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In the formal power series ring $\mathbb{F}[[x]]$ over a field $\mathbb{F}$ of characteristic $p>0$, consider an element of the form $f=\sum_{i=0}^\infty a_ix^{p^i}$. Let $R$ denote the unitary subalgebra of $\mathbb{F}[[x]]$ generated by $x$ and $f$.

In my recent work I came across the following problem:

QUESTION. When is $R$ a principal ideal domain?

Of course, this is trivial whenever $x$ belongs to the subalgebra generated by $f$ (or vice versa).

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  • $\begingroup$ Surely then $a_0\neq0$. $\endgroup$ – T. Amdeberhan Dec 8 '16 at 17:58
  • $\begingroup$ @T. Amdeberhan: As $R$ contains the unity, I think that this is not so relevant. $\endgroup$ – Salvatore Siciliano Dec 8 '16 at 18:23
  • $\begingroup$ Oh yeah. I forgot unitary. $\endgroup$ – T. Amdeberhan Dec 8 '16 at 18:31
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    $\begingroup$ What kind of conditions are you expecting? Even to say that $x,f$ are algebraically dependent (which is necessary to have a pid) is a condition. That is not in general enough of course. $\endgroup$ – Mohan Dec 11 '16 at 17:09
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    $\begingroup$ If $R$ is generated by one element then $R$ is clearly a PID, as it is isomorphic to a polynomial ring in one indeterminate. I would expect that the converse is also true. $\endgroup$ – Rocky Smith Dec 11 '16 at 18:18
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More generally, let $f=\sum_{i}^\infty a_i x^{p^i}$ and $g=\sum_{i}^\infty b_i x^{p^i}$ two formal power series involving $p$-powers of $x$ only. Then the subalgebra $R$ generated by $f$ and $g$ is a PID iff there exist $\alpha_0, \alpha_1,\ldots,\alpha_m, \beta_0, \beta_1,\ldots,\beta_n \in \mathbb{F}$ with $\alpha_0,\beta_0 \neq (0,0)$ such that the element $$ h=\sum_{i=0}^m\alpha_i f^{p^i}+ \sum_{j=0}^n\beta_j g^{p^j} $$ generates $R$ as an algebra.

Indeed, the condition is clearly sufficient, as in that case $R$ is isomorphic to a polynomial algebra in one indeterminate. For the converse, observe first that $f$ and $g$ cannot be algebraically independent (as already remarked by Mohan), otherwise $R$ would be isomorphic to a polynomial algebra in two variable and so it cannot be a PID. Let $\Omega$ denote the ideal consisting of all elements of $R$ with zero constant term. As any quotient ring of a PID is also a PID, it is easy to see that, for every $k>0$, $R/ \Omega^{p^k}$ is isomorphic to a truncated polynomial algebra in one variable. Consequently, there are $\alpha_0, \alpha_1,\ldots,\alpha_{k-1}, \beta_0, \beta_1,\ldots,\beta_{k-1} \in \mathbb{F}$ with $\alpha_0,\beta_0 \neq (0,0)$ such that the element $$ \bar{h}=\sum_{i=0}^{k-1}\alpha_i f^{p^i}+ \sum_{j=0}^{k-1}\beta_j g^{p^j} $$ is a generator for $R/ \Omega^{p^k}$ as an algebra. As $f$ and $g$ are algebraically dependent, using the fact that $\bigcap_{n=0}^{\infty} \Omega^n=\{ 0\}$ one can now see that there exist an element $h$ with the required property.

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  • $\begingroup$ I think there are some steps missing in this answer. How is the fact that the power series are $p$-powers used? The statement is certainly false for arbitrary polynomials. $\endgroup$ – Will Sawin Dec 11 '16 at 20:06
  • $\begingroup$ Regard $F[[x]]$ as a restricted Lie algebra via the ordinary p-power map and let $L$ be the restricted subalgebra generated by $f$ and $g$. My argument is based on the fact that $R$ is isomorphic to the restricted enveloping algebra of $L$. $\endgroup$ – Salvatore Siciliano Dec 11 '16 at 22:07
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This happens if and only if the coefficients $a_i$ satisfy a simple Frobenius-linear recurrence $a_i = \sum_{j=1}^n a_{i-j}^{p^j} c_j$ for all $i \geq n$.

First assume that $R$ has dimension $1$.

$R$ is an integral domain generated by $x$ and $f$, so it is $\mathbb F[x,f]/I$ for some prime ideal $I$. Because $x$ and $f$ satisfy some relation, $I$ is nonempty, and because $x$ satisfies no relation all on its own, $I$ is maximal, so $I$ is generated by a single element $g(x,f)$.

