Why is an extension $\bar \phi : F(x) \rightarrow F(a)$ an isomorphism if $\phi : F[x] \to F(a)$ is injective?

Question

Why is $\bar \phi : F(x) \rightarrow F(a)$ an isomorphism given $\phi : F[x] \rightarrow F(a)$ satisfy $\ker \phi = \{0\}$ ?

I've been trying to figure out why $\bar \phi$ is an isomorphism, and why $\phi$ is not an isomorphism.

I know that since $\ker \phi = \{0\}$ it follows that $\phi$ is one-to-one (please correct me if I'm wrong).

Now we only need to show $\overline{\phi}$ is surjective in order to show $\overline{\phi}$ is an isomorphism $F(x) \rightarrow F(a)$.

How can I see that $\bar \phi$ is an isomorphism $F(x) \rightarrow F(a)$? And why can I extend a one-to-one homomorphism on the polynomial ring to achieve this result?

OBS: $F(a)$ denote the smallest subfield of $E$ that contains $F$ and $\{a\}$.

Answer 1

The map you're looking at is evaluation at a point, $a$. There are two cases:

Case 1: $a$ is algebraic over $F$, in which case, there is a polynomial , $p(x)\in F[x]$ such that $p(a)=0$ showing that $\phi$ is not injective since it sends $F[x]\ni p(x)\to 0$.

Case 2: $a$ is transcendental. Then let ${p(x)\over q(x)}\in\ker \overline{\phi}$ be in reduced form for some polynomials $p,q\in F[x]$. This implies

$$\overline{\phi}\left({p(x)\over q(x)}\right)={\phi(p(x))\over\phi(q(x))}={p(a)\over q(a)}=0\iff p(a)=0$$

Since $a$ is transcendental, $p(x)=0$, the zero polynomial, hence the kernel is trivial. On the other hand, this part of the proof has given us a vital clue:

Let $y={p(a)\over q(a)}\in F(a)$*, since $a$ transcendental, $q(a)\ne 0$ when $q\ne 0$ (here's where we use injectivity), but then $\overline{\phi}(p(x)/q(x))={\phi(p(x))\over\phi(q(x))}=y$ so that $\overline{\phi}$ is surjective.

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*Note that since $F(a)$ is the smallest field containing $F$ and $a$, we can see that we can get all rational functions in $a$ with coefficients in $F$ in $F(a)$ because those are the things we can make using $F$, $a$, and the field operations, $\cdot, +, -, \div$, but also those are the only operations we're allowed, so all elements of $F(a)$ are rational functions in $a$ with coefficients in $F$, so that every $y$ has the claimed form.

Answer 2

If $\phi$ is injective, then $\bar{\phi}$ simply exchanges $x$ for $a$ in any rational function of $x$. This has an obvious inverse map: exchange $a$ back for the element $x$. Note that $F[x]\to F(a)$ cannot be an isomorphism if $a$ is transcendental, since e.g. $a^{-1}$ has no preimage. A noninjective $\phi$ fails to induce a map $F(x)\to F[a]$ in the first place, since $\phi(1/f(x))$ would be ill-defined if $f(x)\in\ker\phi$.