Consider the fundamental unit $\varepsilon$ of a real quadratic number field $k = {\mathbb Q}(\sqrt{p})$ for primes $p \equiv 1 \bmod 4$, and let $h$ denote its class number. By Dirichlet's work on class number formulas, $\varepsilon^h$ is a norm of a cyclotomic unit in the maximal real subfield $K^+$ of the field $K = {\mathbb Q}(\zeta\_p)$ of $p$-th roots of unity. In particular, $\varepsilon$ is the norm of a cyclotomic unit if $h = 1$.


If $h > 1$ (in our case, $h$ is odd), we know that $h$ divides the class number $h^+$ of $K^+$, and that the cyclotomic units form a subgroup of index $h^+$ in the group of all units of $K^+$. Thus it might well be possible that $\varepsilon$ is the norm of a cyclotomic unit even in this case, and my question is: Is the fundamental unit of $k$ (for prime values of $p$) always the norm of a unit from $K^+$?



Are quadratic units cyclotomic norms?
The answer is no, and it fails in the very first example. Let $L/\mathbf{Q}$ be the degree six field inside $M = \mathbf{Q}(\zeta\_{229})^{+}$.
The unit group has rank five, and can be computed explicitly via pari. Both you and I will have no difficulty computing that the norm of each unit in $L$ to $K = \mathbf{Q}(\sqrt{229})$ is a power of $\varepsilon^3$, where $\varepsilon$ is the fundamental unit of $K$. Since the norm of any unit of $M = \mathbf{Q}(\zeta\_{229})^{+}$ to $K$ is certainly the norm of a unit from $L$, we are done.


On the other hand, if $h$ is prime to $p-1$, then both $\varepsilon^{h}$ and $\varepsilon^{(p-1)/2}$ are norms from $M$, so $\varepsilon$ is as well. This happens, for
 example, if $p$ is a Fermat prime. The next quadratic field of prime discriminant and non-zero class number is indeed $\mathbf{Q}(\sqrt{257})$, so in this case, $\varepsilon$ is a norm even though $h\_K = 3$. But this also happens for $p = 577$, $761$, $1093$, $1229$, etc.


