Understanding the relationship between $F[\alpha]$ and $F(\alpha)$ where $\alpha$ is algebraic I'm new to abstract algebra, however I don't have many resources, only my teachers' handout (I'm no longer their students). There's several place I don't know what is happening, and I have tried to figure it out by myself, but it's so hard. Thus I ask here. Let $E,~F$ are fields and $F\leq E$. Let $\alpha\in E$ and $\phi_{\alpha}$ be its evaluation homomorphism. Suppose $\alpha$ is algebraic over $F$. We know that $F[x]/\langle \textbf{irr}(\alpha,F)\rangle\cong \phi_{\alpha}(F[x])$ by the first isomorphism theorem. (Since $\langle \textbf{irr}(\alpha,F)\rangle$ is a maximal ideal, $\phi_{\alpha}(F[x])$ is a field.)


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*Q1: Do we literally define $F[\alpha]$ (rather than finding it is equal to or isomorphic to) by $\phi_{\alpha}(F[x])$? Is $F[\alpha]$ a custom 
symbol? Is there a name for such thing?

*Q2: Is the symbol $F(\alpha)$ again a custom one? Is there a name for such thing?

*Q3: I somehow feel that the definition of $F(\alpha)$ is exactly the quotient field of $F[\alpha]$. Is it?

*Q4: If what I guessed in Q3 is correct, then why do we need to put so many effort to prove that $F[\alpha]=F(\alpha)$ in this case (see my teacher's proof in the second picture)? I mean, since $F[\alpha]$ is a field, then its quotient field is equal to it by the trivial reason.




 A: Q1. $F[\alpha]$ is defined as the smallest subring of $E$ containing $F$ as subring and $\alpha$.
In other words, $F[\alpha]$ is the intersection of all subrings of $E$ containing $F$ as subring and $\alpha$.
Equivalently, $F[\alpha]$ is equal (not only isomorphic) to the image of your evaluation homomorphism $\phi_\alpha$.
Note that $F[\alpha]$ is not, in general, a field, but it's always a ring.
We concern with the ring $F[\alpha]$ because it's property of being image of $\phi_a$ permits us to define ring homomorphisms with domain $F[\alpha]$.
Indeed, recall the following universal property of polynomial rings:

Let $\xi:F\to E$ be a field homomorphism, $X$ indeterminate
  (transcendental) over $F$ and $\alpha\in E$. Then there exists a unique
  ring homomorphism $\hat\xi:F[X]\to E$ extending $\xi$ and sending $X$
  to $\alpha$.

This is the key property which is used to define the evaluation homomorphism $\phi_\alpha$.
Note that a similar property doesn't holds for $F(\alpha)$ and $F(X)$.
Q2. $F(\alpha)$ is defined as the smallest subfield of $E$ containing $F$ as subfield and $\alpha$.
Then $F(\alpha)$ is always a field which contains $F[\alpha]$ as subring.
Q3. Yes, $F(\alpha)$ is the field of fractions of $F[\alpha]$.
Q4. As you said, $F[\alpha]=F(\alpha)$ if and only if $F[\alpha]$ is a field and this happens if and only if $\alpha$ is algebraic over $F$.
On the other hand, when $\alpha$ is not algebraic (and it is said to be transcendent) over $F$, then $F[\alpha]$ is not a field and it's a proper subring of $F(\alpha)$.
In this case, $F[\alpha]$ is isomorphic to the ring of polynomial in one indeterminate while $F(\alpha)$ is isomorphic to the field of rational function in one indeterminate.
