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Proposition 5.38 of the book Cyclotomic Fields by Lawrence Washington is as follows:

Let $m\geq 1$ be square free and assume 3 does not split completely in $\mathbb{Q}(\sqrt{-m})$. If 3 divides the class number of $\mathbb{Q}(\sqrt{3m})$ then 3 divides the class number of $\mathbb{Q}(\sqrt{-m})$.

The proof uses the 3-adic class number formula and the congruence of integer values of 3-adic L-functions to get $$\big(1-\frac{\chi\omega(3)}{3}\big)\frac{2h\log_3{\epsilon}}{\sqrt{D}}\equiv-(1-\chi(3))B_{1,\chi}\mod 3.$$ Here $\chi$ is the character for $\mathbb{Q}(\sqrt{-m})$, $\chi\omega$ is the character for $\mathbb{Q}(\sqrt{3m})$, $\epsilon,h,D$ are the fundamental unit, class number, and discriminant for $\mathbb{Q}(\sqrt{3m})$.

Then he checks that the left hand side is $h$ times something integral and therefore that if $3|h$, $3|-B_{1,\chi}=h(\mathbb{Q}(\sqrt{-m}))$. This last step uses that 3 isn't split so that $1-\chi(3)\neq 0$.

Working through the proof, I don't see where any special properties of 3 are used besides it being prime. Am I missing something, or is it true for any prime $p$ in place of 3 with the exact same proof?

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You're missing the fact that the cubic unramified extension becomes a Kummer extension over the field you get by adjoining $\sqrt{-3}$, which fails to hold for primes $p > 3$.

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  • $\begingroup$ Thank you very much for this, it (and your answer here mathoverflow.net/a/75027) have helped me understand why 3 is needed for this theorem. The proof you give is different from Washington's, he uses $p$-adic $L$-functions. I edited my post to sketch his proof. Do you know if your proof can be rephrased in this language? Or where his proof might break down for $p\neq 3$? $\endgroup$ – Peter Dec 28 '17 at 3:02
  • $\begingroup$ I think I see how this affects Washington's proof now. The class number of $\mathbb{Q}(\sqrt{-m})$ is $-B_{1,\chi}$ where $\chi$ is the Dirichlet character corresponding to $\mathbb{Q}(\sqrt{-m})$. We can understand $B_{1,\chi}\mod 3$ via the 3-adic $L$-function corresponding to $\chi\omega$, and because $\mathbb{Q}(\zeta_3)$ is quadratic the character $\chi\omega$ corresponds to the quadratic field $\mathbb{Q}(\sqrt{3m})$ to which we can then apply the $3$-adic class number formula. For any larger prime, $\chi\omega$ will come from some more complicated field, so things will be messier. $\endgroup$ – Peter Dec 29 '17 at 23:59

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