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If $f, g \in k[x]$, where $k$ is a field, then $k(f, g) = k(h)$ for some rational function $h \in k(x)$ (this is a special case of Lüroth's theorem).

Question 1: Under what conditions does the above hold with $h = x$, i.e. $k(f, g) = k(x)$?

This question seems to be very natural in the context of Lüroth's theorem. In the terminology used in the literature on rational parametrizations of algebraic varieties, it asks for conditions under which a given polynomial parametrization of an affine plane curve is also “proper” (i.e. generically one-to-one).

Recently I typed up (mainly for myself) a constructive proof of Lüroth's theorem from Schinzel's Selected Topics on Polynomials, and unless I made a mistake, the algorithm from the proof implies a simple sufficient condition when characteristic of $k$ is zero:

Assume the characteristic of $k$ is zero. If $\deg(f)$, $\deg(g)$ are relatively prime, then $k(f,g) = k(x)$.

It clearly is so simple that it must be well known if it is true. However I managed neither to locate it in the literature (e.g. in Rational Algebraic Curves), nor to find a mistake in my elementary proof (which can be found in the “Applications” section of this post).

Question 2: Is there a reference which studies this (provided it is correct), and other conditions which guarantee that the field generated by a given set of polynomials is the field of rational functions itself?

To clarify: I am looking for conditions in terms of coefficients, Newton polygons, etc - something that can be used to choose $f, g$ such that $k(f,g) = k(x)$. Conditions which say "compute a sequence of gcds (or Gröbner basis), and the last element of the sequence should have degree one" are easy to check, but does not immediately give a recipe to pick polynomials satisfying these conditions.

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    $\begingroup$ There is a simple proof of the sufficient condition that you give, and characteristic 0 is not needed: we always have $[k(x):k(f)] = \mathrm{deg}(f)$, so the degree of $k(x)$ over $k(f,g)$ must divide both $\mathrm{deg}(f)$ and $\mathrm{deg}(g)$, hence $k(f,g)=k(x)$. Regarding Question 1, could you be more precise for what kind of conditions you are looking for? As you already worked out, there is an algorithm which gives the solution. Another result: if $\operatorname{deg}(f) = \operatorname{deg}(g) = p$ is prime, then either $k(f,g) = k(x)$ or $g$ is of the form $(af+b)/(cf+d)$. $\endgroup$
    – François Brunault
    Commented Dec 31, 2021 at 8:58
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    $\begingroup$ For an algorithm, another method would be to write $f=p(x)/q(x)$, $g=r(x)/s(x)$, and apply Gröbner basis methods to the ideal $I=\langle qf-p, sg-r \rangle$ in $k[x,f,g]$. This will give you (basically by eliminating $x$, but the resultant is not enough I think) the minimal polynomial of the pair $(f,g)$, that is, the unique irreducible 2-variable polynomial $P$ such that $P(f,g)=0$, which you can then just test for the degree. Note that $P$ can be an interesting additional information. $\endgroup$
    – François Brunault
    Commented Dec 31, 2021 at 9:43
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    $\begingroup$ @FrançoisBrunault: thanks for the neat proof of the condition. I clarified my request in the question. And, happy new year! $\endgroup$
    – pinaki
    Commented Jan 1, 2022 at 4:43

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