Question: For a prime $p$, is every involution in $\mathbb{F}_p[[x]]$ with a zero constant term a reduction modulo $p$ of some involution in $\mathbb{Z}[[x]]$?

Here involution in $A[[x]]$ means $f\in A[[x]]$ such that $f\circ f=x$.

Obviously, a reduction mod $p$ of an involution in $\mathbb{Z}[[x]]$ yields an involution in $\mathbb{F}_p[[x]]$. Moreover, at least one involution in at least one $\mathbb{F}_p[[x]]$ is not obtainable by reducing mod $p$, namely, $1+x\in\mathbb{F}_2[[x]]$, exactly because its constant term is nonzero. (Update: The second sentence in this paragraph confuses the rings of polynomials and power series, see comments and answers for corrections.)

But do we get all involutions in $\mathbb{F}_p[[x]]$ with the zero constant term this way?

If there is a reference where this is discussed, that would be very helpful, too.

  • $\begingroup$ @YCor Involution = compositional self-inverse, i.e. $f(x)$ such that $f(f(x))=x$. $\endgroup$ – Alexander Burstein May 5 '19 at 20:18
  • 2
    $\begingroup$ @YCor I don't think we are considering involutions of $\mathbb F_p[[x]]$. Rather, we are considering involutions $f\in\mathbb F_p[[x]]$. $\endgroup$ – Wojowu May 5 '19 at 20:22
  • 2
    $\begingroup$ @Wojowu Exactly. As the title says, involutions in $\mathbb{F}_p[[x]]$, not on $\mathbb{F}_p[[x]]$. Perhaps, that was a source of confusion. $\endgroup$ – Alexander Burstein May 5 '19 at 20:24
  • 3
    $\begingroup$ Involutions in $\mathbb F_p[[x]]$ with zero constant term are involutions of $\mathbb F_p[[x]]$, or, well, the associated topological ring automorphisms $x \to f(x)$ are. $\endgroup$ – Will Sawin May 5 '19 at 20:38
  • 4
    $\begingroup$ Involutions with nonzero constant term don't really make sense, as composition of power series with nonzero constant term isn't well-defined. If you do it in the ring of polynomials, $f(x)=1+x$ does in fact lift, to $f(x)=1-x$. $\endgroup$ – Will Sawin May 6 '19 at 16:57

If $p \neq 2$, an involution in $\mathbb F_p[x]$ with zero constant term must have $\pm 1$ as the coefficient of $x$. If the leading coefficient is $1$, we can inductively see that each coefficient is zero (writing $f(x) = x + a x^n + \dots, f(f(x))= x+ 2a x^n + \dots $ so $a=0$.) Of course this trivial involution can be lifted.

On the other hand, if the leading coefficient is $-1$, then we can take $x f(x)$ a power series with leading coefficient $-x^2$. We can look at the unique algebra homomorphism $\mathbb F_p[[x]] \to \mathbb F_p[[x]]$ that sends $x$ to $f(x)$ and preserves the topology. Under this automorphism, $-x f(x)$ is fixed, so $\sqrt{-x f(x)}$ (unique up to $\pm 1$) is sent to itself or minus itself. Since its leading term is nonzero, it is sent to minus itself. So up to reparameterization by the invertible power series $\sqrt{-x f(x)}$, this is the standard involution $\pm 1$. We can lift $\sqrt{-x f(x)}$ to the integers, where the power series of the inverse function will also be integral, and use those to lift the involution.

In characteristic two, the situation is worse as there are more involutions. For instance we can take $ f(x)= \left( x^n/(1+x^n)\right)^{1/n}$ for any odd $n$. Some of these can be lifted (for instance if $n=1$ we can do $f(x) = -x/ (1-x)$ but I don't know if all can be.


In characteristic $0$, let $f(x) = - x/ (1 + c x^n)^{1/n}$ for any odd $n$ and $c$ in $\mathbb Z$, sufficiently divisible as to make that power series integral. Then $f$ is an involution: $$ 1 + c f(x)^n = 1 + \frac{-c x^n}{ (1+ cx^n)} = \frac{1}{ 1+ cx ^n} $$ so $$f(f(x)) = - \frac{ f(x) }{ (1+ cx^n)^{1/n} } = -\frac{ -x }{ (1+cx^n)^{1/n} } (1+ cx^n)^{1/n}=x.$$ According to the theorem of Klopsch that Lubin cites, it seems tha this gives all involutions mod $2$.

