3
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Jyrki Lahtonen Jun 24 '15 at 5:38
  • $\begingroup$ @JyrkiLahtonen I usually say $q(\alpha)=0$ and $\alpha\ne 0$ gives you minimal polynomial candidates, and enforcing minimal degree gives you one outright. $\endgroup$ – Adam Hughes Jun 24 '15 at 5:48
  • $\begingroup$ 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? $\endgroup$ – Cay Jun 24 '15 at 14:40
2
$\begingroup$

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 $\alpha$.

$\endgroup$
  • 1
    $\begingroup$ I don't understand the part that goes: "and hence this homomorphism can be extended to its field of fractions". $\endgroup$ – Cay Jun 24 '15 at 10:53
  • $\begingroup$ 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. $\endgroup$ – P Vanchinathan Jun 24 '15 at 12:14
  • $\begingroup$ 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. $\endgroup$ – David Wheeler Jun 24 '15 at 12:24
  • $\begingroup$ 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. $\endgroup$ – Cay Jun 24 '15 at 14:44
  • $\begingroup$ 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. $\endgroup$ – Sophia Mar 30 at 5:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.