Having promised a longtime collaborator that I would clear my plate to finish up some joint work of ours, I am swallowing my pride and tossing up the following technical point of function field arithmetic to the experts:

Let $k_0 = \mathbb{F}_2(t)$, and let $k(P)$ be a finite separable extension of $k_0$. Let $S$ be a finite set of places $v$ of $k_0$ (thus corresponding to monic irreducible polynomials in $\mathbb{F}_2[t]$ together with possibly the place at infinity). Consider the set of monic irreducible polynomials $w \in \mathbb{F}_2[t]$ with the following properties:

(i) Every place $v$ in $S$ splits completely in the quadratic Artin-Schreier extension of $k_0$ defined by the polynomial $X^2 + X = \frac{1}{w}$, and
(ii) $w$ splits completely in $k(P)$.

Problem: Show that there are infinitely many such $w$.

The natural strategy is to show that the set of such $w$ satisfies some Chebotarev condition -- presumably it even contains the set of all primes splitting in a certain finite Galois extension of $k_0$ -- and then apply the Chebotarev Density Theorem in this context (e.g. Theorem 9.13A in Rosen's Number Theory in Function Fields).

I have already proved similar statements when $k_0$ is a number field or is $\mathbb{F}_p(t)$ with $p$ an odd prime: for the latter I used the Quadratic Reciprocity Law in such fields. In the present case I was trying to use Hasse's characteristic 2 Quadratic Reciprocity Law -- for which my only exposure is this nice note of K. Conrad -- but after some hours of fiddling around, I am having trouble making this work: in the above formulation at least, what is given is not exactly a reciprocity law -- i.e., it doesn't directly compare $[w,\ell_v)$ to $[\ell_v,w)$ but only establishes a certain periodicity relation which is, in the classical cases, equivalent to QR.

Many thanks if you can help me out with this! I really need it in order to complete revisions on a long overdue paper, so a solution will be worth an acknowledgment in the paper at the least.

  • $\begingroup$ Why write $k(P)$ and not $k_0(P)$? $\endgroup$ – KConrad May 18 '11 at 4:47
  • $\begingroup$ You can ignore the infinite place in $k_0={\mathbf F}_2(t)$ since it splits in the extension of $k_0$ defined by a root of $X^2+X−1/h$ for any nonconstant $h$ in ${\mathbf F}_2[t]$. Equivalently, $X^2+X-1/h$ splits over ${\mathbf F}_2((1/t))$. Letting $h$ have degree $d > 0$, write $1/h = u/t^d$, where $u$ is in $1 + (1/t){\mathbf F}_2[[1/t]]$. Setting $Y = t^dX$, we want to show $f(Y) := (1/t^d)Y^2+Y-u$ in ${\mathbf F}_2[[1/t]][Y]$ splits over ${\mathbf F}_2((1/t))$. Since $f(1) = 1/t^d+1-u$ is in the maximal ideal of ${\mathbf F}_2[[1/t]]$ and $f'(1) = 1$, we can use Hensel's lemma. $\endgroup$ – KConrad May 18 '11 at 5:21
  • $\begingroup$ A similar argument as in the previous comment goes through with ${\mathbf F}_2$ replaces by any finite field ${\mathbf F}$ of characteristic 2: the infinite place on ${\mathbf F}(t)$ splits in the quadratic extension of ${\mathbf F}(t)$ defined by a root of $X^2+X-1/h$ for any monic nonconstant $h$ in ${\mathbf F}[t]$. (The polynomial $X^2+X-1/h$ is irreducible over ${\mathbf F}(t)$ since it's quadratic and one can check by contradiction that this polynomial has no root in ${\mathbf F}(t)$ because $h$ is nonconstant in ${\mathbf F}[t]$.) $\endgroup$ – KConrad May 18 '11 at 5:27
  • $\begingroup$ @K: I write $k(P)$ just because that's the notation of my paper: there is a finite separable extension $k/k_0$ (i.e., an arbitrary function field), a curve $C_{/k}$, and $P \in C(k^{\operatorname{sep}})$. As for your second comment: thanks. I was pretty sure that things worked out nicely at infinity (that's why I took the reciprocal!), but seeing the details is helpful. $\endgroup$ – Pete L. Clark May 18 '11 at 6:03
  • 2
    $\begingroup$ That the quadratic reciprocity law in ${\mathbf F}_2[t]$ does not let you relate $[w,\ell_v)$ and $[\ell_v,w)$ doesn't mean it is "not exactly a reciprocity law". After all, the Hilbert reciprocity law and Artin reciprocity law (in its general form) have absolutely nothing to do with reciprocating symbols but they're still called reciprocity laws. $\endgroup$ – KConrad May 18 '11 at 6:28

