galois theory - Showing Multiplicative Inverses of Polynomials of Algebraic Elements Are Also Polynomials

1. Let $E$ be a field extension of $F$.


2. Let $\alpha \in E$ be algebraic over $F$.


3. Consider $a_0 + a_1 \alpha + \ldots + a_{n} \alpha^n = f(\alpha) \in F[\alpha]$.

How can I show that $\frac{1}{a_0 + a_1 \alpha + \ldots + a_{n} \alpha^n} \in F[\alpha]$?

I have tried to think about the simpler case of just showing that $1/\alpha \in F[\alpha]$ but haven't been able to come up with anything. I feel like I'm missing something really obvious.

Answer

This follows from the fact that if $F$ is a field then $F[x]$ is a Euclidean domain with respect to the degree map.

Let $\beta = a_0 + a_1\alpha + ... + a_n\alpha^n\neq 0$ be the element we are interested in finding the inverse of. Then there is a corresponding polynomial $\beta(x) = a_0 + a_1 x + ... + a_n x^n$ in $F[x]$.

Let $m(x)$ be the minimal polynomial of $\alpha$ over $F$. Now we may assume the degree of $\beta(x)$ is less than the degree of $m(x)$ since if $\beta(x)$ contained higher powers of $\alpha$ we could use the minimal polynomial to replace these powers with smaller powers of $\alpha$.

Now the fact that $F[x]$ is a Euclidean domain, along with the fact that $\beta(x)$ and $m(x)$ must be coprime ($m(x)$ is irreducible and $\beta(\alpha)\neq 0$) tells us that there exist polynomials $a(x),b(x)\in F[x]$ such that:

$a(x)\beta(x) + b(x)m(x) = 1$.

Substituting $x=\alpha$ tells us that $a(\alpha)\beta = 1$ and so we have found our multiplicative inverse $a(\alpha)\in F[\alpha]$.