Observe that $g(x_1+x_2, f(x_1) + f(x_2))= g(x_1+x_2, f(x_1+x_2))$ vanishes as an element of $\mathbb F[[x]] \otimes \mathbb F[[x]]$. So it vanishes as an element of $R \otimes R = F[x,f]/g(x,f) \otimes F[x,f]/g(x,f)$.

Hence $g(x_1+x_2,y_1+y_2)$ lies in the ideal generated by $g(x_1,y_1)$ and $g(x_2,y_2)$. Because these polynomials all have the same degree, and $g(x_1,y_1)$ and $g(x_2,y_2)$ are in independent variables, we have $g(x_1+x_2,y_1+y_2)=c_1 g(x_1,y_1)+ c_2 g(x_2,y_2)$. We can clearly see that $c_1 = c_2 =1$.

In particular $g(x,y) =g(x,0)+g(0,y)$, so $g$ splits into a polynomial in $x$ and a polynomial in $y$. Furthermore $g(x_1+x_2,0)=g(x_1,0)+g(x_2,0)$, so both these are additive. In other words, the only monomials appearing in $g(x,y)$ are $x^{p^i}$ and $y^{p^i}$.

Let $g(x,y) = \sum_{k=1}^m b_k x^{p^k} + \sum_{j=1}^n c_j y^{p^k} $.

Expanding $g(x,\sum_i a_i x^{p^i})=0$, we get for all $i> \max(n,m)$ that $\sum_j c_j a_{i-j}^{p^j} =0$. Prune the coefficients $c_j$ so that the first one is nonzero, then divide the other coefficients by it. Then add additional $0$s to the end until the condition $i> \max(n,m)$ is no longer necessary. This gives the desired recurrence.


Conversely if $a_i = \sum_{j=1}^n a_{i-j}^{p^j} c_j$ for all $i \geq n$, then $f(x) - \sum_{j=1}^n f(x)^{p^j} c_j$ is a $p$-power polynomial in $x$ of degree at most $p^{n-1}$, i.e. $$f(x) - \sum_{j=1}^n f(x)^{p^j} c_j = \sum_{k=1}^{n-1} b_k x^{p^k}$$

The relation $$y- \sum_{j=1}^n y^{p^j} c_j = \sum_{k=1}^{n-1} b_k x^{p^k}$$ defines a finite etale cover of $\mathbb A^1$ - etale because the coefficient of $f(x)$ is nonzero. (We actually may want to work with a slightly smllaer finite etale cover, the connected component of $(0,0)$. To check this is a PID, it is sufficient to check that this smooth curve is isomorphic to $\mathbb A^1$. But there is a clear group structure on this curve: $(x_1,y_1) +(x_2,y_2)=(x_1+y_1,x_2+y_2)$. So the smooth affine curve has a transitive group of automorphisms and hence is isomorphic to $\mathbb P^1$ minus one or two points. It can't be $\mathbb P^1$ minus two points as there is no group homomorphism from that to $\mathbb A^1$.

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  • $\begingroup$ I think that your argument could be used to prove the stronger statement that I wrote in my answer, as well. $\endgroup$ – Salvatore Siciliano Dec 11 '16 at 22:24
  • $\begingroup$ @SalvatoreSiciliano I agree, although I wouldn't say it is stronger. Just different - yours gives an algebraic construction of the possible $f$, while mine gives a numerical description of the coefficients. $\endgroup$ – Will Sawin Dec 11 '16 at 22:36
  • $\begingroup$ Sure, I agree. In any case, as my area is not algebraic geometry, it is more natural for me to think in terms of restricted Lie algebras and their restricted enveloping algebras. $\endgroup$ – Salvatore Siciliano Dec 11 '16 at 22:42
  • $\begingroup$ @ Will Sawin: By the way, are you aware of any characterization of when the coordinate ring K[G] of a connected reduced algebraic group scheme G over the field K is a PID? Of course, as G is one-dimensional, the answer is easy when K is algebraically closed. What about the general case? Probably, this could be suitable for a possible question for MO, as well... $\endgroup$ – Salvatore Siciliano Dec 11 '16 at 22:54
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    $\begingroup$ @SalvatoreSiciliano You are right, descent to imperfect fields is not classified by cohomology, obviously, so some of what I've said is wrong. Instead one can show that the only form of $\mathbb G_a$ that is a PIDs is the standard one. Consider a generator of the ideal of the identity element. I t remains a generator of this ideal after base change to an algebraically closed field. The only generators of this ideal on $\mathbb G_a$ are multiples of the standard coordinate. Any multiple of the standard coordinate defines an isomorphism with $\mathbb G_a$ over the base field. $\endgroup$ – Will Sawin Dec 17 '16 at 11:51

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