Question on $F[a]=F(a)$ Let $E:F$ be a field extension and $a\in K$. Then, 
$$a\text{ is algebraic over F }\iff F(a)=F[a]$$
Proof: Let's take the evaluation epimorphism $\epsilon:F[x]\longrightarrow F[a],f(x)\mapsto \epsilon(f(x)):=f(a)$. We have:
$a\text{ is algebraic over F }\iff \exists f(x)\in F[x]:f(a)=0\iff \ker \epsilon\neq (0)$. Also, $F[x]/\langle \ker \epsilon \rangle \cong F[a]$ (from 1st Isomorphism Theorem). But, $F[x]$ is a PID, so, $F[a]$ is a domain $ \iff \ker \epsilon$ prime ideal $\iff \ker \epsilon$ is a maximal ideal $\iff  F[a]$ is a field. But, the quotient field, $Q(F[a])=F(a)$ is the smallest field which containts $F[a]$. So, $F[a]=F(a)$.
Now my questions are:


*

*Can we move in this syllogism conversely (are we ok for the $\Leftarrow$ direction)? 

*If yes, we proove this direction with another way? 

*And the most important: What $F[a]=F(a)$ means in fact? How could these sets be equal, from the time that we now that the second one contains equivalevalent classes in the form $\frac{f(a)}{g(a)},\ g(a)\neq 0$?


Thank you in advance.
 A: Yes, the statement is an "if and only if".
I can understand why you might think that $F[a] \neq F(a)$, but here's an example that will convince you otherwise. Consider the element 
$$  \frac 1 {\sqrt{2} - 1} \in \mathbb Q (\sqrt{2}).$$
As written, this element looks like it is not in $\mathbb Q[\sqrt{2}]$, because it is a fraction.
But this is an illusion! By "rationalising the denominator", you can write it as
$$ \frac 1 {\sqrt{2} - 1} = \frac {\sqrt{2} + 1} {(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac {\sqrt{2} + 1} {2 - 1} = \sqrt{2} + 1.$$
So this element is in $\mathbb Q[\sqrt{2}]$.

Now, I'll give you an alternative proof that $f(a)/g(a)$ is in $ F[a]$ if $a$ is algebraic over $F$:
Since $a$ is algebraic, it has a minimal polynomial $m(x) \in F[x]$. The minimal polynomial $m(x)$ is irreducible, so the greatest common divisor of $m(x)$ and $g(x)$ is $1$. Therefore, by Euclid's algorithm, there exist polynomials $u(x) , v(x) \in F[x]$ such that $$u(x)g(x) + v(x)m(x) = 1.$$
Substituting $x = a$ into this equation, and using the fact that $m(a) = 0$, we find that $$\frac 1 {g(a)} = u(a),$$
and hence,
$$\frac{f(a)}{g(a)} = f(a)u(a),$$
which is now manifestly an element of $F[a]$.
On the other hand, if $a$ is transcendental over $F$, then you can easily check that $F(a)$ is isomorphic to the field of rational functions $F(t)$, whereas $F[a]$ is isomorphic to the ring of polynomials $F[t]$, and these are different rings.
A: I may be missing something, but I don’t see any proof of the “$\Leftarrow$” direction.
Here it is: If $\alpha$ is not algebraic over $F$, then it’s transcendental, and the evaluation $F[T]\to F[\alpha]$ has zero kernel (that, in effect, is the definition of transcendentality) and is onto by construction. So $F[\alpha]$ is isomorphic to a polynomial ring, which is not a field. But by definition, $F(\alpha)$ is a field. So $F[\alpha]\ne F(\alpha)$.
