# Prove that $F[a]$ is a field if and only if $a$ is algebraic over $F$

This is really just a yes/no question, but I need to make sure that I am write about the following.

If $\alpha$ is algebraic over $F$, then $F[\alpha]$ is a field. To show this, I let $f(x)$ be the minimal polynomial for $F$ for $\alpha$ over $F$. Then the ideal generated by $f$ is maximal. Then one can show that there is an isomorphism $$F[x] / \langle f(x)\rangle$$ and $$F[\alpha].$$ And this shows that $F[\alpha]$ is a field.

Is this a correct way to do this?

How might I prove that $F[\beta]$ is a field only if $\beta$ is algebraic over $F$?

Yes you are correct that that is a way to show $F[\alpha]$ is a field. To show that $F[\beta]$ is a field implies $\beta$ is algebraic, note that $\frac{1}{\beta}$ must be in $F[\beta]$, and thus is a polynomial in $\beta$. What can you do from here?
If it's a field you must have an equation of the form $$\frac{1}{a}=c_0+\cdots +c_na^n$$