Proof that $F(a)\approx F[x]/\langle p(x) \rangle$

Question

I'm reading a proof of the theorem that $F(a)$ is isomorphic to $ F[x]/\langle p(x) \rangle$ in the book called Contemporary Abstract Algebra by Gallian, and I don't understand some key elements of the proof [I think this book is too confusing in numerous places in the sections on rings and fields, but it was clear and straightforward on groups]. Here are the theorem and the proof:

Here's what I do not understand:

(1) If $\langle p(x) \rangle$ is a maximal ideal in $F[x]$ and $\ker \phi\ne F[x]$, then how does it follow from the First Isomorphism Theorem for Rings and the corollary (below) that $\phi(F[x])$is a subfield of $F(a)$?

(2) When the author writes that

$\phi(F[x])$ contains both $F$ and $a$ and recalling that $F(a)$ is the smallest such field ...

what does he refer to by "the smallest such field"? The smallest what field?

(3) Can't one immediately see the image of $\phi$ is indeed $F(a)$, just by definition? $\phi$ takes all polynomials in $F[x]$ and evaluates them at $a$. I don't understand the author's argument at all.

(4) The author also writes that

But isn't the basis for $F(a)$ one-dimensional, after all?

Sorry for the long message. Would really appreciate your help in understanding this proof. In my opinion the author is skipping too many details. In my university, I'd get reduced marks for unproved statements like "clearly, [...]".

Answer 1

(1) $\langle p(x)\rangle$ is a maximal ideal in $F[x]$. According to Theorem 17.5, $p(x)$ is irreducible over $F$. According to Corollary 1, $F[x]/\langle p(x)\rangle$ is a field. But $\ker{\phi}=\langle p(x)\rangle$. According to the First Isomorphism Theorem of Rings, $\phi(F[x])\cong F[x]/\langle p(x)\rangle$, showing that the image of $\phi$ is a subfield of $F(a)$.

(2) $F(a)$ is the smallest field containing both $F$ and $a$ by definition (see the discussion above the Definition in page 340).

(3) Yes, and the author proved it easly as you say (end of first paragraph). But since the image of $\phi$ is $F(a)$, it follows from the result in (1) that $F[x]/\langle p(x)\rangle\cong F(a)$

(4) Why do you think that $F(a)$ is one-dimensional? Is it because you see it as a vector space over itself? $\mathbb{C}$ is one dimensional as a $\mathbb{C}$-vector space, but $2$-dimensional as an $\mathbb{R}$-vector space. Even $\mathbb{R}$ can be infinite dimensional if you consider it as a vector space over $\mathbb Q$. $F(a)$ is surely one dimensional as a vector space over itself, like any field. But over $F$, that's not the case because $a\notin F$ (otherwise $p(x)$ wouldn't be irreducible), but given the way any element can be expressed in $F(a)$ as shown in the theorem, the dimension of $F(a)$ as a vector space over $F$ is exactly the degree of $p(x)$.

Answer 2

(1) The First isomorphism theorem tells you that $\;F[x]/\ker\phi\cong\phi(F[x])\;$ , and corollary 1 tells you that $\;F[x]/\langle p(x)\rangle\;$ is a field whenever $\;p(x)\;$ is irreducible. Both things, together with $\;\ker\phi=\langle p(x)\rangle\;$ gives you the result you ask about.

(2) Smallest such field" means here " the smallest field containing both $\;F\;$ and $\;a\;$" .

(3) Well, by its mere definition, $\;\phi(g(x)):=g(a)\;$ , so $\;\phi(F[x])=F[a]\;$ . Yet, as $\;a\;$ is algebraic over $\;F\;$ , we have that $\;F[a]=F(a)\;$ and we're done.