This is the exact wording of a proposition from a note in commutative algebra.
Let $E/F$ be a field extension such that $E=F(x)$ for some element $x \in E$ (meaning that $E$ is the smallest field containing $F$ and $x$.) TFAE:
(i) $E/F$ is a finite extension
(ii) $x$ is algebraic over $F$.
(iii) $E$ is generated by $x$ as an $F$-algebra.
My question:
I am confused with what the differences between $F(x)$ and $F[x]$ is. It seems equality holds when $x$ is algebraic.
How would we argue the equivalences? In particular, I couldn't get from (iii) back to (ii) or (i).
Below are my attempts for 1. (can ignore, but please check if you have time):
$(1) \Rightarrow (2) $ Suppose $\dim (E_F) =k < \infty$, then note that $1,x,\ldots ,x^k$ form a dependent set, hence $\sum f_i x^i = 0 $. (So this shows that any element of $E$ is algebraic over $F$)
$(2) \Rightarrow (1)$ If $x$ is algebraic over $F$, then we should have $F(x) = F[x]$. This could be obtained by considering the quotient of the minimal polynomial. By divison algorithm, we can show $|F[x]:F|=n$ where $n$ is degree of minimal polynomial.
$(2) \Rightarrow (3)$ As above $(2)$ implies $F(x) = F[x]$, and so is an $F$-algebra, and is clearly the smallest $F$-algebra containing $x$.
EDIT: After more research, I updated my attempts.
(i) $\implies$ (ii) is correct.
(ii) $\implies $(iii): all you have to prove is $P(x)^{-1}\in F[x]$, ($P(x)\ne 0$) if $x$ is algebraic over $F$. To see this, consider the (linear) map of multiplication by $P(x)$ in $F[x]$: it is an injective endomorphism of $F[x]$, hence it is surjective since $F[x]$ is finite-dimensional, so $1$ is attained, i.e. there exists a polynomial $Q(X)$ such that $P(x)Q(x)=1$: this means $P(x)^{-1}=Q(x)$.
(iii) $\implies$ (i): If $E=K[x]$, the ideal of polynomials in $K[X]$ which vanish at $x$ is a maximal ideal, generated by a polynomial $P(X)$ of minimal degree $d>0$.
$E\simeq K[X]/(P(X))$ is a finite dimensional $K$-vector space, generated by $1, x, \dots , x^{d-1}$. Indeed, any element in $E$ is a polynomial $Q(x)$ in $x$. By Euclidean division in $K[X]$, we have $$Q(X)=P(X)Q_1(X)+R(X), \quad\deg R<d,$$ whence $$Q(x)=P(x)Q_1(x)+R(x)=R(x).$$