Transcendental element and algebraic closure

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?

• 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 ? – reuns Feb 5 '17 at 20:58
• 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 $\overline{K}\cap K(t) = K(t)$!) – Slade Feb 5 '17 at 21:03
• I am sorry, $\alpha$ and $t$ are the same thing in that problem. I am going to edit it. – bttmbrcelo Feb 5 '17 at 21:05
• @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 ? – reuns Feb 5 '17 at 21:07
• I am sorry @user1952009 I did not understand. :( – bttmbrcelo Feb 6 '17 at 1:38

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 $1$.