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This is perhaps a trivial question, but I've asked a few colleagues and they couldn't answer. For a given abelian group $M$, is it possible to have several different actions of the ring of $p$-adic numbers $\mathbb{Z}_p$? Or is it necessarily unique when it exists?

If the action were continuous of some sort then it would have to be unique: integers are dense in $\mathbb{Z}_p$, and for $n \in \mathbb{Z}$ one has $n \cdot x = x + \dots + x$ ($n$ times) so the action (if it exists) cannot be changed. But what if there are no assumptions about continuity?

(To be completely precise: by an action, I just mean a bilinear map $\mathbb{Z}_p \otimes_{\mathbb{Z}} M \to M$ satisfying $a \cdot (b \cdot m) = ab \cdot m$.)

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    $\begingroup$ Let $R$ be a commutative ring which admits two distinct ring homomorphisms $\mathbb Z_p\to R$, e.g. $R=\mathbb C$. Now take an abelian group $M$ such that $R$ embeds into the endomorphisms of $M$, e.g. $M=R$ viewed as a group. $\endgroup$
    – Wojowu
    Commented Jun 25, 2020 at 8:24
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    $\begingroup$ No you won't explicitly exhibit any such homomorphism at all. But for $x$ transcendental in $\mathbf{Z}_p$, for every transcendental $t$ in $\mathbf{C}$ there exists a ring homomorphism $\mathbf{Z}_p\to\mathbf{C}$ mapping $x$ to $t$ (it makes use of AC). $\endgroup$
    – YCor
    Commented Jun 25, 2020 at 8:32
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    $\begingroup$ An alternative is to consider the two obvious homomorphisms $\mathbf{Z}_p\to\mathbf{Z}_p\otimes_{\mathbf{Z}}\mathbf{Z}_p=:B$. While these homomorphisms are explicit, the fact that they are distinct (i.e., that $B$ is not reduced to $\mathbf{Z}_p$) makes use of some choice as far as I know. $\endgroup$
    – YCor
    Commented Jun 25, 2020 at 8:35
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    $\begingroup$ In fact I think YCor's example is in some sense "universal" (for lack of a better word), in that for a map of commutative rings $A\to B$, if the two maps $B\to B\otimes_A B$ agree, then the restriction $\mathrm{Mod}_B\to \mathrm{Mod}_A$ is fully faithful; so an $A$-module either is or isn't a $B$-module, but there is only one way to make it so. $\endgroup$
    – Maxime Ramzi
    Commented Jun 25, 2020 at 10:40
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    $\begingroup$ Thinking twice, I tend to guess that "the two canonical maps $\mathbf{Z}_p\to \mathbf{Z}_p\otimes_\mathbf{Z} \mathbf{Z}_p$ are distinct", holds in ZF. Namely choose $x$ transcendental: the claim is that $x\otimes 1$ and $1\otimes x$ are distinct. Otherwise we have some "relation" saying that they are equal, and this relation holds in some finitely generated subring of $\mathbf{Z}_p$ containing $x$, and in the latter I guess that we distinguish by making an explicit pair of homomorphisms. $\endgroup$
    – YCor
    Commented Jun 25, 2020 at 11:01

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