Htykuut

I don't even know where to start in this exercise. Let $E$ be a field extension of $F$. An element $alpha in E$ is algebraic iff $F[alpha] = { f(alpha): f(x) in F[x] }$ is a field. Can anyone give some tips on how to prove it? It seems kind of counterintuitive for me.

edit: I posted an answer to the exercise. If anyone wants to comment to see if it sounds right, I would be very glad.

Hint: $F[alpha]$ is a field if and only if $alpha^{-1}in F[alpha]$. Show that the latter is equivalent to $alpha$ is algebraic.

How about this for an answer?
Suppose that $F[alpha]$ is a field. Then, $alpha^{-1} in F[alpha]$, so there is a polynomial $g(x) in F[x]$ such that $g(alpha) = alpha^{-1}$. Then we have that $alpha g(alpha) = 1$, which implies that $alpha$ satisfies $xg(x) - 1 in F[x]$. So we have that $alpha$ is algebraic over $F$.

Conversely, let $p(x) in F[x]$ be the minimal polynomial (that is, the unique monic irreducible polynomial) such that $p(alpha) = 0$ and consider the evaluation map $phi: F[x] to E$ defined by $phi (f(x)) = f(alpha)$. Then we have $kerphi = {f(x) in F[x]: f(alpha) = 0} = langle p(x) rangle$. Because $p(x)$ is irreducible, $F[x]/langle p(x) rangle$ is a field and, by the isomorphism theorem, is isomorphic to $phi(F[x]) = {f(alpha): f(x) in F[x]}$, implying that the latter is a field.

Does it sound right?

The theorem you want to use is Bézout's Lemma:

If $f, g in F[x]$ are coprime then there exists $lambda, mu in F[x]$ such that $$lambda f + mu g = 1. $$

Generally speaking, this is the theorem you should look for for finding inverses modulo something. For instance if $f(alpha) = 0$ then this gives

$$ mu(alpha) g(alpha) = 1. $$

Or if instead you looked at $F[x]/(f)$ then you get

$$ mu g equiv 1 pmod f $$

(These are related ideas.)

The other direction, Eclipse Sun already hinted at.