By my comments to the question, we can assume the infinite place is not in $S$ (because its splitting in the indicated quadratic extensions is automatic). Let the elements of $S$ be $\pi_1,\dots,\pi_r$, where the $\pi_i$'s are distinct (monic) irreducibles in ${\mathbf F}_2[t]$. For each $\pi_i$, the Artin-Schreier map $\wp(x) = x^2 + x$ on the field ${\mathbf F}_2[t]/(\pi_i)$ has image equal to half of that field. Fix a nonzero element of the image, say $g_i \bmod \pi_i$ (why? you'll see shortly). For a monic irreducible $\pi$ in ${\mathbf F}_2[t]$, if $\pi \equiv 1/g_i \bmod \pi_i$ then the congruence $x^2 + x \equiv 1/\pi \bmod \pi_i$ is solvable in ${\mathbf F}_2[t]/(\pi_i)$ by the choice of $g_i$, and then by Hensel's lemma the polynomial $X^2 + X - 1/\pi$ splits in the completion ${\mathbf F}_2(t)$ at $\pi_i$, so $\pi_i$ splits in the quadratic extension of ${\mathbf F}_2(t)$ obtained by adjoining a root of $X^2 + X - 1/\pi$.

So what you would like is to find infinitely many monic $\pi$ such that $\pi \equiv 1/g_i \bmod \pi_i$ for $i = 1,\dots,r$ and $\pi$ splits completely in a predetermined finite separable extension $L/{\mathbf F}_2(t)$. We can assume $L/{\mathbf F}_2(t)$ is Galois by replacing $L$ with its Galois closure over ${\mathbf F}_2(t)$, and then the hypothesis that $\pi$ splits completely in $L$ is the same as saying $\pi$ is unramified in $L$ with trivial Frobenius conjugacy class in the Galois group of $L/{\mathbf F}_2(t)$.

At the same time, the congruence condition $\pi \equiv 1/g_i \bmod \pi_i$ for monic irred. $\pi$ is also a Frobenius constraint using Carlitz extensions. The group $({\mathbf F}_2[t]/(\pi_i))^\times$ is naturally the Galois group of the splitting field over ${\mathbf F}_2(t)$ of the Carlitz polynomial associated to $\pi_i$. Write this splitting field as $K_{\pi_i}$. The natural isomorphism of the (abelian) Galois group of $K_{\pi_i}/{\mathbf F}_2(t)$ with $({\mathbf F}_2[t]/(\pi_i))^\times$ identifies the Frobenius element attached to $\pi$ with $\pi \bmod \pi_i$ for any monic irred. $\pi$ in ${\mathbf F}_2[t]$ not equal to $\pi_i$. (This is analogous to the way $({\mathbf Z}/(m))^\times$ is identified with the Galois group of a cyclotomic field with the Frobenius at a prime $p$ being $p \bmod m$.) Therefore asking that the monic irred. $\pi$ satisfy $\pi \equiv 1/g_i \bmod \pi_i$ for all $i$ is a bunch of simultaneous Frobenius conditions on $\pi$ in the fields $K_{\pi_1},\dots,K_{\pi_r}$. (Warning: these Frobenius conditions for $\pi$ are usually not conditions for $\pi$ to split completely in $K_{\pi_i}$ because we're working with reciprocals of nonzero elements mod $\pi_i$ that are in the image of the Artin-Schreier map, and the Artin-Schreier map has no good multiplicative properties. Maybe you could get a split completely interpretation using the reciprocal of a nonzero element in the image of the Artin-Schreier map mod $\pi_i$ which is itself in the image of the Artin-Schreier map mod $\pi_i$. A count shows it is very likely that there is such an overlap, but I haven't checked it for certain.)

The Carlitz extensions $K_{\pi_1},\dots,K_{\pi_r}$ are linearly disjoint over $F_2(t)$, so there is no problem finding infinitely many $\pi$ with given Frobenius elements in each of the Galois groups of $K_{\pi_i}/{\mathbf F}_2(t)$. (This is also a consequence of the Chinese remainder theorem and Dirichlet's theorem modulo $\pi_1\cdots\pi_r$.) If the field $L$ is linearly disjoint from all the fields $K_{\pi_i}$ over ${\mathbf F}_2(t)$ then we can express all of Pete's original conditions as a single Frobenius condition in the Galois group of their composite field over ${\mathbf F}_2(t)$ and that has infinitely many solutions by Chebotarev. On the other hand, if $L$ is not linearly disjoint from those Carlitz extensions over ${\mathbf F}_2(t)$ then there could be some subtle compatibility issues to be sure Chebotarev can be applied. I'll stop at this point since I've shown the basic plan for how to interpret everything as a Frobenius condition in a Galois group over ${\mathbf F}_2(t)$.


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