Re: field notation what is an integer and how to I put it in %way?

From: carol claxton (houw2love_at_yahoo.com)
Date: 02/27/05 

Date: Sun, 27 Feb 2005 18:26:23 +0000 (UTC)

On 02 Oct 2001, Nico Benschop wrote:
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>Tom St Denis wrote:
>>
>> "Nico Benschop" <n.benschop@chello.nl> wrote in message
>> news://3BB9B125.5BD19201@chello.nl...
>> > Steven Taschuk wrote:
>> > >
>> > > "Tom St Denis" <tomstdenis@yahoo.com> wrote:
>> > > > "Richard Carr" <carr@cpw.math.columbia.edu> wrote
>> > > > > On Tue, 2 Oct 2001, Tom St Denis wrote:
>> > > ...
>> > > > > :What is the correct notation for a
>> > > > > : "finite field with q elements generated by a ...[*]
>> > > > > : irreducible element p"
>> > > > >
>> > > > > There aren't any irreducibles in a field as every element is
>> > > > > either 0 or a unit so what you ask doesn't make sense.
>> > > ...
>> > > > Irreducible. Not reducible. I.e p is from some larger field or
>> > > > ring and it cannot be reduced with multiplicative factors.
>> > > > What am I missing?
>> > >
>> > > In a field, every nonzero element is a unit. This means that every
>> > > element has a multiplicative inverse. Write a^-1 for the
>> > > multiplicative inverse of a; thus a.a^-1 = 1. But now every element
>> > > is divisible by every other nonzero element; for to show that a is
>> > > divisible by b, ie., to show that there is some field element k such
>> > > that a = bk, take k = (b^-1).a (which is easily seen to be an element
>> > > of the field). So, in a field, every element can be factored.
>> > > No field elements are irreducible.
>> >
>> > Re[*]:
>> > Maybe Tom means a primitive root, cyclic units group G with G = {p^*}.
>>
>> [funny stare using glazed eyes at screen] gah?
>>
>> I semi-recall that primitive elements generate the entire
>> multiplicative group under unique powers? Like 45^x mod 257
>> generates all of the multiplicative group mod 257.
>> What is a "cyclic units group", does that refer to the same idea?
>> -- Tom
>
>Precisely, although I would have to check if G_257 = 45* :
>
> Yes it is, says PARI, by: znorder(Mod(45,257)) = 256,
>
> and the smallest primroot is 3, by: znprimroot(257) = Mod(3,257).
>
>A smaller example: F_7 = {0, 3*} (mod 7), where 3* = {3,2,-1,-3,-2,1}
>
>The cubic roots a^3 == 1 mod 7 are: Cub = {2,-3,1} = {2,4,1}
>and mod 7^2 : Cub = {42,24,66} a=42 (=30 decim)
>
> Notice: a + 1/a == -1 (a^2 == 1/a) mod p^k,
>
> where a^p == a, (1/a)^p == 1/a, (-1)^p == -1 (mod p^k, p=1 mod 6)
>
> so : (a+b)^p == a+b == a^p + b^p (with a^p=a, b^p=b, a+b=-1)
>
> hence an example of FLT mod p^k (for residues, not for
>inetegers;-)
>
>And such group mod p^k (odd prime p) is also cyclic for any k>1 ...(!)
>In fact, for the "most important & practical prime 2" this is not
>true, but 'nearly so': you only have to adjoin -1 (mod 2^k) to
>generate the full group of units. Actually semi-generator 3 will
>do "it" everytime: G = {-1,3}* mod 2^k for all k>2 (skip -1 if k<3).
>
>Check out my US patent # 5923888 (13jul1999) described at
>-- http://arxiv.org/abs/math.GM/0105029 -- NB
>
>
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