Are quadratic units cyclotomic norms? 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 $K^+$?
 A: Edit:Let $K=\mathbb Q(\zeta_p)^+$
Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a fact about  Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether  $|\mathrm{H}^1(K/k,\mathcal O_K^\times)|=n$, where $n=[K:k]$.
Apply the exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. We have the following exact sequence
$$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$
where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. 
Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$.  If $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling said. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$.  $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.
References:[1] Topics in Iwasawa theory. Greenberg.
[2] Visibility of ideal classes. Schoof and Washington.
A: 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.
