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Are there any definitions for (ir)regular primes which do not use class number divisibility or Bernoulli numbers? For reference, Wikipedia gives both the first definition (their primary one) and the second (i.e., Kummer’s criterion, using Bernoulli numbers).

I'm looking for a third (or more) fundamentally different characterization — bonus points if it can be stated and proven in an elementary (if not necessarily simple) way. Here's a conjectured example: A prime $p$ is irregular if and only if there exists some even integer $2 \le n \le p-3$ such that $$ p^2 \mid (1^n + 2^n + \dotsb + (p-1)^n + p^n). $$

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    $\begingroup$ Are you looking for something other than "$p$ is regular if $p$ does not divide the class number of $\mathbb{Q}(\zeta_p)$, where $\zeta_p$ is a $p$-th root of unity"? $\endgroup$
    – David E Speyer
    Commented Mar 30, 2012 at 20:59
  • $\begingroup$ Hi David, Yes, I'm looking for something else -- I’m sorry I wasn’t more clear in my original posting. Wikipedia <en.wikipedia.org/wiki/Regular_prime> gives both your definition (their primary definition) and Kummer’s criterion (with Bernoulli numbers). I was wondering if anyone since Kummer has found a third (or more) equivalent criterion or definition. Thanks! Kieren. $\endgroup$
    – Kieren MacMillan
    Commented Mar 31, 2012 at 0:20
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    $\begingroup$ Are you looking for a connection with some other circle of ideas? The fact that the $p$-part of the class group of the $p$th cyclotomic field is related to values of the Riemann zeta function is already pretty cool. Also, I would define $p$ to be regular as David Speyer, has, and everything else would be a "theorem" not a "definition". $\endgroup$
    – Rob Harron
    Commented Mar 31, 2012 at 0:52
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    $\begingroup$ We can just keep translating into different language. How about "The reduced $K_0$ of the ring of integers of $\mathbb{Q}(\zeta_p)$ is $p$-divisible"? $\endgroup$
    – S. Carnahan ♦
    Commented Mar 31, 2012 at 3:02
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    $\begingroup$ It seems to me that, using the summation formula for consecutive powers (Faulhaber's formula), your "conjectured example" is equivalent to the definition of the divisibility of Bernoulli numbers. I would not call this a new definition. $\endgroup$
    – Chris Wuthrich
    Commented Sep 9, 2014 at 13:27

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David Frank, Irregular primes and integrality theorems for manifolds, Comment. Math. Helv. 50 (1975), no. 3, 321–326, MR0407856 (53 #11626): from the review, by K H Mayer, we have,

For any $p$ the following are equivalent: (i) $p$ is an irregular prime and (ii) for some $k>1$ there exists a manifold $M$ of dimension $4k$ satisfying the integrality conditions with order $\Sigma_{M^b}$ divisible by $p$.

I'm afraid you'll have to read the review to see what "the integrality conditions" are.

Also of possible interest is Tauno Metsänkylä, Maillet's matrix and irregular primes, Studies in honour of Arto Kustaa Salomaa on the occasion of his fiftieth birthday, Ann. Univ. Turku. Ser. A I No. 186 (1984), 72–79, MR0748520 (85j:11145). Here the rank of a certain matrix is given in terms of the irregularity index of a prime.

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