Galois invariants in a ring of fractional power series over a finite field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:45:03Z http://mathoverflow.net/feeds/question/95105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95105/galois-invariants-in-a-ring-of-fractional-power-series-over-a-finite-field Galois invariants in a ring of fractional power series over a finite field Jared Weinstein 2012-04-25T01:55:27Z 2012-04-25T01:55:27Z <p>Let $\mathbf{F}_q$ be a finite field, and let $A=\mathbf{F}_q [[ x^{1/q^\infty} ]]$ be the completion of $\mathbf{F}_q[x^{1/q^\infty}]$ with respect to the $x$-adic topology. Then the $q$th power Frobenius map $\pi\colon A\to A$ is an automorphism. If $F$ is the local field $\mathbf{F}_q((\pi))$, there is an action $F^\times\to \text{Aut} A$ by continuous automorphisms; an element of $F$ sends $x$ to the corresponding linear fractional power series. Let $\mathcal{O}_F=\mathbf{F}_q[[\pi]]$. </p> <p>My question is: Can anyone produce an explicit nonconstant element of $A$ which is $\mathcal{O}_F^\times$-invariant?</p> <p>I assure you that such elements exist! Here's why: Let $f(X)=\pi X+X^q\in F[X]$. Let $x_1,x_2,\dots$ be a compatible family of roots of iterates of $f$, with $x_1\neq 0$. Let $F_n=F(x_n)$. Then by Lubin-Tate theory, <code>$\text{Gal}(F_n/F)\cong (\mathcal{O}_F/\pi^n)^\times$</code>. If $F_\infty$ is the union of the $F_n$, then $F_\infty/F$ is a maximal totally ramified abelian extension of $F$, with Galois group <code>$\mathcal{O}_F^\times$</code>. Let $K$ be the $\pi$-adic completion of $F_\infty$.</p> <p>I claim that $\mathcal{O}_K$ is isomorphic to $A$. Indeed, it isn't hard to see that the sequence $x_1^q,x_2^{q^2},\dots$ converges in $\mathcal{O}_K$ to an element $x$, all of whose $q$th power roots lie in $\mathcal{O}_K$. Observe that $x^{1/q^n}-x_n$ is divisible by $\pi$, hence by $x^{1/q}$, and therefore (since $\mathcal{O}_K$ is topologically generated by the $x_n$) we have $\mathcal{O}_K=\mathbf{F}_q[x^{1/q^\infty}]+x^{1/q}\mathcal{O}_K$. This shows that $\mathbf{F}_q[x^{1/q^\infty}]$ is dense in $\mathcal{O}_K$, which proves the claim. </p> <p>The Galois action of <code>$\mathcal{O}_F^\times$</code> on $F_\infty$ extends to the completion $K$. It's a fun exercise to see that the isomorphism $\mathcal{O}_K\cong A$ respects the action of $\mathcal{O}_F^\times$. Meanwhile, we have the element $\pi\in \mathcal{O}_K\cong \mathbf{F}_q[[x^{1/q^\infty}]]$. This element is invariant under the Galois group $\mathcal{O}_F^\times$. What is its power series development in $x$? </p>