Re: mth powers in GF(p^n)
- From: quasi <quasi@xxxxxxxx>
- Date: Thu, 29 Sep 2005 01:32:49 -0700
On Thu, 29 Sep 2005 15:24:42 +1000, Gerry Myerson
<gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>In article <af6nj1dlr2f8t3n409h1ghsohjit5j398n@xxxxxxx>,
> quasi <quasi@xxxxxxxx> wrote:
>
>> On 28 Sep 2005 21:07:44 -0700, "Wolff" <daniel.wolff@xxxxxxxx> wrote:
>>
>> >Hello, any hints on showing that (p^n-1,m)=1 implies each member of
>> >GF(p^n) is an mth power?
>> >Thanks.
>> >DMW
>>
>> Hints:
>>
>> Consider the map f defined by f(x)=x^m. Can you verify that f is a
>> homomorphism from the multiplicative group of GF(p^n) to itself?
>>
>> What is the kernel of f?
>>
>> Note: The condition (p^(n-1),m)=1 is clearly equivalent to the much
>> simpler condition (p,m)=1, and probably should have been stated that
>> way. However if it had been stated as (p,m)=1, then you would need to
>> push it back up to (p^(n-1),m)=1 in order to compute the kernel of f.
>
>Maybe p^n-1 means (p^n) - 1, not p^(n - 1).
Yep -- p^n-1, thanks. After all, it's a field, so the multiplicative
group is all but 0.
So then ignore the "Note:" part of my reply -- sorry, but the hints
should still be ok.
quasi
.
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