0
$\begingroup$

I am trying to solve the following problem:

Let $\alpha$ be a transcendental element in some extension field of $H$. If $\frac{f(\alpha)}{g(\alpha)} \in H(\alpha)$ is algebraic over $H$ then this quotient is a constant in $H$.

I know that this is the same (or at least very similar) to prove that the algebraic closure of $H(\alpha)$ in $H$ is $H$. What did I?

Since $\frac{f(\alpha)}{g(\alpha)}$ is algebraic over $H$ then there exist some polinomial $h \in H[x]$ such that $h(\frac{f(\alpha)}{g(\alpha)})=0$.

Then if we multiply this by $g(\alpha)^n$ where $n$ is the degree of $h$ we can conclude that $g(\alpha)$ divides $f(\alpha)^n$ and $f(\alpha)$ divides $g(\alpha)^n$. Then, I know that somehow I can conclude from this that $g|f$ and $f|g$ and by this that they are associates and therefore $f(\alpha)=cg(\alpha)$ for some constant in $H$. But I need help in two things:

  1. I know that somehow I am using that $H[x]$ is a Princial Ideal Domain (or an Unique Factorization Domain) to conclude that any irreducible factor of $g$ divides $f$ and vice-versa. But I dont understand clearly the argument, so if someone can help me I would be very grateful.

  2. I cannot see why we need the hyphotesis that $\alpha$ is transcendental over $H$. Am I forgetting something?

Thank you everybody in advance!

$\endgroup$
  • $\begingroup$ You want to show that if $c \in \overline{K} - K$ and $t$ is transcendental over $K$, then $c \not \in K(t)$, that's it ? $\endgroup$ – reuns Feb 5 '17 at 20:58
  • $\begingroup$ I think $\alpha$ and $t$ are supposed to be the same thing. I think he wants to show that $\overline{K} \cap K(t) = K$ when $t$ is transcendental. (Note: if $t$ is algebraic, then $\overline{K}\cap K(t) = K(t)$!) $\endgroup$ – Slade Feb 5 '17 at 21:03
  • $\begingroup$ I am sorry, $\alpha$ and $t$ are the same thing in that problem. I am going to edit it. $\endgroup$ – bttmbrcelo Feb 5 '17 at 21:05
  • $\begingroup$ @Slade Well if $t$ is transcendental over $K$, and $u \in K[t], u \not \in K$ then $\phi(\sum_{n=0}^d c_n t^n) = \sum_{n=0}^d c_n u^n$ is clearly an injective morphism $K[t] \to K[t]$, since otherwise $\sum_{n=0}^d c_n u^n = \sum_{n=0}^d c_n \sum_{k=0}^m a_k t^{nk} = 0$ i.e. $t$ is algebraic. Then extend this to $K(t)$ in some way, and we get the claim ? $\endgroup$ – reuns Feb 5 '17 at 21:07
  • $\begingroup$ I am sorry @user1952009 I did not understand. :( $\endgroup$ – bttmbrcelo Feb 6 '17 at 1:38
1
$\begingroup$

Suppose that we have a rational function $f(X) = p(X)/q(X)$, written in lowest terms with $p(X),q(X)\in H[X]$, and suppose that $f$ satisfies a polynomial equation:

$$a_n (f(X))^n + a_{n-1} (f(X))^{n-1} \cdots + a_1 (f(X)) + a_0 = 0$$

Here $a_i \in H$. Clearing denominators:

$$a_n (p(X))^n + a_{n-1} (p(X))^{n-1} q(X) + \cdots + a_1 p(X) (q(X))^{n-1} + a_0 (q(X))^{n-1} = 0$$

From this equation, we can see that $q(X)$ is a factor of $a_n (p(X))^n$. Since $p$ and $q$ are relatively prime, $q(X)$ is a factor of $a_n$, and the only way this can happen is if $a_n=0$, or $q$ is a constant.

If $a_n=0$, it follows that $f$ cannot satisfy any nontrivial polynomial equation, and is therefore transcendental over $H$. If $q$ is a constant, then $f$ is a polynomial and it is easy to see it is transcendental.

Facts we need to make this work:

  • It is possible to write a rational function in lowest terms. This follows from the existence of greatest common divisors in $H[X]$.
  • If $p$ and $q$ are relatively prime, then $p^n$ and $q$ are relatively prime. This follows from unique factorization in $H[X]$, but it also follows from the simpler Bézout's identity: if $a$ and $b$ are polynomials such that $ap + bq=1$, then taking the n-th powers gives a linear combination of $p^n$ and $q$ that equals $1$.
$\endgroup$

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.