So I've been trying to prove that, for a field $F$ and an element $\alpha$ in a field extension of $F$, $F[X]/(f(X)) \cong F(\alpha)$ where $f(X) \in F[X]$ is the minimal polynomial over $\alpha$. Since I'm still in high school I don't have a certain textbook about field theory that I can follow along with so I've been searching for various PDFs here and there and hope that I learn something from them. I've seen this theorem in a lot of PDFs but I can't seem to fully understand the proof. Here's what I know so far:
$\varphi: F[X] \rightarrow F(\alpha)$ defined by $g(X) = g(\alpha)$ is a homomorphism. Furthermore, $F[X]$ is a PID and since $ker(\varphi)$ is an ideal to $F[X]$ then $(f(X)) = ker(\varphi)$. What's left to prove is that $Im(\varphi) = F(\alpha)$ by the first isomorphism theorem. What I can't prove is that $Im(\varphi)$ is a field, i.e it has inverses. If I manage to prove that then it's quite trivial that $Im(\varphi) = F(\alpha)$ by the definition of $F(\alpha)$. I hope that someone is kind enough to help me!
Best Answer
The key in proving that the image is a field is Bézout's Identity.
Another way to write this is $b \equiv g^{-1} \pmod{f}$.
So if $g(\alpha) \in \operatorname{Im}(\varphi)$, and is non-zero, then by division, we can find a quotient and remainder such that $g = qf + r$ with $\deg r < \deg f$. So now $g(\alpha) = r(\alpha)$ and since $\deg r < \deg f$, we know that $r$ and $f$ are relatively prime.
Now, by Bézout's Identity, we find that $$a(\alpha)f(\alpha) + b(\alpha)r(\alpha) = b(\alpha)r(\alpha) = 1.$$ So $b(\alpha) = g(\alpha)^{-1}$.