# Transcendental Extensions. $F(\alpha)$ isomorphic to $F(x)$

Let $E$ be an extension field of $F$ and $\alpha \in E$. Then $\alpha$ is transcendental over $F$ if and only if $F(\alpha)$ is isomorphic to $F(x)$, the field of fractions of $F[x]$.

This was a theorem in an abstract algebra textbook with a very brief proof. Can someone please explain why this theorem holds? I'm having difficulty grasping the concepts at hand.

Thanks!

• If $\alpha$ is transcendental over $F$, then $p(x)/q(x)\mapsto p(\alpha)/q(\alpha)$ is an isomorphism. If $\alpha$ is algebraic over $F$, then $[F(\alpha):F]<\infty$. Hmm. That may have been "a very brief proof". Thinking what else there is to say... Basically the point is that $q(\alpha)=0$ if and only if $q(x)=0$ (added later: when $\alpha$ is transcendental). – Jyrki Lahtonen Jun 24 '15 at 5:38
• @JyrkiLahtonen I usually say $q(\alpha)=0$ and $\alpha\ne 0$ gives you minimal polynomial candidates, and enforcing minimal degree gives you one outright. – Adam Hughes Jun 24 '15 at 5:48
• Thanks Jyrki Lahtonen, I understand your explanation as you actually gave the isomorphism, unlike the explanation in my textbook. Adam Hughes, I don't understand what you're getting at, please can you elaborate? – Cay Jun 24 '15 at 14:40

Evaluating every polynomial $f(x)\in F[x]$ at at the fixed point $\alpha\in E$ is a ring homomorphism from the integral domain $F[x]\to$E that is identity on the constants (the polynomials of degree 0).

This evaluation homomorphism not being injective is the same as saying $\alpha\in E$ is algebraic over $F$. (Any polynomial in the kernel gives a relation for $\alpha$, making it algebraic).

That means for transcendental elements it is injective, and hence this homomorphism can be extended to its field of fractions.

EDIT: So by basic theorems of ring isomorphism the domain ring, in fact a field $F(x)$ is isomorphic to the image the subfield of $E$ generated by $F$ and $\alpha$.

• I don't understand the part that goes: "and hence this homomorphism can be extended to its field of fractions". – Cay Jun 24 '15 at 10:53
• Elements of the fraction field of an integral domain $R$ are of the form $f/g$ with $f,g\in R$, and so a homomorphism $\phi$ from there to a field $E$ can extended by $\phi(f)/\phi(g)$. This should be explained in text books where the define fraction fields. – P Vanchinathan Jun 24 '15 at 12:14
• Explicitly, the homomorphism $\psi: F(x) \to E$ is given by $\psi(f(x)/g(x)) = \phi(f(x))\cdot\phi(g(x))^{-1} = f(\alpha)\cdot (g(\alpha))^{-1}$. Note that this "only makes sense" if $\alpha$ is transcendental, for otherwise we cannot define $(g(\alpha))^{-1}$ for any $g$ where $\alpha$ is a root. – David Wheeler Jun 24 '15 at 12:24
• So the solution in my textbook says that the kernel is the empty set precisely when the evaluation homomorphism is one-to-one. I understand that part. However, it then goes: "Hence, $E$ must contain a copy of $F[x]$. The smallest field containing $F[x]$ is the field of fractions $F(x)$." Then it concludes that $E$ must contain a copy of this field. My problem is that I undersatnd the explicit isomorphisms constructed by the people answering my question, which I believe is the best way to prove the theorem. However, I just don't understand the solution in the textbook. – Cay Jun 24 '15 at 14:44
• In Fraleigh's abstract algebra, 21.8 Corollary states that "Every field L containing an integral domain D contains a field of quotients of D". Field of quotients and field of fractions have same meaning. I hope this may help you. – Sophia Mar 30 at 5:25