| cite | improve this answer | |
  • $\begingroup$ Why is $\sqrt{xf(x)}$ well-defined (up to $\pm$)? I'd understand it in characteristic zero using a series for $\sqrt{1+t}$, but I'm not sure in finite odd characteristic. $\endgroup$ – YCor May 5 '19 at 21:37
  • $\begingroup$ @YCor The denominators in the series are all powers of two. Alternately you can build it step by step. If the coefficients of the square root are $a_n$, the equation for the degree $n$ term looks like $2 a_n = $ a polynomial in $a_1,\dots, a_{n-1}$, so has a unique solution in any field of characteristic not $2$. $\endgroup$ – Will Sawin May 5 '19 at 21:58
  • 2
    $\begingroup$ @YCor this carries over to the coefficients of $(1+t)^{1/n}$ for all $n \geq 1$: the prime factors of the denominators of the coefficients are exactly the prime factors of $n$. See my answer at math.stackexchange.com/questions/696669/…. $\endgroup$ – KConrad May 5 '19 at 23:02
  • 1
    $\begingroup$ @darijgrinberg since your $a_n$ is invertible, your $\mathbf a$ is $(b_1x)^ng(x)$ where $g(x) \in 1 + xA[[x]]$, so showing $\mathbf a$ has a unique $n$th root once $b_1$ is chosen with $b_1^n = a_n$ is equivalent to showing each $u(x) \in 1 + xA[[x]]$ has a unique $n$th root in $1 + xA[[x]]$. For that you can basically use Hensel's lemma for $f(t) = t^n - u(x) \in A[[x]][t]$: from $f(1) = 1 - u(x) \equiv 0 \bmod x$ and $f'(1) = n \in A^\times \subset A[[x]]^\times$, there is a unique $y(x) \in A[[x]]$ such that $f(y(x)) = 0$ and $y(x) \equiv 1 \bmod x$, meaning $y(x) \in 1 + xA[[x]]$. $\endgroup$ – KConrad May 5 '19 at 23:12
  • 1
    $\begingroup$ @darijgrinberg I said "basically use Hensel's lemma" rather than "use Hensel's lemma" since an $x$-adic absolute value on $A[[x]]$ is generally submultiplicative rather than multiplicative if $A$ is a commutative ring and not specifically an integral domain: $|A(x)B(x)|_x \leq |A(x)|_x|B(x)|_x$. But we still have $|A(x)U(x)|_x = |A(x)|_x = |A(x)|_x|U(x)|_x$ when $U(x) \in A[[x]]^\times = A^\times + xA[[x]]$. $\endgroup$ – KConrad May 5 '19 at 23:15

This is algebraic dynamics, a big field that I’m outsider to, though I have made some peripheral contributions to it.

First, I’m assuming that you’re asking about ring automorphisms of $\Bbb Z[[x]]$ and of $\Bbb F_p[[x]]$, and the involutions from among these. Under this assumption, $x\mapsto1+x$ is not an involution of $\Bbb F_2[[x]]$, since it wants to take the nonunit $x$ to the unit $1+x$.

Second, and more generally, in view of the fact that all automorphisms of $\Bbb F_q[[x]]$ are necessarily continuous, we don’t need to specify that our involution $x\mapsto u(x)$ have no constant term in $u$.

Third, though Will has described an infinite set of nonconjugate involutions (in the group of all invertible series over $\Bbb F_2$), as I recall, according to a theorem of Klopsch, these are all that there are out there, in this $\Bbb F_2$-case.

Finally, all the involutions of $\Bbb Z[[x]]$ that I know about are (naturally) the $[-1]$-automorphisms of $\Bbb Z$-formal group laws, and these look like $-x+cx^q\pmod{x^{q+1}}$ for a prime power $q$. Surely there are other involutions in this situation, but I don’t know them. In particular, if you try $n=5$ here, you get a series $x+x^6+\cdots\in\Bbb F_2[[x]]$ that doesn’t seem to come from $[-1]_F(x)\in\Bbb Z[[x]]$ for any $\Bbb Z$-formal group law $F$. But globally-defined formal group laws I’m rather ignorant about. Nobody’s saying, either, that a $\Bbb Z$-involution has to belong to a formal group over $\Bbb Z$.

I think your question has to remain unanswered, at least with the expertise exhibited here so far.

EDIT — Addendum:

All thanks to Will Sawin for his persistence in the face of my repeated careless computational errors. The principle involved is that for an integral domain $R$, the map on the semigroup of formal series under substitution $u(x)\in R[[x]]$ with no constant term and $u'(0)\ne0$, namely $$ u(x)=xw(x)\mapsto \bigl(u(x^m))^{1/m}=x\bigl(w(x^m)\bigl)^{1/m}\,, $$ is a homomorphism, as long as it’s defined. For definedness, you need $u'(0)=w(0)$ to have an $m$-th root in the base, and more broadly, you need some sort of convergence theorem to tell you that you can take the $m$-th root of the series $w(x)$.

In the comments, I pointed out that $(1+p^2x)^{1/p}\in\Bbb Z[[x]]$, and this means that $(1+m^2x)^{1/m}\in\Bbb Z[[x]]$ as well, for any positive integer $m$.

Will says, start with the fundamental involution $u(x)=-x/(1+x)=\sum_1^\infty(-x)^n$, first apply $u(x)\mapsto u(m^2x)/m^2=U(x)$, and apply the “stretching-out” operation above, which will be defined as long as $U'(0)=-1$ has an $m$-th root in $\Bbb Z$, namely as long as $m$ is odd. Since $U(X^m)=-x^m+m^2x^{2m}-m^4x^{3m}+\cdots=-x^m[1-m^2x^m+m^4x^{2m}-\cdots]$, the conditions for the existence of the $m$-th root are satisfied. Thus $\bigl(U(x^m)\bigr)^{1/m}$ is a good lifting to $\Bbb Z[[x]]$ of the characteristic-two involution of the form $-x(1+g(x^m))$ for $m$ odd.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.