Let $\langle x\rangle$ be an infinite cyclic group and $V$ the additive group of an infinite vector space over $\mathbb{Z}_p$.

Is it possible to make $\langle x\rangle$ act irreducibly on $V$?

If we have $V=\dots\times\langle x_{-1}\rangle\times\langle x_0\rangle\times\langle x_1\rangle\times\dots$, I was thinking that the action by left (or right) translation could be the one, but I cannot prove it.

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    $\begingroup$ This won't work, as the set $W = \{ \sum a_{i} x_{i} : \sum a_{i} = 0 \}$ is a proper, non-trivial subspace, which is left invariant by $\langle x \rangle$. $\endgroup$ – Andreas Caranti Mar 5 '15 at 11:54
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    $\begingroup$ And $W$ is not irreducible, because the submodule generated by $x_0-x_2$ does not contain $x_0-x_1$. $\endgroup$ – Derek Holt Mar 5 '15 at 11:58

$\newcommand{\Z}[0]{\mathbb{Z}}$I am sure some basic theoretical fact easily yields that the answer is no, but here's an elementary argument.

Suppose you have such an action. Let $0 \ne v \in V$. Then the $v x^{i}$ for a basis of $V$, for $i \in \Z$.

For, suppose there is a non-zero (finite) linear combination $$ \sum_{i} a_{i} x^{i} v = 0. $$ Applying a suitable $x^{k}$, we may assume this is $$ \sum_{i=0}^{n} a_{i} x^{i} v = 0,\tag{key} $$ with $a_{0} \ne 0 \ne a_{n}$. This means that $$ U = \langle v, x v, \dots, x^{n-1} v \rangle $$ is an invariant subspace: just check using (key) that $$ x (x^{n-1} v) = x^{n} v = - a_{n}^{-1} (\sum_{i=0}^{n-1} a_{i} x^{i} v) \in U, \quad \text{and} \quad x^{-1} v = -a_{0}^{-1} (\sum_{i=0}^{n-1} a_{i+1} x^{i} v) \in U. $$

Now that the $v_{i} = v x^{i}$ are a basis, just note that $$ W = \{ \sum_{i} a_{i} v_{i} : \sum_{i} a_{i} = 0 \} $$ is a non-trivial, proper subspace.

  • $\begingroup$ Thanks! (also for the comments) $\endgroup$ – W4cc0 Mar 5 '15 at 14:04
  • $\begingroup$ @W4cc0, you're welcome! $\endgroup$ – Andreas Caranti Mar 5 '15 at 14:05

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