Timeline for Find $\mathbb{Z}$-basis of module over Dedekind domain provided its pseudobasis
Current License: CC BY-SA 4.0
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Sep 15, 2022 at 14:47 | history | edited | LSpice | CC BY-SA 4.0 |
Deleting spurious space in title, while this is on the front page
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Sep 15, 2022 at 9:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Aug 16, 2022 at 8:38 | answer | added | Александр Каренин | timeline score: 0 | |
Aug 9, 2022 at 19:24 | history | edited | Александр Каренин | CC BY-SA 4.0 |
edited title
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Aug 9, 2022 at 18:39 | comment | added | Александр Каренин | If $K = \mathbb{Q}[\sqrt{-5}]$ and $\mathfrak{a}=(2,1+\sqrt{-5})$ then the $\mathfrak{a}$ as a module is isomorphic to $2\cdot\mathbb{Z} \oplus (1+\sqrt{-5})\cdot\mathbb{Z} \cong \mathbb{Z}^2$ and the elements in its integral basis are $2$ which is a vector $(2,0)^T$ and $1+\sqrt{-5}$ which is $(1,1)^T$. So basis of $\mathfrak{a}$ is $\begin{pmatrix} 2 & 1\\ 0 & 1 \end{pmatrix}$. | |
Aug 9, 2022 at 18:39 | comment | added | Александр Каренин | Yes! The $\mathbb{Z}$ is typo. I meant that $\mathfrak{a}$ is isomorphic to $\bigoplus_{i} a_i \cdot \mathbb{Z}$ because it is $\mathcal{O}_K$-module (for $\mathcal{O}_K$ - ring of integers of $K$) and therefore a $\mathbb{Z}$ module itself. I'll send an example in the next comment. | |
Aug 9, 2022 at 17:29 | comment | added | LSpice | $\mathbb{ZZ}$ in your title should be just $\mathbb Z$, right? Also, I am confused by your writing a fractional ideal as $\bigoplus_i a_i\cdot\mathbb Z$, where $a_i$ are integers … if you mean $a_i \in \mathbb Z$, then this can only give ideals of $\mathbb Z$; but, if you mean $a_i \in \mathcal O_K$, then it can only give ideals of $\mathcal O_K$, not fractional ideals, right? | |
S Aug 9, 2022 at 15:51 | review | First questions | |||
Aug 9, 2022 at 16:46 | |||||
S Aug 9, 2022 at 15:51 | history | asked | Александр Каренин | CC BY-SA 4.0 |