1
$\begingroup$

Let $R$ be a locally compact, separably metrizable ring (commutative with an identity) and let $M$ be a closed submodule of $R \oplus R$. Is the projection of $M$ onto the first coordinate closed?

$\endgroup$

1 Answer 1

3
$\begingroup$

Not necessarily.

Start from $A=\mathbf{R}[t]/(t^2)$ and $R$ its closed cocompact unital subring $\{a+bt:a\in\mathbf{Z},b\in\mathbf{R}\}$. Write $J=tA$.

Then every additive subgroup of $R^2$ contained in $J^2$ is an $R$-submodule. Hence, to get a counterexample, it is enough to pick a lattice in the plane $J^2$ whose projections are not closed.

$\endgroup$
1
  • $\begingroup$ Thank you for this simple, clear example! $\endgroup$
    – Nik Weaver
    Commented Sep 28, 2023 at 1:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .