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)

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 $u=t^4, v=t^6$.

• Thank you, this worked. The equations did indeed have very large coefficients. – Philoi Jan 26 '11 at 17:52

Dan's right, the inclusion $K(Y)\subset K(X)$ would give a map $X\to Y$. However, I think you meant to write $K(X)\subset K(Y)$, which is easier to see is the correct formulation if you write it as $K(u,v)\subset K(t)$. Also, you meant to say $s$ as a rational function of $u$ and $v$, not of $X$ and $Y$.

You can get an equation (probably singular) for $X$ by computing the resultant of $(1+4t)^5u-t(4+t)^5$ and $(1+4t)v-t^5(4+t)$ with respect to the variable $t$. This gives an equation in $u$ and $v$. After dividing by a common factor, I get that $X$ is given by the following rather horrible affine equation in the $(u,v)$-plane: \eqalign{ -u^6 &+ (281474976710656 v^5 - 24739011624960 v^4 + 685517045760 v^3 \cr &- 6305218560 v^2 + 11887110 v) u^5 + (-24739011624960 v^5\cr &+ 301150356111360 v^4 - 292164004085760 v^3 - 7564564531215 v^2\cr &- 6305218560 v) u^4 + (685517045760 v^5 - 292164004085760 v^4 \cr &+ 114773655178260 v^3 - 292164004085760 v^2 + 685517045760 v) u^3 \cr &+ (-6305218560 v^5 - 7564564531215 v^4 - 292164004085760 v^3 \cr &+ 301150356111360 v^2 - 24739011624960 v) u^2 + (11887110 v^5 \cr &- 6305218560 v^4 + 685517045760 v^3 - 24739011624960 v^2 \cr &+ 281474976710656 v) u - v^6 = 0. } Then $t\to(u(t),v(t))$ gives the map from $\mathbb{P}^1=Y$ to $X$. Someone who is more adept than I am using computer algebra systems can probably figure out the degree of this map, and write $k(u(t),v(t))$ as $k(s(u(t),v(t)))$ for an explicit $s(u,v)$.

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 $s=t$.