Convergence of a product in $\mathbb Q_2[[X]]$

Question

I thought it would be very easy to prove, but in fact, I did not manage to prove or disprove this fact: the sequence of polynomials $$\left(\prod_{j=0}^k\big(1-2^{2^j}X\big)\right)_{k\in\mathbb N}$$ converges in $\mathbb Q_2[[X]]$. When considered as function in $\mathbb Q_2$ it is very easy to see that the sequence converges towards a function defined over $\mathbb Q_2$, but I do not any idea when the sequence is considered as formal object.

Answer 1

The coefficients you use are all in $\mathbf Z_2$, so I advise working in $\mathbf Z_2[[x]]$ rather than in $\mathbf Q_2[[x]]$.

You are using the wrong topology on the power series, as mentioned already in a comment to your question. The right one to use on $\mathbf Z_p[[x]]$, where $p$ is prime, comes from the $p$-adic absolute value $|\sum c_nx^n|_p := \max |c_n|_p$, with respect to which $\mathbf Z_p[[x]]$ is complete. (That $|\cdot|_p$ is multiplicative on $\mathbf Z_p[[x]]$ follows from the condition $|f(x)|_p = 1$ being equivalent to $f(x) \bmod p$ being nonzero in $(\mathbf Z/(p))[[x]]$.) Inside $\mathbf Z_p[[x]]$, the subring $\mathbf Z[[x]]$ is dense with respect to $|\cdot|_p$ but the subring $\mathbf Z[x]$ is not, e.g., when $f(x) = \sum_{n \geq 0} x^n$, we have $|f(x)-g(x)|_p = 1$ for all $g(x)$ in $\mathbf Z[x]$ since $f(x) - g(x)$ has $n$-th coefficient 1 when $n > \deg g$. So the completion of $\mathbf Z[[x]]$ with respect to $|\cdot|_p$ is $\mathbf Z_p[[x]]$, while the completion of $\mathbf Z[x]$ with respect to $|\cdot|_p$ is smaller: that completion is the ring $\mathbf Z_p\langle x\rangle$ of power series whose coefficients tend to $0$. See here.

In any case, when $A$ is a domain complete with respect to an absolute value $|\cdot|$ and $\{a_n\}$ is a sequence in $A$ such that $|a_n - 1| < 1$ for all $n \geq 0$, the infinite product $\prod_{n \geq 0} a_n$ converges if $a_n \to 1$ in $A$. Apply this to $A = \mathbf Z_p[[x]]$ with the $p$-adic absolute value and $a_n = 1 - p^{p^n}x$, for which $|a_n-1|_p = 1/p^{p^n} < 1$ (you used $p=2$ but there is nothing special about $p=2$).

Remark. The ring $\mathbf Z_p[[x]]$ has multiple topologies defined by powers of different ideals, with three useful ones being the $p$-adic topology, the $x$-adic topology, and the $(p,x)$-adic topology. (This ring has $(p,x)$ as its unique maximal ideal.) It is $p$-adically complete, $x$-adically complete, and $(p,x)$-adically complete, where the last completeness implies the first two since the ideals $(p)$ and $(x)$ are contained in $(p,x)$, so $(p)^n$ and $(x)^n$ are contained in $(p,x)^n$. To appreciate the difference between these three topologies, show $(1+x)^{p^n} - 1 \in (p,x)^n$, so $(1+x)^{p^n} \to 1$ in the $(p,x)$-adic topology but not in the $p$-adic or $x$-adic topologies. That it lies in $(p,x)^n$ is worked out in the answer here.