# Is the intersection of two function fields over finite fields again a function field?

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.

• 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. – Raju Dec 14 '18 at 17:59
• 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. – Raju Dec 14 '18 at 18:03
• 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. – diddy Jan 11 at 14:18
• 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? – diddy Jan 11 at 14:18
• 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. – Raju Jan 11 at 20:22