Timeline for Under What assumptions on $p$, $\mathcal{O}_K^* \simeq \mathbb{Z}_p^{*} \oplus \mathbb{Z}_p^{*}$
Current License: CC BY-SA 4.0
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Sep 20, 2018 at 0:40 | comment | added | efs | Please see Corollary 4.3.14 in Cohen's book, and the link that @KConrad posted below your question. The free part of the unit group of the ring of integers of $K$ is always isomorphic to the product of $n$ copies of the $p$-adic integers, where $n$ is the degree of the extension. | |
Sep 19, 2018 at 15:56 | comment | added | user89236 | If we take the torsion free part of $\mathcal{O}_K^*$, which will be $1+\pi\mathcal{O}_K$ where $\pi$ is the uniformizer of $\mathcal{O}_K$ then under some assumptions on $p$, Is it possible to prove an isomorphism between the $1+\pi\mathcal{O}_K$ and $(1+p\mathbb{Z}_p)\oplus(1+p\mathbb{Z}_p)$. | |
Sep 18, 2018 at 22:53 | comment | added | efs | Well that's what math tells... | |
Sep 18, 2018 at 14:20 | comment | added | user89236 | @ EFinat you mean to say that $\mathcal{O}_K^*$ can never be isomorphic to $\mathbb{Z}_p^*\oplus \mathbb{Z}_p^*$ | |
Sep 18, 2018 at 9:57 | comment | added | efs | $\mathcal{O}_K^*$ is never torsion free, it contains $-1$. | |
Sep 18, 2018 at 7:12 | comment | added | user89236 | If we assume that $\mathcal{O}_K^*$ is torsion free then is it possible $\mathcal{O}_K^*\simeq \mathbb{Z}_p^*\oplus \mathbb{Z}_p^*$. Because I want to know under what assumptions $\mathcal{O}_K^*$ is isomorphic to $\mathbb{Z}_p^*\oplus \mathbb{Z}_p^*$ | |
Sep 17, 2018 at 21:19 | comment | added | KConrad | Actually, even the case $p=2$ is problematic. The torsion part of $\mathbf Z_2^\times$ is $\{\pm 1\} \cong \mathbf Z/2\mathbf Z$ so the torsion subgroup of $\mathbf Z_2^\times \oplus \mathbf Z_2^\times$ is the noncyclic group $\{\pm 1\}^2$ and thus can't be isomorphic to $\mathcal O_K^\times$ in the 2-adic case. | |
Sep 17, 2018 at 20:46 | comment | added | efs | @KConrad Yes, in the comment that I wrote and deleted, I wrote that condition, and I forgot to put it here. Now it remains the case $p=2$. I'm busy now. I'll check in a little, or please answer the question and I delete this answer. | |
Sep 17, 2018 at 20:44 | history | edited | efs | CC BY-SA 4.0 |
added 57 characters in body
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Sep 17, 2018 at 20:18 | comment | added | KConrad | That group is cyclic if $p = 2\ldots$ | |
Sep 17, 2018 at 17:59 | history | answered | efs | CC BY-SA 4.0 |