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I am interested in the question above. I know that the answer is NO if the base field is for instance $\mathbb{Q}$ (the intersection of $\mathbb{Q}(x^2)$ and $\mathbb{Q}((x-1)^2)$ is $\mathbb{Q}$ where $x$ is a $\mathbb{Q}$-transendental element).

But how about two function fields over a finite field $\mathbb{F}_q$?

Let $F$ be a function field over $\mathbb{F}_q$ and $F_1$ and $F_2$ two sub function fields of $F$ such that both extensions are separable. The counterexample from above doesn't work here. More generally: If $F/F_1$ and $F/F_2$ are Galois with the corresponding subgroups $U_i$ of the finite (here one needs a finite field) automorphism group $G$ of $F$, then, by finite Galois theory, $F_1 \cap F_2$ is the fixed field of the subgroup which is generated by $U_1$ and $U_2$. Since the subgroup is finite, the fixed field must be a function field.

But, what if only one of the extensions is Galois (the other case (both non-Galois) should follow from this case by considering the Galois closure of one of the extensions)? Or is the claim not true and somebody can provude a counterexample?

Ansatz 1: Is it possible to construct an extension $E/F$ such that $E/F_1$ and $E/F_2$ are Galois?

Ansatz 2: Starting with the easiest case: Let $F=\mathbb{F}_q(x)$ be a rational function field and $F_i = \mathbb{F}_q(f_i(x))$ with polynomials $f_i$. It boils down to the question whether there are polymomials $g_i$ such that $$ g_1(f_1(x)) = g_2(f_2(x)). $$

I tried out some specific examples, but couldn't find an extendable pattern. For $q=3$ and $f_1 = x^2$ and $f_2 = (x-1)^2$ (example from above), we can choose the $g_i$ in the following way $$ (x^2)^3 + (x^2)^2 + x^2 = x^6 + x^4 + x^2 = ((x-1)^2)^3 + ((x-1)^2)^2 + (x-1)^2. $$

Thank you.

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    $\begingroup$ Consider the Hecke correspondence of modular curves $Y(1)\leftarrow Y_0(l)\rightarrow Y(1)$ over $\mathbb{F}_p$ and take the associated diagram of function fields. This gives a counterexample to the question in the statement. $\endgroup$ – Raju Dec 14 '18 at 17:59
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    $\begingroup$ Self-promotion: see sections 3 and 4 of <math.columbia.edu/~raju/Papers/dynamics.pdf>. Cribbing from Mochizuki, I say that correspondence of curves $X\leftarrow Z\rightarrow Y$ over $k$ has a core if the transcendence degree of $k(X)\cap k(Y)$ over $k$ is 1. The claim is that most Hecke correspondences will yield correspondences without a core. To prove the above example is a correspondence without a core, use Exercise 3.13 with Lemma 4.10. $\endgroup$ – Raju Dec 14 '18 at 18:03
  • $\begingroup$ Sorry for the late reaction. Thank you very much. It is nice to know that the intersection doesn't have to be a function field. Moreover, I would also like to have an example which I can show somebody who only knows the elementary basics of function field theory. $\endgroup$ – diddy Jan 11 at 14:18
  • $\begingroup$ Something like the intersection of the function fields $\mathbb{F}_2(x^3)$ and $\mathbb{F}_2(x^2(x+1))$ where one can maybe show that there is no non-constant polynomial $f$ such that $f(x^2(x+1))\in\mathbb{F}_2(x^3)$. But this doesn't seem as easy as I expected. There is no $f$ up to degree 300. This can be verified by using a computer algebra system. But I can not find a general reason. Maybe, you have an idea or another elementary example? $\endgroup$ – diddy Jan 11 at 14:18
  • $\begingroup$ Here's an idea. Consider the associated correspondence of curves $X\leftarrow Z\rightarrow Y$. This yields a many-valued self-function $F$ on $X$ (go up and down along the correspondence). If the intersection of function fields is again a function field, then there exists $N$ so that for any point $x$, the iterated orbit under $F$ has no greater than $N$ distinct elements. So you just need to find $\mathbb F$ points with arbitrarily large orbit size. $\endgroup$ – Raju Jan 11 at 20:22

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