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 $f$.