so they do not split. Moreover, $p=(q-1)/2$ is prime and $[L:{\mathbb Q}]=p$. Thus, the hypotheses of Theorem \[thm-garbanati\] are satisfied for $K$ and $L$.

If $2$ is inert in ${\mathbb Q}(\zeta_p)^+/{\mathbb Q}$, then Theorem \[thm-estes\] shows that $h_K$ is odd, and therefore $h_K^-$ is odd as well, since $h_K^- = h_K/h_K^+$ by definition. Since the discriminant of $K$ is a power of $q$, Theorem \[thm-garbanati\] part (2) shows that $[C_L^+:C_L^2]=1$ and therefore $[{\mathcal{O}}_L^{\times,+}:({\mathcal{O}}_L^\times)^2]=1$ as well by part (1). We conclude that $\rho_\infty=0$.

There is in fact a conjecture of Davis and Taussky that says that $\rho_\infty=0$ in the case of $L={\mathbb Q}(\zeta_q+\zeta_q^{-1})$, where $p=(q-1)/2$ is a Sophie Germain prime. For more on the Davis–Taussky conjecture see [@davis], [@edgar], [@estes], [@kim], and [@stevenhagen].

\[conj-taussky\] Let $q$ and $p$ be primes such that $q=2p+1$, and let $L={\mathbb Q}(\zeta_q+\zeta_q^{-1})$, where $\zeta_q$ is a primitive $q$-th root of unity. Then, $C^+_L=C_L^2$. (Thus, it follows that $\rho_\infty = \dim_{{\mathbb F}_2} {\mathcal{O}}_L^{\times,+}/({\mathcal{O}}_L^\times)^2 = 0.$)

In the next result we note that the Davis–Taussky conjecture is equivalent to the class number of ${\mathbb Q}(\zeta_q)$ being odd. (We thank David Dummit for pointing out the following equivalence to us.)

\[thm-davistausskyclassnumberodd\] Let $q$ and $p$ be primes such that 