Timeline for Hasse principle for rational times square
Current License: CC BY-SA 3.0
15 events
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| Jun 10, 2016 at 18:25 | history | edited | Mikhail Borovoi | CC BY-SA 3.0 |
III to Ш
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| Jun 5, 2016 at 16:17 | comment | added | user42024 | For me it looks more natural because then you can relate it to cohomological obstructions and use the machinery of algebraic geometry/algebraic number theory afterwards. Also from algebraico-geometric point of view the "base change" of $K^\times/\mathbb Q^\times$ to $\mathbb Q_p$ is the diagonal quotient by $\mathbb Q_p^\times$, not the sum of quotients of each summand | |
| Jun 5, 2016 at 16:08 | comment | added | user42024 | Yes, i guess it is weaker indeed. I was following the suggestion in the comment of GH from MO to choose one rational q for all places above fixed p, and also replace Q with Q_p. Sorry for not mentioning that | |
| Jun 5, 2016 at 13:02 | comment | added | David E Speyer | The phrase "where $k$ is square" is an artifact of bad editing. Cannot write it as $q k^2$ where $q \in \mathbb{Q}_5$ and $k \in K^{\times}_{2+i} \times K^{\times}_{2-i}$. | |
| Jun 5, 2016 at 12:30 | comment | added | David E Speyer | For example, let $K = \mathbb{Q}[i]$, and choose $a \equiv 1 \bmod 2+i$ and $\equiv 2 \bmod 2-i$. Then $a$ is of the form $q k^2$ in both $K_{2+i}$ and $K_{2-i}$, but it is not zero in $\left( (K_{2+i}^{\times}/(K_{2+i}^{\times})^2) \times (K_{2-i}^{\times}/(K_{2-i}^{\times})^2) \right)/\mathbb{Q}_5$. In other words, we can't write it as $qk^2$ where $k$ is square at both $2+i$ and $2-i$, since $1$ is a QR and $2$ is a non-QR modulo $5$. Probably the OP wanted to ask for the condition you answered about. | |
| Jun 5, 2016 at 12:25 | comment | added | David E Speyer | I didn't follow every detail, but I am worried that your translation of the original question into cohomological language may be flawed. You write "if you want to prove that $a \in K^{\times}/((K^{\times})^2 \mathbb{Q})$ is zero if and only if it is zero in $\left( \prod_{\mathfrak{p}|p} K_{\mathfrak{p}}^{\times}/(K_{\mathfrak{p}}^{\times})^2 \right)/ \mathbb{Q}_p$. But the OP asks for it to be zero in $\prod_{\mathfrak{p}|p} K_{\mathfrak{p}}^{\times}/\left( (K_{\mathfrak{p}}^{\times})^2 \mathbb{Q}_p \right)$ and this seems like a weaker condition to me. (continued) | |
| Jun 5, 2016 at 8:09 | comment | added | Mikhail Borovoi | Yes, it is well known that for any finite Galois module $M$ over a number field $k$, the Tate-Shafarevich kernel $ш^1(k,M)$ is finite. | |
| Jun 4, 2016 at 21:54 | comment | added | user42024 | I deleted the wrong part from the answer But I suppose it still should be true that $Sh(T[n])$ is finite | |
| Jun 4, 2016 at 21:53 | history | edited | user42024 | CC BY-SA 3.0 |
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| Jun 4, 2016 at 21:50 | comment | added | user42024 | Ah, I had a simple argument in my head, with writing down long exact sequences of cohomology groups for $0\rightarrow T[n] \rightarrow T \rightarrow T\rightarrow 0$ for absolute Galois groups of $\mathbb Q$ and sum over $\mathbb Q_p$ correspondingly and looking on the natural map from one to another, but it appears to be wrong | |
| Jun 4, 2016 at 21:19 | comment | added | Mikhail Borovoi | Could you please explain, why $Ш^1(T)=0$ implies $Ш^1(T[n])=0$? | |
| Jun 4, 2016 at 21:10 | comment | added | Mikhail Borovoi | It is known that in general $Ш^1(T[4])\neq 0$, see Section 2 of Sansuc's paper, however it is not immediately clear how to construct a counter-example for $T[2]$. | |
| Jun 4, 2016 at 19:55 | comment | added | user42024 | One additional comment: Shafarevich-Tate group of a torus is known to be finite, so if you replace degree 2 by some prime p in your question, this kind of argument shows that for a fixed field K the statement is true for p big enough (bigger than the order of Sh(T)) | |
| Jun 4, 2016 at 19:41 | history | edited | user42024 | CC BY-SA 3.0 |
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| Jun 4, 2016 at 19:31 | history | answered | user42024 | CC BY-SA 3.0 |