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Consider an odd prime $p$ and a positive integer $a$ as close to 2 as possible that is not a quadratic residue $mod$ $p$.

If we extend the ring $mod$ $p$ with the element $b = a^{\frac{1}{2}}$ then we get a finite abelian ring of order $p^2$.

Every integer element $t$ of this ring different from 0 satisfies $t^{p-1} = 1$ ( Fermat's little ).

Likewise every non-integer element $s$ of this ring satisfies $s^{(p-1)^2} = 1$.

The question now becomes if we put the elements in a matrix-like-square with the rules

1) 0 at bottom left

2) +1 means go to right

3) +$b$ means go up

Then is there a pattern in the $k$ th powers of the elements s i.e. $s^k$ ?

With pattern i mean something like Fermat's little or similar algebra or a geometric pattern such as knight moves on a chess board.

I do not believe these patterns to be random.

I have been thinking about the distributive property to get an answer but with no succes sofar.

Sieving also comes to my mind.

How to handle this ?

Also Im not sure about the name abelian ring... I can only find texts about abelian groups most of the time.

Also Im unconfident about notation.

$mod$ $p$ $mod$ $pb$ or $mod$ $pb$ $mod$ $p$ seems weird.

Maybe I just did too many magic knight tours :)

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This is going to be the finite field with $p^2$ elements. –  Berci Sep 20 '12 at 15:36
    
Right these rings are actually fields. No zero-divisors. –  mick Sep 20 '12 at 15:47
    
I think "commutative ring" is preferable to "abelian ring". Quite rarely, they are used as synonyms. Nobody is confused about what "commutative ring" means, though. Some algebraists use "abelian ring" for rings whose idempotents are all central. –  rschwieb Sep 20 '12 at 16:09
    
@rschwieb : Ah thank you , that explains alot about the terminology. –  mick Sep 20 '12 at 16:12
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@mick The $p$ and $b$ will not really mix in any sort of "mod" notation. The only way I can think of expressing the new field using a quotient is $\mathbb{F}[x]/(x^2-a)$, where $\mathbb{F}$ is in turn $\mathbb{Z}/(p)$. To write it in terms of $b$, the go-to notation would be $\mathbb{F}[b]$. –  rschwieb Sep 20 '12 at 16:19

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