Timeline for Surjectivity of the norm of units in Galois extensions ramified exactly at one finite prime
Current License: CC BY-SA 4.0
6 events
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| Jul 11 at 20:40 | comment | added | Ehsan Shahoseini | Also, what we discussed about irregular primes is also true. This is because if in a finite extension of number fields we have a totally ramified prime, then capitulation map is injective. Also we need to use the BRZ exact sequence of my paper "ostrowski quotient for...". It is good to have a look at example 3.1 in my mentioned paper. | |
| Jul 11 at 20:35 | comment | added | Ehsan Shahoseini | I can prove Arnold Scholz's Theorem that you mentioned for any cyclic extension with one totally ramified prime and with no restriction on class number of base field $K$. So I need to make a citation to Scholz's work. Please let me know the reference. | |
| Feb 28, 2023 at 16:45 | comment | added | Ehsan Shahoseini | 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. | |
| Feb 27, 2023 at 14:42 | comment | added | Franz Lemmermeyer | 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. | |
| Feb 26, 2023 at 15:08 | comment | added | Ehsan Shahoseini | 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 | |
| Feb 26, 2023 at 11:43 | history | answered | Franz Lemmermeyer | CC BY-SA 4.0 |