Construct add/mul tables for the cong classes of the ideal $(x^2+1)$ in the ring $\mathbb{Z}_3[x]$.

Question

Construct the addition & multiplication tables for the congruence classes of the ideal $(x^2+1)$ in the ring $\mathbb{Z}_3[x]$.

Additionally, the question also asks if the factor ring $\mathbb{Z}_3[x]/(x^2+1)$ is a field.

What I did was list the congruence classes of ideal $(x^2+1)$ in the ring $\mathbb{Z}_3[x]$, which are represented by the polynomials $0, 1, 2, x, x+1, x+2, 2x, 2x+1, 2x+2$. But how do I proceed further in trying to add and multiply the congruence classes? Do I add and multiply them $\mod{3}$ and if so, how does that work for polynomials?

Also, would this be a field?

Answer 1

Indeed, to add/multiply them you add/multiply them modulo $3$. Additionally, you have to find the representant of the congruence class modulo $(x^2+1)$, which you do by replacing $x^2$ by $-1 = 2$.

So, for example, $(x + 1)(x + 2) = x^2 + 3x + 2 = x^2 + 2 = 2 + 2 = 1$ in ${\mathbb Z}_3[x]/(x^2+1)$.

From the multiplication table you will see that every non-zero element has an inverse, so this is indeed a field. This can be more easily seen without computing the whole multiplication table by noting that $x^2 + 1$ has no roots in ${\mathbb Z}_3$. Because it is of degree $\leq 3$, that makes it a maximal ideal and hence the quotient ring ${\mathbb Z}_3[x]/(x^2+1)$ is a field.

Answer 2

You do as in all congruence classes: you perform the operations in $\mathbf Z/3\mathbf Z[x]$, then reduce mod. $x^2+1$.

Example:

I prefer to denote the elements in $\mathbf Z/3\mathbf Z$ as $0,\,1,\,-1$. Thus the elements of $\mathbf Z/3\mathbf Z[x]$ are, denoting $\omega$ the congruence class of $x$ (it satisfies the equation $\omega^2=-1$ by construction): $$\{\,0,1,-1, \omega,-\omega, 1+\omega, -1+\omega, 1-\omega,-1-\omega\,\}$$ and, say: $$(-1+\omega)(-1-\omega)=-(\omega^2-1)=-(-1-1)=2=-1.$$

On the other hand, the polynomial $x^2+1$ is irreducible in $\mathbf Z/3\mathbf Z[x]$, since it is a polynomial with degree $\le 3$ with no root in $\mathbf Z/3\mathbf Z$. Thus the ideal it generates is maximal, and the quotient ring is indeed a field.