Transcendental extension, rational functions and field tower

I would like to solve this exercise (Lang, Algebra):

1. Let $E=F(x)$ where $x$ is transcendental over $F$. Let $K \neq F$ be a subfield of $E$ which contains $F$. Show that $x$ is algebraic over $K$.

2. Let $E=F(x)$. Let $y=f(x)/g(x)$ be a rational function, with relatively prime polynomials $f, g \in F[x]$. Let $n= \max(\deg f, \deg g)$. Suppose $n > 1$. Prove that $[F(x) : F(y)]=n$.

The first point is quite simple and is solved here: For field extensions $F\subsetneq K \subset F(x)$, $x$ is algebraic over $K$

In fact, it is sufficient to consider the polynomial $$p(t):=f(t)-g(t) \frac{f(x)}{g(x)} \in K[t]$$

The second point essentially asks to prove that $p(t)$ is irreducible, but I can't see how to do that. I think that it is possible to use Bezout's lemma, but I don't know how.

Thank you

First I begin with a lemma:

Let $f,g \in F[t]$ be two coprime polynomials. Let $P_n \in F[t]$ be polynomials of degree strictly below $m = \max(\deg(f), \deg(g))$. If

$$\sum_{n=0}^N P_n f^n g^{N-n} = 0$$ for some integer $N$, then all $P_n$ are zero.

Indeed, one can assume that $\deg(g) = m$. Then $g$ divides $\sum_{n=0}^{N-1} P_n f^n g^{N-n}$, hence it also divides $P_N f^N$. By coprimality, $g$ divides $P_N$ which shows that $P_N$ is zero by the condition on degrees. It follows that $\sum_{n=0}^{N-1} P_n f^n g^{(N-1)-n}$ = 0 and one argues by infinite descent.

Now take $y = \frac{f(x)}{g(x)}$, with $f$ and $g$ coprime in $F[t]$. We want to show that the minimal polynomial of $x$ in $F(y)$ is of degree $m = \max(\deg(f), \deg(g))$. Let $p$ be a polynomial in $F(y)[t]$, such that $p(x) = 0$. Write $p(t) = \sum_{k=0}^r a_k t^k$, with $a_k \in F(y)$ and suppose that $r < m$. We can write $a_k = \frac{P_{k}(y)}{Q_{k}(y)}$, with $P_{k}, Q_{k} \in F[t]$. So we have an equality of the form

$$\sum_{k=0}^r \frac{P_{k}(y)}{Q_{k}(y)} x^k = 0.$$

We thus get

$$\sum_{k=0}^r P_{k}(y) \prod_{k' \neq k} Q_{k'} (y) x^k = 0,$$

which we rewrite as

$$\sum_{k=0}^r \widetilde{P_{k}(y)} x^k = 0,$$ with $\widetilde{P_{k}} =: \sum_l a_{kl} t^l \in F[t]$.

Then

$$\sum_{k=0}^r \sum_l a_{kl} \frac{f(x)^l}{g(x)^l} x^k = 0.$$

We multiply by $g(x)^L$ for large $L$ (that is $L$ is larger than all $l$ such that an $a_{kl}$ is not zero) and get

$$\sum_{k=0}^r \sum_l a_{kl} f(x)^l g(x)^{L - l} x^k = 0,$$ which we can rewrite as

$$\sum_{l} (\sum_{k= 0}^r a_{kl} x^k) f(x)^l g(x)^{L-l} = 0.$$

By the lemma (and the transcendentality of $x$), all $a_{kl}$ are zero. Hence $\widetilde{P_{k}}$ is zero too and this shows that $a_k$ is zero. Hence, $p$ itself is zero, which concludes the proof.

• Wow, thank you! The proof seems correct to me... Sorry for answering so late! – aerdna91 Jan 1 '14 at 20:25