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Let $p$ be a prime number, let $x$ be a variable, and consider two power series over the ring $\mathbb{Z}_p$ of $p$-adic integers:

$a(x):=\underset{n\geq 1}{\sum}{\frac{p^n}{n!}x^n}=px+\frac{p^2}{2}x^2+\frac{p^3}{6}x^3+\cdots$

$b(x):=\underset{n\geq 1}{\sum}{\frac{p^n}{n!}x^{2n}}=px^2+\frac{p^2}{2}x^4+\frac{p^3}{6}x^6+\cdots$

My question is, can we find a power series $0\neq f(u,v)\in\mathbb{Z}_p[[u,v]]$ with coefficients in $\mathbb{Z}_p$, such that $f(a,b)=0$, or in other words are $a(x)$ and $b(x)$ topologically algebraically independent (TAI) over $\mathbb{Z}_p$?

It is not true that $a$ and $b$ are TAI over $\mathbb{Q}_p$, which we can observe simply by choosing a sequence of polynomials with rational coefficients which remove successively higher and higher powers of $x$. For example:

$0=a(x)^2-pb(x)-pa(x)b(x)-\frac{(7p-12)p}{12}b(x)^2+\cdots$

In fact, we can apply the same argument to say that any distinct pair of univariate power series over $\mathbb{Q}_p$ are not TAI over $\mathbb{Q}_p$.

Unfortunately, I can think of no way of ensuring that the coefficients of this power series lie in $\mathbb{Z}_p$, or even that the series can be scaled by a power of $p$ so that they will.

If anyone has any ideas or suggestions, I'd be very interested to hear them. Thanks.

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This question has an open bounty worth +50 reputation from AdJoint-rep ending in 4 days.

Looking for an answer drawing from credible and/or official sources.

I need a proof that such a power series cannot exist, or else an example of one.

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It’s late at night, and I hope I’m not getting egg all over my face here, in this argument tailored to your particular example.

First, I’m going to define $\log(x)=-\sum_{n\ge1}(-x)^n/n=x-x^2/2+x^3/3-\cdots$ and $\exp(x)=\sum_{n\ge1}x^n/n!$, so that this log and exp are inverse power series of each other, defined over $\Bbb Q_p$.

Next, your $a(x)$ is $\exp(px)$ and your $b(x)$ is $\exp(px^2)$, both of them landing in $\Bbb Z_p[[x]]$. Thus we can say, by taking logs, that $\log\bigl(a(x)\bigr)=px$ and $\log\bigl(b(x)\bigr)=px^2$, giving $$ \bigl(\log\bigl(a(x)\bigr)\bigr)^2=p\log\bigl(b(x)\bigr)\,, $$ a manifest statement of topological algebraic dependence, but over $\Bbb Q_p$. Now, $a(x)$ and $b(x)$ have all coefficients divisible by $p$, and $\log(px)\in\Bbb Z_p[[x]]$, so that the displayed series actually has $\Bbb Z_p$-coefficents.

This seems to be telling me that if you had only asked for the topological algebraic dependence over $\Bbb Z_p$ of $A(x)$ and $B(x)$ where $A(x)=a(x)/p$ and $B(x)=b(x)/p$, we would have it. But I’m not seeing the desired result at this point.

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  • $\begingroup$ Thanks for your answer @Lubin. I did think that it would help to use the logarithm, but as you observed, this only solves the problem for $\frac{a(x)}{p}$ and $\frac{b(x)}{p}$, and unfortunately this isn’t enough. Also, in order for $exp$ and $log$ to be mutually inverse, we need the $exp$ series to start at 0, not 1, whereas the series $a$ and $b$ both start at 1. Thanks for your help anyway. $\endgroup$ – AdJoint-rep Apr 16 at 14:43
  • $\begingroup$ I’m not sure about the meaning of the sentence beginning with “Also…”. You’re not saying that the two series presented here fail to be inverses of each other, are you? $\endgroup$ – Lubin Apr 16 at 15:19
  • $\begingroup$ Actually no, I reread your answer, and if you define $exp$ and $log$ the way you do, they are mutually inverse, it’s just that we usually define the exponential series by $exp(x)=\underset{n\geq 0}{\sum}{\frac{x^n}{n!}}$ as opposed to $exp(x)=\underset{n\geq 1}{\sum}{\frac{x^n}{n!}}$. $\endgroup$ – AdJoint-rep Apr 16 at 15:48
  • $\begingroup$ Of course. Actually, these are the logarithm and exponential for the multiplicative formal group $\mathcal M(x,y)=x+y+xy$. $\endgroup$ – Lubin Apr 16 at 17:52

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