# Topological Algebraic Independence of power series

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.

## 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.

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.
• 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. – AdJoint-rep Apr 16 at 14:43
• 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 $exp(x)=\underset{n\geq 1}{\sum}{\frac{x^n}{n!}}$. – AdJoint-rep Apr 16 at 15:48
• Of course. Actually, these are the logarithm and exponential for the multiplicative formal group $\mathcal M(x,y)=x+y+xy$. – Lubin Apr 16 at 17:52