# Multiplication structure for finite abelian rings of order $p^2$.

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.

$mod$ $p$ $mod$ $pb$ or $mod$ $pb$ $mod$ $p$ seems weird.
This is going to be the finite field with $p^2$ elements. – Berci Sep 20 '12 at 15:36
@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