Let $V$ be a countably infinite dimensional $K$-vector space over a local field $K$ (nontrivially discretely valued with finite residue field). Let $o$ be the ring of integers of $K$.

Given two free o-submodules $L_1,L_2$ in $V$, does $L_1+L_2 \subsetneqq V$ hold? (Here L_1+L_2 denotes the smallest $o$-module inside $V$ containing $L_1$ and $L_2$)

  • $\begingroup$ 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? $\endgroup$ – Alex B. Nov 3 '10 at 12:15
  • $\begingroup$ Sure, thanks for pointing out and sorry for the typo. $\endgroup$ – Tiffy Nov 3 '10 at 12:42
  • 2
    $\begingroup$ 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? $\endgroup$ – Kevin Buzzard Nov 3 '10 at 13:59

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