Isomorphism of the function field of the projective line with $\mathbf{C}(s)$ Suppose I chose two rational functions, say,
$$u = \frac{t(4+t)^5}{(1+4t)^5}, \qquad
v = \frac{t^5(4+t)}{(1+4t)}.$$
Then I know that $K(X) = \mathbf{C}(u,v)$ is
the function field of the projective line
(Proof: If $K(Y) = \mathbf{C}(t)$, then
there is an inclusion $K(X) \subseteq K(Y)$ and
hence a surjection
$\mathbf{P}^1 \simeq Y \rightarrow X$, and
so $X$ must have genus $0$). 
From this it follows that
$K(X) \simeq \mathbf{C}(s)$ for some
$s \in \mathbf{C}(t)$.
Is there a practical easy algorithm to
explicity construct such an $s$ and,
moreover, write $s$ as a rational function
of $u$ and $v$?
Is there an easy way at least to determine the
degree of the map $Y \rightarrow X$? 
(EDIT: There's some ambiguity here: I mean $X$ and $Y$ to be the unique smooth curves (i.e. $\mathbf{P}^1$) with function fields $K(X)$ and $K(Y)$).
In case you were wondering, the
specific choice of $u$ and $v$ where motivated
by the question:
Deciding whether a given power series is modular or not
(This question was posted 
on math.stackexchange a week ago; I am cross posting it here because it did not receive any replies:
https://math.stackexchange.com/questions/17960/function-field-of-the-projective-line)
 A: In the specific example, $k(u)\subset k(u,v)\subset k(t)$ and the degree $[k(t):k(u)]=6$.
Joe Silverman's equation shows that $[k(u,v):k(u)]=6$, so $k(u,v)=k(t)$ and you can take $s=t$.
A: The usual algebraic proof of L\"uroth's theorem gives the following procedure for finding a single generator of the subfield $L= K(u_1,\dotsc,u_r)$ of $K(t)$: let $a$ be any non-constant coefficient of the minimal monic polynomial of $t$ over $L$. Then $K(a) = L$.
Perhaps one can concoct a fast algorithm to compute these sorts of things using linear algebra. A lazy-person's way to do it would be to use Groebner bases. For example, in the case in question, consider the ring $R = \mathbf{C}[s,t,u,v]$ equipped with the lexicographic order where s>t>u>v. (Funny to call that "lexicographic.") Let $J$ be the ideal
$$(s(1+4t)-1, u - t(4+t)^5s^5, v-t^5(4+t)s)
$$
of $R$. If you compute a reduced Groebner basis for $J$, you will find elements of degrees (in the order $(s,t,u,v)$)
$$
(1,1,5,23), (0,2,5,23), (0,1,5,23), (0,1,5,24), (0,0,6,6),
$$
whose coefficients are too enormous for me to include here. (Despite the enormous coefficients, the calculation took only a fraction of a second on my run-of-the-mill laptop.) The element of degree $(0,0,6,6)$ is the relation between $u$ and $v$ that Joe Silverman calculated using resolvents. Each of the third and fourth elements provides a way to write $t$ as a rational function of $u$ and $v$.
In the general setting (of $L=K(u_1,\dotsc,u_r)$ in $K(t)$), if you perform a similar Groebner-basis calculation, you will find the minimal polynomial (but not in monic form) of $t$ over your subfield $L$ among the elements of your Groebner basis. Dividing that relation by the coefficient of the highest power of $t$, you get the minimal monic polynomial of $t$ over $L$. Take a non-constant coefficient $a$ of this monic polynomial, and you will have $K(a) = L$. A similar Groebner-basis calculation will provide expressions for the $u_i$ in terms of $a$. Thoughtful examination of the proof of L\"uroth's theorem may allow one to speed up the calculation (and to state things in a less roundabout way). To see how this goes in practice, you could try the simple case $u=t^4, v=t^6$.
