Bernoulli Number Congruence

Question

In a paper by L. Carlitz entitled A Property of the Bernoulli Numbers, it is written

The Bernoulli numbers may be defined by the symbolic relation $$(B+1)^n-B^n=0 $$ with $n>1$ and $B_0=1$, where after expansion the exponents are replaced by subscripts. By the Staudt-Clausen theorem, the denominator of $B_{2n}$ contains the prime $2$ to the first power. It is proved (in another article) that $$2B_{2n}\equiv 4n+1 \pmod{8}$$ for $(n\ge 2)$

Perhaps I am missing something, but I'm confused with the congruence. The bernoulli numbers are fractions, so how can these fractions have congruences to natural numbers? Or am I assuming that the $B_{2n}$ is the denominator?

Answer 1

As the modulus is $2^3$, for any number $n$ relatively prime to $2$, one can interpret $a/n \bmod 8$ as $a \cdot \overline{n} \bmod 8$, where $\overline{n}$ is the modular inverse of $n \bmod 8$.

The claimed congruence holds in this sense.

Answer 2

If $x \in \mathbb Q$ has denominator prime to $p$, then it makes sense to speak of the residue class of $x$ mod $p$. Simply write $x=a/b$ with $b$ prime to $p$, and put $\overline{x} = \overline{a}/\overline{b}$ where $\overline{b}$ is the inverse of $b$ mood $p$.

To be pedantic, one might say that the residue class map $\mathbb Z \to \mathbb F_p$ extends to the local ring of $\mathbb Z$ at $p$, i.e. it extends to a ring homomorphism

$$\mathbb Z_{(p)} \to \mathbb F_p$$

where $\mathbb Z_{(p)}$ is the subring of $\mathbb Q$ in which we have inverted every prime but $p$.

Now, since $2B_{2n}$ has odd denominator, the congruence which you have written actually makes sense. (You're working mod $8$ rather than mod $2$ but the principle is the same.)