Elements of a field when $p(x)$ = $x^2$ + $x$ + $2$ is irreducible in $Z_3[x]$.

Question

So we have that $p(x)$ = $x^2$ + $x$ + $2$ is irreducible in $Z_3[x]$.

Let $I$ = $<p(x)>$ and let A = $Z_3[x]$/I.

Let $\sigma$ = $x+1$.

a) Write every element of A in terms of $\sigma$ and powers of $\sigma$ and the elements of $Z_3$.

APPROACH:

From what I can see the elements of A have this structure: $r(x)$ = $bx$ + $a$

So let $A$ = { $a$ + $bx$ + $I$ | $a$,$b$ $\in$ $Z_3$ }

Therefore, the elements of A are:

0+I   1+I  2+I  x+I  x+1+I  x+2+I  2x+I  2x+1+I  2x+2+I

Is this correct?

b) Finding the inverse of $\sigma$ and $2\sigma$ + 1 in A:

This part I really have no approach, unless it also works by doing GCD.

Answer 1

So the field has 9 elements as you have written down. Now observe that the multiplicative group has 8 elements and is a cyclic group. The question has already given you a generator of this multiplicative group. write each element in its power and then you can easily find the inverse. For example you have $\sigma^{8}=1$. So inverse is $\sigma^{7}$.