Timeline for Hasse principle for rational times square
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 6, 2016 at 15:13 | comment | added | Mikhail Borovoi | @nfdc23: Yes, the quotient splits off because it is isomorphic to the kernel of the norm! | |
| Jun 6, 2016 at 14:40 | comment | added | nfdc23 | @MikhailBorovoi: I meant to write the Tate-Shafarevich group over $\mathbf{Q}$, not $K$; a typo. The reason for the vanishing with odd degree (no Galois hypothesis needed!) is that in such cases the quotient map ${\rm{R}}_{K/\mathbf{Q}}(\mu_2) \rightarrow T$ modulo $\mu$ splits off as a direct summand (as a $\mathbf{Q}$-group), so the Tate-Shafarevich group for $T[2]$ over $\mathbf{Q}$ is thereby a direct summand of that of ${\rm{R}}_{K/\mathbf{Q}}(\mu_2)$ over $\mathbf{Q}$, or equivalently of $\mu_2$ over $K$. That in turn is trivial by Grunwald--Wang (as for $\mu_n$ whenever $8\nmid n$). | |
| Jun 6, 2016 at 1:32 | comment | added | David E Speyer | An earlier version of this post concluded by asking about a group theoretic lemma; I have now found a counter-example to the requested result. I'll post more about it if I get the chance later. | |
| Jun 6, 2016 at 1:32 | history | edited | David E Speyer | CC BY-SA 3.0 |
deleted 734 characters in body
|
| Jun 5, 2016 at 21:04 | comment | added | Mikhail Borovoi | @user42024: Right, this is clear from the answer by David Speyer. | |
| Jun 5, 2016 at 20:17 | comment | added | user42024 | I suppose nfdc23 assumed $K$ to be Galois as he considers $Ш^1(K,T)$ | |
| Jun 5, 2016 at 19:10 | comment | added | Mikhail Borovoi | @nfdc23: It seems that you claim that $Ш^1(\mathbb{Q},T[2])=0$ when $[K:\mathbb{Q}]$ is odd. If this is what you mean, could you sketch a proof? I can prove this assertion assuming that the degree of a normal closure $N$ of $K$ over $\mathbb{Q}$ is odd. | |
| Jun 5, 2016 at 16:41 | comment | added | nfdc23 | The vanishing of a specific Tate-Shafarevich group as in user42024's answer is equivalent to the sufficiency of the finer necessary conditions in the comment of "GH from MO" (due to the vanishing of $Ш^1(K,T)$ that follows from Hilbert 90 and the local-to-global sequence for Brauer groups). That vanishing always (!) holds when $[K:\mathbf{Q}]$ is odd, so your "No Hasse principle" example shows that the necessary conditions as stated by "GH from MO" (and used by user42024) are genuinely stronger than the OP's necessary conditions; the stronger ones should be used. | |
| Jun 5, 2016 at 12:57 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 734 characters in body
|
| Jun 5, 2016 at 12:32 | comment | added | David E Speyer | I now think that my "No Hasse principle" example, although it literally responds to what the original question asked, does not respond to its intent. See my comments on user42024's question for what I suspect was intended. | |
| Jun 5, 2016 at 1:28 | history | edited | David E Speyer | CC BY-SA 3.0 |
deleted 4 characters in body
|
| Jun 5, 2016 at 0:15 | history | answered | David E Speyer | CC BY-SA 3.0 |