#kurt hensel

[farouche-landau]

Delighted to share the news that Kurt Hensel was Fanny Mendelssohn Hensel's grandson with someone else who appreciates the work of both! I am very fond of Kurt Hensel's bookplate, which shows his love of music as well as mathematics:

(note the elliptic curve on the paper to the left of the violin!) And yes, German academia was a small world back in the day, and the Mendelssohn family was particularly well-connected: you may already know that Fanny and Felix's younger sister Rebecka married Dirichlet, and their cousin Ottilie married Kummer. And extending the family tradition, one of Kurt Hensel's grandsons, Walter Hayman, also became a mathematician, after leaving Germany for England at age 12 to escape the Nazis. He had a pretty remarkable life (his autobiography is on my to-read list, but it's not available as an e-book, so I will actually have to go to the university library to get it...): among his many accomplishments, he helped his wife Margaret found the British Mathematical Olympiad and send the first British team to the International Math Olympiad. And one of Walter's daughters, Sheila Hayman, became a filmmaker, and directed two films about her family: I still haven't seen Fanny: the Other Mendelssohn, but her earlier film Mendelssohn, the Nazis, and Me covers her family history and the impact of the Nazis in a moving way.

[thetwomeatmeal]

See if you can spot the mistake in the following proof, an attempt by Kurt Hensel to prove that e is transcendental, as an early application of p-adic numbers.

Suppose that $e$ is algebraic over $\mathbf{Q}$; then it also defines a finite extension $\mathbf{Q}_p(e)$ of the p-adic numbers. Better yet, the series $$e^p = \sum_{n \ge 0} \frac{p^n}{n!}$$ converges in $\mathbf{Z}_p$, for any $p$. Convergence is a consequence of the calculation of the $p$-adic valuation of a factorial, due originally to Legendre: for $n > 0$, we have $$v_p(n!) = \left\lfloor \frac{n}{p}\right\rfloor + \left\lfloor \frac{n}{p^2}\right\rfloor + \dotsb < \sum \frac{n}{p^i} = \frac{n}{p-1} \le n.$$ So all the terms are $p$-adically integral, and $$v_p\left(\frac{p^n}{n!}\right) > n - \frac{n}{p-1},$$ which goes to $\infty$ with $n$ if $p > 2$. (You can prove convergence for $p = 2$ by a tighter inequality, but it won’t actually matter.) In fact, the only term with valuation 1 is when $n = 1$, so that $$e^p = 1 + p + \dotsb \in \mathbf{Z}_p.$$ An analysis of the multiplicative group of $\mathbf{Z}_p$ shows that this number has no $p$th roots, so $[\mathbf{Q}_p(e):\mathbf{Q}_p] = p$. Thus $[\mathbf{Q}(e):\mathbf{Q}] \ge p$, and letting $p$ go to infinity, we get a contradiction, proving that $e$ is transcendental.