Timeline for $\mathbb{Z}$-module structure of the subring generated by an algebraic number
Current License: CC BY-SA 3.0
10 events
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| Apr 25, 2017 at 18:23 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Apr 24, 2017 at 23:24 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Apr 24, 2017 at 22:15 | history | edited | Qfwfq | CC BY-SA 3.0 |
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| Apr 24, 2017 at 22:08 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Apr 24, 2017 at 21:48 | comment | added | Will Sawin | @oxeimon: Sorry, I wrote this very quickly as I had to leave. I will edit it shortly. I do want $\mathcal O_L'$ to be as you say, but the condition that it is contained in $L$ should be implied by the condition on the tensor product. Yes, I need $L$ proper. Still, the condition of the second paragraph always obtains when $a$ is integral as then one can take $L=\mathbb Q$, $\mathcal O_L'=\mathbb Z$. The problem is trivial when $a$ is integral. I should say locally free rather than free, and then it is true about rings of integers. | |
| Apr 24, 2017 at 21:22 | comment | added | Will Chen | Okay so you want $\mathcal{O}_L'$ to be a subring of $L$ containing $\mathcal{O}_L$? I'm sorry, I still don't understand the purpose of the condition you stipulate in your second paragraph. Surely you need more conditions, like $L$ a proper subfield of $K$? (or else, if $a\in\mathcal{O}_K$, then you could take $L = K$, $\mathcal{O}_L' = \mathcal{O}_L$). Or, are you assuming that $a$ is not integral over $\mathbb{Z}$? Also, in your second paragraph how do you deduce that $\mathcal{O}_K[a]$ is free over $\mathcal{O}_L'$? (Are rings of integers always free over smaller rings of integers?) | |
| Apr 24, 2017 at 20:55 | comment | added | Will Sawin | @oxeimon $\mathbb Z[\alpha] \times _{\mathbb Z} \mathcal O_K$ is not isomorphic to $\mathbb Z[\alpha]$. The condition forces $\mathcal O_L' \otimes_{\mathcal O_L} L = L$. | |
| Apr 24, 2017 at 20:54 | comment | added | Will Sawin | @oxeimon The key is that it is isomorphic to a specific copy of $K$, embedded by the obvious scalar multiplication action. The only elements of $K$ acting by scalar multiplication on $\mathbb Z[\alpha]$ that preserve $\mathbb Z[\alpha]$ are themselves in $\mathbb Z[\alpha]$ (look at wehre they send $1$.) | |
| Apr 24, 2017 at 20:51 | comment | added | Will Chen | Forgive my denseness, but why is it sufficient to show that the $\mathbb{Z}$-module endomorphisms of $\mathbb{Z}[a]$, when tensored with $\mathbb{Q}$, is isomorphic to $K$? Is this linked to the condition in your second paragraph? I also don't really understand this condition. Is $L$ a subfield of $K$? Can't you always set $L = \mathbb{Q}$ and $\mathcal{O}_L' = \mathbb{Z}[a]$? | |
| Apr 24, 2017 at 20:21 | history | answered | Will Sawin | CC BY-SA 3.0 |