- #1
mebassett
- 5
- 0
Hi folks, long time reader, first time poster.So I'm trying to prove that $\hat{Z} = \varprojlim Z/nZ$ is isomorphic to Hom(Q/Z, Q/Z).
Problem is, I have a lot of trouble describing even a single endomorphism of that torsion group (except, of course, the trivial one, that sends everything to 0.)
Q/Z can be considered as the set of all roots of unity. and I should be able to find an endomorphism on that group, just by permuting key roots of unity (simple example: {1,-1,i,-1}) But I haven't been able to generalize this for the whole thing.
I also tried a pretty naive thing, taking a \in \hat{Z}, say a=(a_1, a_2, ...) and trying to define some \theta : Q/Z \to Q/Z, with \theta (m/n) = (m * a_n) / n. which I believe defines a map from \hat{Z} into Hom(Q/Z, Q/Z), but I don't have much in the way of an inverse, especially since I can't describe anything but trivial members of Hom(Q/Z, Q/Z)
I'm aware that \hat{Z} is also the product of all the p-adic integers Z_p, but I'm having trouble proving that as well.
In any case, I'm stuck, I don't want to spend much more time on this (Because its just one small bit of a much larger project), so I thought I'd try asking the kind and clever folks here for some help.so, Hom(Q/Z, Q/Z) is isomorphic to \hat{Z}. Any ideas on how to prove it?
Thanks.
Problem is, I have a lot of trouble describing even a single endomorphism of that torsion group (except, of course, the trivial one, that sends everything to 0.)
Q/Z can be considered as the set of all roots of unity. and I should be able to find an endomorphism on that group, just by permuting key roots of unity (simple example: {1,-1,i,-1}) But I haven't been able to generalize this for the whole thing.
I also tried a pretty naive thing, taking a \in \hat{Z}, say a=(a_1, a_2, ...) and trying to define some \theta : Q/Z \to Q/Z, with \theta (m/n) = (m * a_n) / n. which I believe defines a map from \hat{Z} into Hom(Q/Z, Q/Z), but I don't have much in the way of an inverse, especially since I can't describe anything but trivial members of Hom(Q/Z, Q/Z)
I'm aware that \hat{Z} is also the product of all the p-adic integers Z_p, but I'm having trouble proving that as well.
In any case, I'm stuck, I don't want to spend much more time on this (Because its just one small bit of a much larger project), so I thought I'd try asking the kind and clever folks here for some help.so, Hom(Q/Z, Q/Z) is isomorphic to \hat{Z}. Any ideas on how to prove it?
Thanks.
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