Timeline for Cyclic cubic extensions and Kummer theory
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2018 at 4:10 | comment | added | Pound Sterling | Given that (as agreed in the comments) the only possible interpretation of the notation $M^{G = \chi}$ is to mean $\{m \ | \ m \in M, gm = \chi(m)g\}$, then not only is $H^1(\mathbf{Q},\mathbf{Z}/3\mathbf{Z}) = (k^{\times}/k^{\times 3})^{G = \chi}$ correct, it is absolutely the entire point, and if you don't understand that then you don't understand the answer. | |
| Oct 7, 2018 at 19:55 | comment | added | Daniel Loughran | BTW: I think the stated equation $H^1(\mathbb{Q},\mathbb{Z}/3 \mathbb{Z}) = (k^{\times}/k^{\times 3})^{G = {\chi}}$ is not correct. Either the first term should be $H^1(k,\mathbb{Z}/3 \mathbb{Z})$, or in the second term one should take $G$-invariants. This, compounded with the non-standard notation, added a lot to my initial confusion. | |
| Sep 21, 2018 at 5:25 | comment | added | Felipe Voloch | @DanielLoughran Going from the name, she probably teaches at Hogwarts. And Pound Sterling must be the reincarnation of quid. | |
| Sep 20, 2018 at 21:31 | comment | added | Pound Sterling | Let $\sigma \alpha = \beta$. If $k(\alpha^{1/3})$ is Galois over $\mathbf{Q}$, then $k(\alpha^{1/3}) = k(\beta^{1/3})$ thus $\beta = \alpha$ or $\alpha^{-1} \mod k^{*3}$. So $\beta = \alpha x^3$ or $\alpha^{-1} x^3$ with $x \in k^*$. In case 1 the order $2$ element (sending $\alpha$ to $\beta$ and $\zeta_3$ to $\zeta^2_3$) does not commute with the order $3$ element (sending $\alpha^{1/3}$ to $\zeta_3 \alpha^{1/3}$ and fixing $\zeta_3$), and in case 2 it does. So abelian extensions are given by $k(\alpha^{1/3})$ with $\alpha \in k^{*}/k^{*3}$ where $\sigma \alpha = \alpha^{-1}$ in this group. | |
| Sep 20, 2018 at 20:28 | comment | added | Daniel Loughran | @Margerie Mumblecrust: Thanks again for the answer. BTW, I would love to know which University you work at, which teaches Galois cohomology and Kummer theory in an introductory undergraduate Galois theory course! | |
| Sep 20, 2018 at 20:24 | vote | accept | Daniel Loughran | ||
| Sep 20, 2018 at 17:20 | comment | added | Pound Sterling | Your comment is ambiguous, but assuming you mean: The action of the non-trivial $\sigma \in G$ on $k^{\times}/k^{\times 3}$ <i>induced from the isomorphism</i> $H^1(k,\mathbf{Z}/3 \mathbf{Z}) \simeq k^{\times}/k^{\times 3}$ is via the map $\alpha \mapsto \chi(\sigma)(\alpha^{\sigma}) = (\alpha^{\sigma})^{-1} \mod k^{\times 3}$, then yes, that is correct. Also, I wouldn't use this notation, but from the context $M^{G = \chi}$ means the elements of $M$ where "$G$ acts by $\chi$" which is also just $M^{\chi}$. | |
| Sep 20, 2018 at 11:06 | comment | added | Daniel Loughran | I think I see where my confusion lies: the point is that the action of $G$ on $k^*/k^{*3}$ is not the one induced from the action of $G$ on $k^*$, but rather $G$ acts on $k^*/k^{*3}$ via $(\sigma,\alpha) \mapsto 1/\alpha^{\sigma}$ (where $\alpha^{\sigma}$ denotes the usual action of $G$ on $k^*$). Is this correct? | |
| Sep 19, 2018 at 22:16 | comment | added | Pound Sterling | But this isomorphism doesn't respect the $G$-action --- $G$ acts on $\mathbf{Z}/3 \mathbf{Z}$ by the trivial character, and on $\mu_3$ by the character $\chi$, which is also equal to $\chi^{-1}$. Hence $H^1(k,\mathbf{Z}/3 \mathbf{Z})^{G} = H^1(k,\mu_3)^{\chi}$. | |
| Sep 19, 2018 at 22:16 | comment | added | Pound Sterling | Inflation-Restriction on cohomology (look it up) gives an isomorphism $H^1(\mathbf{Q},\mathbf{Z}/3\mathbf{Z}) = H^1(k,\mathbf{Z}/3 \mathbf{Z})^G$ where $G = \mathrm{Gal}(k/\mathbf{Q})$. As $\mathrm{Gal}(\overline{\mathbf{Q}}/k)$-modules, $\mathbf{Z}/3 \mathbf{Z} = \mu_3$, so $H^1(k,\mathbf{Z}/3 \mathbf{Z}) = H^1(k,\mu_3)$. | |
| Sep 19, 2018 at 21:17 | comment | added | Daniel Loughran | Whilst I appreciate the answer, I don't think I can accept it until the notation is clarified and also it is clarified why $\alpha^{-1}$ appears, rather than $\alpha$. | |
| Sep 18, 2018 at 18:07 | comment | added | Daniel Loughran | Thanks for the extra details. You have defined what $M^{\chi}$ is, but I still don't understand what the notation $M^{G = \chi}$ means. Normally $M^G$ denotes the $G$-invariants for a module with $G$ action, but it seems that you mean something else. | |
| Sep 18, 2018 at 17:25 | history | edited | Margerie Mumblecrust | CC BY-SA 4.0 |
added 911 characters in body
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| Sep 18, 2018 at 16:06 | comment | added | Daniel Loughran | I was also hoping for an explicit polynomial of degree $3$ which defines the cyclic cubic extension of $\mathbb{Q}$. Does your method give this? | |
| Sep 18, 2018 at 16:06 | comment | added | Daniel Loughran | Thanks for the answer, which I'm currently trying to digest. Can you please clarify what you mean by the notation $(k^*/k^{*3})^{G = \chi}$? Do you mean the set of elements which are invariant under the action of the Galois group $G$? If so, why is this those elements $\alpha$ with $\sigma \alpha \equiv \alpha^{-1} \bmod k^{*3}$, and not $\sigma \alpha \equiv \alpha \bmod k^{*3}$? (I assume that $\sigma$ denotes the non-trivial element of $G$?). | |
| Sep 17, 2018 at 22:57 | history | answered | Margerie Mumblecrust | CC BY-SA 4.0 |