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This answer says that each p-adic number is a ratio of endomorphisms of a subgroup of a quotient of $\Bbb Q$.

If I wanted to realise this ratio of endomorphims for the $2$-adic number $-\dfrac13$ say, then given the context I guess the subgroup of a quotient of the rationals is $G=\Bbb Z[\frac12]/\Bbb Z$ since that's the Prufer 2-group.

Would two endomorphims $e_1,e_2$ on $G$ to yield $-\dfrac13$ simply be $e_1=-1\cdot G$ and $e_2=3\cdot G$, with conjugation to form the group operation?

There is an obvious case for the above, but it feels like the answer should be more complicated. The definition of an endomorphism as opposed to an automorphism or an isomorphism seems to require it's not necessarily reversible, therefore presumably it need not surject.

it's a hire car baby
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1 Answers1

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The "ratio" in the answer there comes in only to make the step from $\mathbb Z_p$ to $\mathbb Q_p$. Every element of $\mathbb Z_p$ is already an endomorphism. Since $-1/3 \in \mathbb Z_2$, it corresponds already to a unique element of the endomorphism ring $\mathrm{End}(\mathbb Z[\frac12]/\mathbb Z)$. Not surprisingly, that element is multiplication by $-\frac13$ which is a well-defined endomorphism (in this case, automorphism)

$$x +\mathbb Z \mapsto -\frac13 x+\mathbb Z$$

of the Prüfer group $\mathbb Z[\frac12]/\mathbb Z$.

It is only elements with $p$-powers in the denominator for which you need a true "ratio" here; i.e. in your context, something like $\frac12$ or $\frac{-3}{8}$ cannot directly be identified with an endomorphism of $\mathbb Z[\frac12]/\mathbb Z$.

Incidentally, when viewed as elements of $\mathrm{End}(\mathbb Z[\frac1p]/\mathbb Z)$, all elements of $\mathbb Z_p$ except $0$ are still surjective. The ones in $p\mathbb Z_p$ are not injective, and of course the units $\mathbb Z_p^*$ correspond to automorphisms. Most of this was already discussed here.

Torsten Schoeneberg
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