Let $L/K$ be a finite Galois extension of number fields that is ramified exactly at one finite prime and is unramified at all infinite primes. Let $U_K$ and $U_L$ denote the units of the ring of integers of $K$ and $L$, respectively. Some examples show that the norm map on units $Norm_{L^*/K^*}: U_L \rightarrow U_K$ is surjective, or equivalently zeroth Tate cohomology of the group $G$ with coefficients in $U_L$ is $0$. Does this surjectivity of the norm map remain true in general for such extensions?
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A theorem due to Arnold Scholz says that if $K$ has odd class number and $L/K$ is a quadratic extension with a single ramified prime, then the norm map on units is onto.
In general, this does not hold. Take $K = {\mathbb Q}(\sqrt{-5})$ and $L = K(\sqrt{11})$. Then $L/K$ is unramified away from $11$, the unit group of $L$ is $\langle -1, 10 + 3\sqrt{11}\rangle$, and $-1$ is a norm of an element in $L^\times$ by Hasse's norm theorem, but not the norm of a unit.
answered Feb 26 at 11:43
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$\begingroup$ Thanks a lot. What about for the special case $L=\mathbb{Q}(\zeta_{p^n})$ and $K=\mathbb{Q}(\zeta_p)$? I know that in this case the norm map is surjective for regular primes. I expect that it be true also for irregular primes, but I do not know how to prove it. In fact, in this example we have $\hat{H^0}=ker(\epsilon)$ where $ker(\epsilon)$ is the capitulation kernel. Now, regularity of $p$ easily implis that capitulation kernel is $0$. But this method does not work for irregular primes. Note that in this example (which is my goal), surjectivity is equivalent to zeroness of capitulation kernel $\endgroup$ Feb 26 at 15:08
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$\begingroup$ The statement for regular primes is the exact analog of Scholz's theorem in the quadratic case. I'd start looking for a counterexample rather than for a proof. $\endgroup$ Feb 27 at 14:42
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$\begingroup$ By a theorem of Iwasawa about cohomology of units of $\mathbb{Z}_p$ -extensions and using direct limit of BRZ exact sequence in my paper on \textit{Ostrowski Quotients for ...}, I am sure that the direct limit over all positive integers $i$ of $ker(\epsilon_i)$ is a finite group for $K=\mathbb{Q}(\zeta_p)$ and $L_i=\mathbb{Q}(\zeta_{p^i})$ (for both regular and irregular primes). Also, I knew the zeroness for regular primes, hence I guessed for irregular primes each $ker(\epsilon_i)$ is zero. $\endgroup$ Feb 28 at 16:45