0$ means $gs=0$ leading to contradiction. Hence, $g$ is a multiple of $f$.

Field extensions-Theorem 1

Notice the following tower of fields.

 Base field is $K=Q$, intermediate field is $L=Q(\sqrt{3})$ and top field is $M=Q(\sqrt{7})$.
Field $L=Q(\sqrt{3})=\{a+b\sqrt{3}|a,b \in Q\}$.
Field $M=Q(\sqrt{3})=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{7x3=21}|a,b,c,d \in Q\}$. 

Note that the field $M$ is a vector space over $L$ and $K$. Indeed, from the expression $M=Q(\sqrt{3})=\{a+b\sqrt{3}+c\sqrt{7}+d\sqrt{7x3=21}|a,b,c,d \in Q\}$ one can see that $M$ as vector space over $K$ has dimension $4$.
That is $[M:K]=4$.
There is another way of expression $M$ as a vector space over $L$. 

Letting $u_1,u_2$ be elements of this form $u_1=a+b\sqrt{3}$ and $u_2=c+d\sqrt{3}$, one can express elements of $M$ as $u_1+\sqrt{7}(u_2)$.
That is $[M:L]=2$. Similarly $[L:K]=2$. Thus,
$[M:K]=[M:L][L:K]=2*2=4$.

First theorem of Field extensions formalizes these observations.

Theorem 1:Let $K,L,M$ be fields with $K\subset L \subset M$. Then $[M:K]$ is finite if and only if both $[M:L]$ and $[L:K]$  are finite and in that case $[M:K]=[M:L][L:K]$.
Proof is a generalization of above observations.
Assume that $[M:K]$ is finite, then consider $L$ as subspace of $M$ over $K$. As a subspace of finite vector space $[L:K]$ is finite. Any basis that spans $M$ over $K$ (for example, $\{a,b,c,d\}$ in above fields), also spans $M$ over $L$; hence, $[M:L]$ is also finite.

Assume the vector space $M$ over $L$ has $[M:L]=m$ and $L$ over $K$ has $[L:K]=n$, then we need to show that $[M:K]=mn$.

Let $z \in M$. $z$ can be expressed as $z=\sum_i u_i h_i$ where $h_i \in L$. And $h_i=\sum_j c_{ij}v_j$. Putting all this together, $z = \sum_i \sum_j c_{ij}u_i v_j$. $i,j$ range over $m$ and $n$. This is the spanning set for $M$ over $K$.

To show this is independent, we need to show that whenever $\sum_{ij} c_{ij} u_i v_j = 0$ we need to have $c_{ij}=0$.
$\sum_{ij} c_{ij} u_i v_j$ can be rewritten as $\sum_i h_i u_i$ where $h_i \in L$. Since $u_i$ form a basis for $M$ over $L$, we must have each $h_i=0$ and since this means $\sum c_{ij} v_j=0$ and independence of $v_j$ leads to each $c_{ij}=0$.

One consequence of this thoerem is that if the order of $M/L$ is a prime, then this precludes any fields between $M$ and $L$.

More examples of Galois groups

Let $F=Q$. Now, this time consider the polynomial $(x^2-2)(x^2-3)$. Clearly the roots of the polynomial are $\pm\sqrt{2},\pm\sqrt{3}$. The splitting field for this polynomials is $E(\sqrt{2},\sqrt{3}) = \{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}|a,b,c,d, \in Q\}$. Notice, that $x^2=\sqrt{6}$ is also a root.

Dimension of $E$ is $4$ and easy to see that $E$ is a vector space over $F$.

To derive Galois group, we start listing out the automorphisms of the extended field $E$ that fix $F$.

Thus we are seeking $\sigma$ that takes a root of $x^2-2$ to a root of $x^2-2$. That is $\sigma(\sqrt{2})=\pm\sqrt{2}$. Similarly $\sigma(\sqrt{3}) = \pm \sqrt{3}$

The value of both these roots can be taken independently. The following lists all the automorphisms of that fix $F$.

Thus $G=Gal(E/F)=\{\sigma_1=Id,\sigma_2,\sigma_3,\sigma_4\}$

It is interesting to note that we have intermediate fields between $F$ and $E$. Naming them as $B_1=F$ and $B_5=E$, we have other fields $B_2=Q(\sqrt{2}),B_3=Q(\sqrt{3}),B_4=Q(\sqrt{6})$.

These nested fields are shown below:

 As vector spaces $Q(\sqrt{2})$ has dimension $2$. Same is true of other intermediate field $Q(\sqrt{3})$.Field $Q(\sqrt{2},\sqrt{3})$ has dimension $4$ over $E=Q$ has dimensions $2$ over $Q(\sqrt{2})$ and $Q(\sqrt{3})$.

These examples lay foundation for understanding theorems about field extensions.

Simple example of a Galois Group:

Let $Q$ be a field of rationals. Let $x^2-2$ be a polynomial with rational coefficients. That is, let $x^2-2 \in Q[x]$.

Since $\sqrt{2}$ is not a rational, the roots of equation $x^2-2$ will not be in $Q$. However, if we extend this field by adjoining $\sqrt{2}$, then we have a new field over which the polynomial $x^2-2$ has roots. This new field $Q(\sqrt{2})$ is called splitting field of the polynomial. Elements of this field have form $\{a+b\sqrt{2}|a,b \in Q\}$.

Let $K=Q,L=Q(\sqrt{2})$. 

We ask for all automorphisms of $L$  that fix $K$, namely $\phi:L \rightarrow L$ that fix $K$. These automorphisms must take a root of above polynomial to a root. Thus, we have two automorphisms.

$Q(\sqrt{2})$ should map to $\sqrt{2}$. Identity automorphism $Id$ meets this criteria. 

Consider automorphism $\phi_1(\sqrt{2})\rightarrow -\sqrt{2}$. Clearly, this will carry elements of $Q(a+b\sqrt{2}) \rightarrow a - b \sqrt{2}$. 

If we collect all these automorphisms into a set, we have a set consisting of $\{Id,\phi_1\}$

Clearly $\phi_1\circ\phi_1 = Id$. Hence, above set forms  a group of order $2$.  This Automorphism group is called Galois group denoted as $Gal(L/K)$.