# Lüroth theorem for $k \subset k(f,g) \subseteq k(x)$

Let $k$ be a field of characteristic zero (I do not mind to assume that $k=\mathbb{C}$, if things are easier in this case). Lüroth theorem says that a field $L$, $k \subset L \subset k(x)$ containing a nonconstant polynomial over $k$ is equal to $k(h)$ for some $h \in k[x]$. More precisely, in Lüroth theorem $h \in k(x)$, and E. Noether has proved that $h \in k[x]$ (see "Selected Topics on Polynomials" by A. Schinzel). Unfortunately, I do not have the book of A. Schinzel (I have learnt about Noether's result from Formanek's paper), so I do not have an available proof for this result.

Given $f=f(x),g=g(x) \in k[x]$, $k \subset k(f,g) \subseteq k(x)$, Lüroth-Noether theorem says that $k(f,g)=k(h)$ for some $h=h(x) \in k[x]$.

(1) What exactly can be said about $h$ in terms of $f$ and $g$? (Except, of course, that $h \in k(f,g)$).

I have played a little with several $f$'s and $g$'s and I have not found $f$ and $g$ with $1<\gcd(\deg(f),\deg(g))<min\{\deg(f),\deg(g)\}$ such that $k(f,g)=k(x)$. For example: (i) $f=x^2,g=x^3$ with $\gcd(\deg(x^2),\deg(x^3))=1$. (ii) $f=x^3,g=x^6+x^2$ with $\gcd(\deg(x^3),\deg(x^6+x^2))=3=min\{\deg(f),\deg(g)\}$.

(2) Therefore, if $k(f,g)=k(x)$, then I conjecture that necessarily $\gcd(\deg(f),\deg(g))=1$ or $\deg(f) | \deg(g)$ or $\deg(g) | \deg(f)$: Is my conjecture true or false? (Perhaps it is possible to find a counterexample to my conjecture with some help from algebraic geometry?).

Any hints and comments are welcome! Thank you very much.

• Since $k(x)=k(x^3,x^6+x^2)$, then we also have $k(x)=k(x^3 \cdot (x^6+x^2), x^6+x^2) = k(x^9+x^5, x^6+x^2)$. Here $\gcd(\deg(f),\deg(g))=3$ and neither degree divides the other. – Zach Teitler May 29 '18 at 21:57
• Great counterexample! Thank you very much! – user237522 May 29 '18 at 22:05
• @ZachTeitler, please, is it possible to find $f$ and $g$ of the following forms: $f=a_nx^n+\cdots+a_0$, $g=b_mx^m+\cdots+b_0$ with $a_na_0b_mb_0 \neq 0$ and $1 < \gcd(m,n) < min\{m,n\}$? – user237522 May 31 '18 at 20:51
• Yes, it is possible. Since $k(x) = k(x^9+x^5,x^6+x^2)$, then $k(x)=k(x+1)=k((x+1)^9+(x+1)^5,(x+1)^6+(x+1)^2)$ which is $k(x^9+\dotsb+2,x^6+\dotsb+2)$. – Zach Teitler Jun 1 '18 at 6:19

The conjecture (2) is not correct. As you note, $k(x) = k(x^3,x^6+x^2)$; and then $$k(x) = k(x^3 \cdot (x^6+x^2), x^6+x^2) = k(x^9+x^5, x^6+x^2),$$ where $\deg(f)=9$ and $\deg(g)=6$, so $\gcd(\deg(f),\deg(g))=3>1$ but neither $\deg(f)$ nor $\deg(g)$ divides the other one. It is also possible to have counterexamples with nonzero constant coefficients. We have $$\begin{split} k(x) &= k(x+1) \\ &= k((x+1)^9+(x+1)^5, (x+1)^6+(x+1)^2) \\ &= k(x^9+\dotsb+2, x^6+\dotsb+2), \end{split}$$ so if $k$ has characteristic $\neq 2$ then $f$ and $g$ have nonzero constant coefficients. In characteristic $2$, we can find a counterexample with similar manipulations, such as $$k(x) = k(x^3, x^9+x^6+x^2) = k(x^3 \cdot (x^9+x^6+x^2), x^9+x^6+x^2)$$ followed by the same substitution of $x+1$ for $x$.
• Please, can we say something interesting in the special case where $k(f,g)=k(x)$ and the ideal generated by $f'$ and $g'$ is the whole ring $k[x]$? By Hilbert's weak Nullstellensatz $f'$ and $g'$ do not have a common zero, and one can also consider resultants (and D-resultants), but is there a more 'accurate' criterion? – user237522 Jun 12 '18 at 21:15