Re: Multiplicative Homomorphic Mapping

In article <ILL7Dq.E4s@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Aldar C-F. Chan" <aldar@xxxxxxxxxxxxxxxx> wrote:

> Suppose I have a multiplicative subgroup S_1 in the extension field
> K/F generated by a q-th root of unity in K/F. I also have a
> multiplicative subgroup S_2 of order q in some other finite field F'.
> I want to find an invertible group homomorphism between S_1 and S_2.
> Is there a way to implement this map?

That depends on what "have" means, and on what "implement" means.

If you "have" a generator x for S_1 and a generator y for S_2
then you can "implement" your isomorphism by mapping x^n to y^n
for all n.

To see how tricky this might be in practice, the following example,
while not directly relevant, might be instructive. Let p be a whopping
big prime. Then the multiplicative group of Z / pZ and the additive
group of Z / (p - 1) Z are isomorphic, but implementing this isomorphism
is thought to be computationally infeasible, owing to the difficulty
of finding a generator for the 1st group.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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