Show that $F[\alpha] = \{ f(\alpha): f(x) \in F[x]\}$ is a field iff $\alpha$ is algebraic over F

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\ker\phi = \{f(x) \in F[x]: f(\alpha) = 0\} = \langle p(x) \rangle$. 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?