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When introducing $p$-adics, Kurt Hensel produced an incorrect proof of the transcendence of $e$: see https://mathoverflow.net/q/416296 for more details. The problem is that the proof relies on the "universality" of series, that is, the assumption that the series $$\sum_{n=1}^\infty\frac{p^n}{n!},$$ which converges in the $p$-adics, would converge to a number with similar properties as $e^p\in\mathbb R$. (This is false.)

My question is: is there a more dramatic way to apply this "proof" idea to show an obviously false result? For instance, using it to show the transcendence of a number that is actually algebraic.

Zongshu Wu
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  • Look at standard examples of binomial series which converge both in reals and $p$-adics, and always to certain roots of a number, but different ones (one positive, one negative; or one rational, one not) in comments to https://math.stackexchange.com/a/3765961/96384 or https://math.stackexchange.com/q/3578793/96384 and links from there. Take note of answers to https://math.stackexchange.com/q/3264177/96384 and twist them to produce series that converge, "at the same time", to your favourite prescribed numbers in the reals and finite sets of $p$-adics. – Torsten Schoeneberg Mar 25 '24 at 00:45

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In Example 6.2 here is a series of rational numbers converging to $0$ in $\mathbf R$ and to $1$ in $\mathbf Q_p$ for whichever prime $p$ you want. Example 9.5 there is an infinite series of rational numbers that converges to $8/7$ in $\mathbf R$ and to $-8/7$ in $\mathbf Q_3$ and $\mathbf Q_5$. Remark 9.7 gives an example of an infinite series of rational numbers that converges to $3$ in $\mathbf R$ and to $-3$ in $\mathbf Q_2$.

KCd
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  • I'm not asking for such simple examples -- I know they exist. What I want is a more complex example of a series, perhaps for which Hensel's proof applies in the p-adics, but instead converges to an algebraic number in R. – Zongshu Wu Mar 27 '24 at 05:27
  • Nothing about Hensel's proof "applies" in the $p$-adics: his invalid reasoning involves a single series converging in each $\mathbf Q_p$ and $\mathbf R$ and trying to somehow equate the limit of the series in different completions and such equality in two different completions. It needs infinitely many completions to have a chance of working, but it really doesn't work. – KCd Mar 27 '24 at 05:39
  • I already understand what's wrong with Hensel's proof. Well, it does have correct bits, doesn't it? What I'm looking for is something like this: I present Hensel's "proof", which seems reasonable at first glance; Then, I proceed to present another similar "proof" (that is, an argument relying on the erroneous assumption, that would work if the assumption was in fact true) that some series gives a transcendental number, only this time, when I evaluate it in R, it actually is algebraic! – Zongshu Wu Mar 27 '24 at 07:56
  • @ZongshuWu: The link in this answer (and the link in my comment) provide you with sequences $a_n$ which converge to $0$ in $\mathbb R$ and to $1$ in $\mathbb Q_p$. Now to get what you want, take some transcendental element $x \in \mathbb Q_p$ together with a sequence of rationals $x_n$ which goes to $x$ in $\mathbb Q_p$ but stays bounded in $\mathbb R$. Then the sequence of products $a_n x_n$ will go to your transcendental $x$ in $\mathbb Q_p$, but to the very algebraic $0$ in $\mathbb R$. – Torsten Schoeneberg Mar 28 '24 at 04:44
  • @TorstenSchoeneberg seems like a good idea. Thanks! – Zongshu Wu Mar 28 '24 at 11:57