degree of K(x) over K(F/G) This is an ungraded assignment from a course on Galois theory on Coursera, so I hope it's OK to ask the question here.
"Let $F(x)/G(x) \in K(x)$ be a rational function over a field $K$. Show that the extension $K(x)/K(F/G)$ is algebraic and compute its degree."
Ok, the first part was easy for me: the polynomial $p(y) = F(x)/G(x) \cdot G(y) - F(y)$ has $x$ as its root. This shows algebraicity.
For the second part, my guess is that $p(y)$ is also the minimal polynomial, so if I assume that $F,G$ are coprime, then the degree is a $\max(\deg F,\deg G)$. Is that right? And if it is, how to prove that $p(y)$ is in fact minimal?
 A: Because of $K(F/G) = K(G/F)$, we may assume $\deg F\ge \deg G$. If $\deg F =
\deg G$, we can write $G(x)/F(x) = \lambda + G_2(x)/F(x)$ for some $\lambda\in
K$ and $G_2(x)\in K[x]$ with $\deg G_2(x) < \deg F(x)$. Because of $K(G/F) =
K(G_2/F)$, we may assume that 
$$
\deg F>\deg G.
$$
Then $p(y) = F(y) - G(y)\cdot F(x)/G(x)$ is a monic polynomial (up to the leading coefficient of $F(y)$, which is a unit in $K[F/G]$) in
$K[F/G][y]$, with $p(x) = 0$, i. e. $x$ is integral over $K[F/G]$. If
$\mu(y)$ denotes the minimal polynomial of $x$ over $K(F/G)$, then it lies
already in $K[F/G][y]$: $K[F/G]$ is a polynomial ring over $K$ and hence
integrally closed. Since $\mu(y)$ divides $p(y)$, it follows that all zeros of
$\mu(y)$ are zeros of $p(y)$ and therefore integral over $K[F/G]$. The
coefficients of $\mu(y)$ are elementary symmetric functions in these (integral)
zeros and hence itself integral over $K[F/G]$. But by integral closedness, they
have to lie in $K[F/G]$, i. e. $\mu(y) \in K[F/G][y]$ as desired.
Write $\mu(y) = y^n + a_{n-1}(F(x)/G(x))y^{n-1} + \dotsb+ a_0(F(x)/G(x)) \in
K[F(x)/G(x)][y]$. We will show that $n\ge \deg
F(x)$.
Let $N\in\mathbb N$ be minimal with $G(x)^Na_i(F(x)/G(x))\in K[x]$ for all
$i=0,\dotsc,n-1$. We write $a_i(y) = \sum_{j=0}^{m_i}a_{i,j}y^j$ so that $N =
\max\{m_0,\dotsc,m_{n-1}\}$. Rearranging terms in
$$
G(x)^Nx^n + G(x)^Na_{n-1}(F(x)/G(x))x^{n-1} + \dotsb+ G(x)^Na_0(F(x)/G(x)) = 0,
$$
we obtain
$$
G(x)^N\cdot (x^n + a_{n-1,0}x^{n-1}+ \dotsb+ a_{0,0}) = -F(x)\cdot
\sum_{i=0}^{n-1} \frac{G(x)^N(a_i(F(x)/G(x))-a_{i,0})}{F(x)}\cdot x^i.
$$
Since $F(x)$ and $G(x)$ are coprime, it follows that $F(x)$ divides $x^n +
a_{n-1,0}x^{n-1}+\dotsb+ a_{0,0}$ and hence $n\ge \deg F(x)$.
