Transcendental extension, rational functions and field tower I would like to solve this exercise (Lang, Algebra):


*

*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 $K$.


*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
 A: Here is an approach that looks different from Jeremy Daniel's one, so may be worth writing:
The idea is simply that $y$ being transcendental over $K$, it acts as a free variable, and by Gauss Lemma, reducibility in $K(y)[t]$ is equivalent to reducibility in $K[y][t]$, the polynomial ring in variables $y,t$.
Let $y := f(x)/g(x) \in K(x)$, with not both $f,g$ in $K$, so that $y\notin K$.
Let $p(t) := f(t) - yg(t) \in K(y)[t]$.
$x$ is transcendental on $K$: 
By construction.
$x$ is algebraic on $K(y)$: 
Since $p(x) = 0$.
$y$ is transcendental on $K$: 
By multiplicativity of degrees of extensions:
$$
    \underbrace{[K(x):K]}_{\infty} = \underbrace{[K(x):K(y)]}_{<\infty}[K(y):K].
$$
The polynomial $p(t)$ has degree $n:=\max(\deg f,\deg g)$: 
Since the coefficients of $f$ and $g$ don't cancel out (if $f(t) =\sum_i a_i t^i$ and $g(t) = \sum_i b_i t^i$, then $f(t) + yg(t) = \sum_i (a_i + yb_i)t^i$, and $y\notin K$ implies that $a_i + yb_i =0$ iff both $a_i,b_i=0$).
The goal is to show that $p(t)$ is irreducible in $k(y)[t]$, which will imply it is the minimal polynomial for $x$, hence the degree of $x$ is equal to $n$.
$p$ is reducible in $K(y)[t]$ iff it is in $K[y][t]$:
By Gauss Lemma, since $K(y)$ is the field of fractions of $K[y]$.
$K[y][t]$ is the ring of polynomial in two variables over $K$:
Obvious since $y$ is transcendental over $K$.
Now, we consider the polynomial $p(y,t) = f(t) + yg(t) \in K[y,t]$.
If it is reducible, say $h(y,t)k(y,t) = p(y,t)$, then necessarily the "$y$-degree" (=highest occurence of $y$ -- I don't know the usual term) of $h$ is $0$, and that of $k$ is $1$ (wlog), so we may write $h(y,t) = h(t)$ and $k(y,t) = k_1(t) + yk_2(t)$.
Multiplying this yields:
$$
    h(t)k_1(t) + yh(t)k_2(t) = f(t) + yg(t),
$$
and by comparing coefficients, 
$$
    h(t)k_1(t) = f(t),\qquad h(t)k_2(t) = g(t),
$$
hence $h$ divides both $f$ and $g$, contradicting coprimality.
A: 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.
