number theory - The existence of primitive algebraic units modulo 3

Consider the problem of computing

$$\sqrt{2} \mod 3 $$

Whereas we seek a number $n$ such that $n^2 \equiv 2 \mod 3$ and furthermore it is known that both $n$ and $2n$ will satisfy this property, thus the smaller, while maintaining positive coefficients of the 'n' should be selected.

It is apparent that this problem has no obvious initial solution however by observing:

$$ 2 \equiv -1 \rightarrow \sqrt{2} \equiv \sqrt{-1} \equiv i \mod 3 $$

and of course

$$ (2i)^2 \equiv 2 \mod 3$$

but based on our initial stipulation

$$ i \equiv \sqrt{2} \mod 3$$

since it features a smaller positive coefficient.

From here we can then consider the complex integers modulo 3 which are numbers of the form

$$ a + bi$$

such that $ a,b \in \lbrace{0,1,2}\rbrace $

Again attempting to compute square roots in this environment reveals

$$ \sqrt{0} \equiv 0 \mod 3$$
$$ \sqrt{1} \equiv 1 \mod 3$$
$$ \sqrt{2} \equiv i \mod 3$$
$$ \sqrt{i} \equiv 1 + 2i \mod 3$$
$$ \sqrt{1 + i} \equiv \sqrt{1 + i} \mod 3$$
$$ \sqrt{2i} \equiv 1 + i \mod 3$$
$$ \sqrt{1 + 2i} \equiv (1 + i)\sqrt{1 + i} \mod 3$$

$$ \sqrt{2 + 2i} \equiv i\sqrt{1 + i} \mod 3$$

Notice how all the square roots can be defined in terms of $0,1,2,i, \sqrt{1 + i} $

This suggests that:

$ \sqrt{1 + i} $ is another 'natural' unit modulo 3 and therefore we can declare

$$ \sqrt{1 + i} \equiv i_2 $$

whereas the numbers mod 3 now take on the form:

$$ (a_1 + a_2 i_1) + (a_3 + a_4 i_1 ) i_2 $$

where

$$i_1^2 \equiv 2 \mod 3 \ \ and \ \ i_2^2 \equiv 1 + i \mod 3$$

I wonder whether by repeatedly taking square roots of these numbers, will an infinite chain of 'unique' units be generated? Or does it stop and there are only a finite number of

$$i_1 , i_2 ... i_n $$

that need to be defined so that:

$$ \sqrt{((...((a_1 + a_2 i_1) + (a_3 + a_4i_1)i_2) + ((a_5 + a_6 i_1) + (a_7 + a_8i_1)i_2)i_3 ... )))} $$

are all defined.

An environment modulo R for a set of units $\lbrace 1,i_1, i_2 ... i_n \rbrace$ is defined as the set of all linear combinations using $\lbrace{0,1,2... R-1} \rbrace$ for coefficients of the set of all product combinations from the set of units.

Note the size of environments are $R^{2^n}$ where $n$ is the number of units present.

Answer

I have determined that what I have been calling an 'environment' is indeed a finite field (as per popular suggestion the comments).

Furthermore given that it is a finite field that means the theorem:

http://en.wikipedia.org/wiki/Finite_field#Algebraic_closure

Proves that it indeed cannot be closed algebraically and therefore if we consider other operations besides square roots but rather the inverse to any polynomial then indeed an infinitely large set of units will be generated.