Proof using p-adic numbers that $\lim_{x \in \mathbb{N} \to \infty} 2^x = 0$? What am I doing wrong?

Question

Approaching an infinite natural x value for the expression $5^{2^x}$, one is left with the following p-adic number: $$...212890625$$ Which is equal to its own square. With basic algebra one can see that, since this value is equal to itself squared, it's equal to 1. Therefore, $$\lim_{x \in \mathbb{N} \to \infty} 5^{2^x} = 1$$ Therefore $$\lim_{x \in \mathbb{N} \to \infty} 2^x = 0$$ This shouldn't be true. What did I do wrong? I am new to the field of p-adics, so bear with me.

Answer 1

To get this off the unanswered list and turn comments into an answer:

You seem to be using the $10$-adic numbers $\mathbb Z_{10} = \projlim \mathbb Z / \mathbb 10^k$.

It is correct that in those, the sequence $a_n := 5^{(2^n)}$ converges to an element $a \in \mathbb Z_{10}$ that, in $10$-adic a.k.a. decimal expansion, matches what you write.

It is also true that that element satisfies $a^2=a$ (almost by construction). Such elements are called idempotents. In domains, the only idempotents are $0$ and $1$. This element visibly being neither shows that $\mathbb Z_{10}$ is not a domain.

In fact, for prime $p$, the $p$-adics $\mathbb Z_p$ are a domain. Now while for prime powers one would still have $\mathbb Z_{p^r} \simeq \mathbb Z_p$, for a general natural number $n$, the ring of $n$-adic integers $\mathbb Z_n := \projlim_k \mathbb Z / n^k$ happens to be isomorphic to the direct product

$$ \mathbb Z_{p_1} \times \dots \times \mathbb Z_{p_r} $$

where the $p_i$ are the prime divisors of $n$. In particular, the $10$-adics are isomorphic to

$$\mathbb Z_2 \times \mathbb Z_5.$$

Such an isomorphism is basically a limit of isomorphisms coming from the Chinese Remainder Theorem, and I leave it to you to make it explicit. I suspect that the element you found there, under this isomorphism, happens to be the obvious idempotent $(1,0) \in \mathbb Z_2 \times \mathbb Z_5$ -- because the sequence $a_n$ has limit $1$ in $\mathbb Z_2$ and limit $0$ in $\mathbb Z_5$. Can you find a $10$-adic expression of the element representing $(0,1)$? Careful, the first naive attempt of looking for a limit of $2^{(5^n)}$ does not work!

Related question with good answers: Why are $p$-adic numbers and $p$-adic integers only defined for $p$ prime?. Also compare e.g comments to What is an example a zero divisor in the 10-adic integers?.