Timeline for Sum of two free o-submodules in a vector space over a local field
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Nov 3, 2010 at 13:59 | comment | added | Kevin Buzzard | How about this: take $V$ with basis $B$ and let $L_1$ be the free $o$-module spanned by $B$. I'll now try to build free $L_2$ with $L_1+L_2=V$. I'll do it recursively. The point is that $V/L_1$ is a countable set, so just enumerate it $c_1,c_2,c_3,\ldots$, and then at step~$n$ choose $v_n\in V$ lifting $c_n$ such that $v_n$ is linearly independent from all the $v_i$, $i<n$. This is clearly possible because the choices for the lifting of $c_n$ span an infinite-dimensional space. Now let $L_2$ denote the lattice generated by the $v_i$. Does that work to give a counterexample? | |
| Nov 3, 2010 at 12:42 | comment | added | Tiffy | Sure, thanks for pointing out and sorry for the typo. | |
| Nov 3, 2010 at 12:41 | history | edited | Tiffy | CC BY-SA 2.5 |
deleted 10 characters in body; deleted 60 characters in body
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| Nov 3, 2010 at 12:15 | comment | added | Alex B. | Isn't this clearly false? If $L_1$ is a lattice in the subspace with only first entry non-zero and $L_2$ is a full rank lattice in the subspace with 0 in the first entry, then $K\cdot(L_1 + L_2) = V$. Or am I missing something obvious? | |
| Nov 3, 2010 at 12:02 | history | asked | Tiffy | CC BY-SA 2.5 |