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When he introduced $p$-adic numbers, Kurt Hensel produced an incorrect local/global proof of the fact that $e$ is transcendental. Apparently, the intended proof goes along the following lines: studying the well-defined $p$-adic number \begin{equation} e^p=\sum_{n=0}^{\infty}\frac{p^n}{n!}, \end{equation} one shows that $e$, if algebraic, should be of degree at least $p$ over $\mathbb Q$. As this holds for all $p$, $e$ cannot be algebraic. The fundamental mistake in the proof, I have read (e.g Éléments d'histoire des mathématiques), is that it incorrectly assumes that a rational power-series whose evaluation at a rational $r$ converges for the $p$-adic topology to an algebraic number and also converges for the real topology to an algebraic number necessarily converge to the same algebraic number for both topologies, and that is certainly not true.

Does anyone know the details of the correct part of the proof, and in particular how Hensel's supposedly bounded below the purported degree of $e$ ?

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    $\begingroup$ Did anyone ever manage to salvage the proof somehow or is it a dead end? $\endgroup$ Mar 7 at 15:34
  • $\begingroup$ My impression is that it is a complete dead end, because the crucial idea of the proof amounts in the end to an equivocation: the limit of the infinite series defining $e^p$ for the real topology and of the same infinite series for the $p$-adic topology (for larger and larger primes) are totally different things, even if it is tempting to use the same notation. $\endgroup$
    – Olivier
    Mar 8 at 0:13

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You can read Hensel's original article here. The argument is really simple: from this expansion, we see that $e$ satisfies an equation of the form $y^p=1+pu$ where $u$ is a $p$-adic unit. But the polynomial $y^p-(1+pu)$ is irreducible over $\mathbb Q_p$ - writing $y=1+z$ and expanding, we get $$z^p+pz^{p-1}+\dots+pz-pu$$ which is irreducible by Eisenstein criterion. Therefore $e$ must have degree at least $p$ over $\mathbb Q_p$ (and, Hensel's argument goes, hence also over $\mathbb Q$).

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  • $\begingroup$ Thanks a lot, that's great! $\endgroup$
    – Olivier
    Feb 16 at 15:19

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