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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.

Alexandra
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    Do you mean to take the limit in the $2$-adics? Or where, exactly? – paul garrett Feb 17 '24 at 23:23
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    Your number is $10$-adic, and $10$ is composite, so all bets are off. $x^2=1$ can have many solutions, if you're not in a field. – Gerry Myerson Feb 17 '24 at 23:53
  • It seems like the implication from the first to the second limit is incorrect if I understand you. For instance, in the real numbers borrowing your notation, $\lim_{x \in \mathbb{N}\to \infty}(-1)^{2^x}=1$ does not imply $\lim_{x \in \mathbb{N}\to \infty}2^x = 0$ either. – Merosity Feb 18 '24 at 00:14
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    If you are going to do explicit calculations you need an explicit $p$. Many calculations will have different results for $p=2$, for $p=5$, for $p=13$, ... – jjagmath Feb 18 '24 at 01:44
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    Any thoughts about the comments, Alexandra? – Gerry Myerson Feb 20 '24 at 10:10
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    @GerryMyerson From what I can gather, it seems that I should only be using prime bases for p-adics. – Alexandra Feb 20 '24 at 14:02
  • Composite bases have their uses, but if you want the usual arithmetic of fields to apply, then you need a prime base. – Gerry Myerson Feb 20 '24 at 19:50

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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?.

Torsten Schoeneberg
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