Finding elements of $\mathbb{Z}_3[x] /\langle x^2+2\rangle$ and writing the multiplication table

Question

I want to list elements of $\mathbb{Z}_3[x] /\langle x^2+2\rangle$ and write a multiplication table. Here is my attempt to finding the elements:

Let $A =\langle x^2+2\rangle= \{(x^2+2)f(x): f(x) \in \mathbb{Z}_3[x]\}$ and $\mathbb{Z}_3[x] / A = \{f(x) + A: f(x) \in \mathbb{Z}_3[x]\}$ by definition.

Let $f(x) \in \mathbb{Z}_3[x]$. By division algorithm, $f(x) = (x^2+2)q(x) + a + bx$ for some $q(x) \in \mathbb{Z}_3[x]$ and $a,b \in \mathbb{Z}_3$. Hence, $f(x) + A = (x^2+2)q(x) + a + bx + A = a + bx + (x^2+2)q(x) + A$. Since $(x^2+2)q(x) \in A$, $(x^2+2)q(x) + A = A$. Thus, $f(x) + A = a + bx + A$. So we have $\mathbb{Z_3}[x] / A = \{a + bx + A: a,b \in \mathbb{Z_3}\}$. Hence, the elements are the following:

1. $A$
2. $1 + A$
3. $x + A$
4. $2 + A$
5. $2x + A$
6. $1 + x + A$
7. $2 + x + A$
8. $2 + 2x + A$
9. $1 + 2x + A$

My question is:

1) Is this the right derivation?

2) How does multiplication table work in $\mathbb{Z}_3[x] / A$? For example, if I have $(x+A)(1+2x+A) = x(1+2x) + A = x+2x^2 + A$ which is not in the same form as $ax+b+A$.

Answer 1

1. Yes, that is the right derivation.
2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2\times(x^2+2).$$So, the remainder is $0$, wich means that, in your ring, $(x+A)(1+2x+A)=0+A=0$. In particular, your ring is not a field (not a surprise, since $x^2+2=(x+1)(x+2)$ in $\mathbb{Z}_3[x]$).

Answer 2

Using mod notation, in the quotient ring we have $\,\color{#c00}{x^2}\equiv -2\equiv\color{#c00}1\,$ which implies every polynomial is congruent to one of degree $\le 1,\,$ because $\,x^{\large 2q+r}\! = (\color{#c00}{x^{\large 2}})^{\large q}\,x^{\large r} \equiv \color{#c00}{1}^{\large q}\,x^{\large r}\equiv x^{\large r}\, $ for $\,r\in \{0,1\}$

Alternatively breaking into even+odd parts $\,f(x) = g(\color{#c00}{x^2}) + x\, h(\color{#c00}{x^2})$ $\Rightarrow\, f(x)\equiv g(\color{#c00}{1}) + x\, h(\color{#c00}{1})$

Or we can apply Division with Remainder: $\,f(x) = q(x) (\color{#c00}{x^2}\!-\!\color{#c00}1) + ax+b\,\Rightarrow\, f(x)\equiv ax+b$

So every $f(x)$ is congruent to $\,(f\,\bmod x^2\!-\!1)\bmod 3\,$ having degree $\le 1$, and these linear reps $\,f,g\,$ are incongruent else $\,x^2-1\,$ divides a lower degree polynomial $\,f - g \not\equiv 0\pmod{\!3}.\,$ Therefore they comprise a complete set of representatives of the quotient ring classes (cosets). In particular, there are $\,3^2 = 9$ such linear reps $\,ax+b$ corresponding to the $3$ choices for the coef's $\,a,b\bmod 3$. Your table correctly lists these $9$ reps.

To multiply these linear normal-form reps compute the polynomial product then replace $\,\color{#c00}{x^2}\,$ by $\,\color{#c00}{1}\,$

$$ (a_1x + a_0)(b_1 x + b_0)\, \equiv\, (a_0 b_1 + a_1 b_0)\, x + a_0 b_0 \color{#c00}{+1}\,a_1 b_1$$

while performing coefficient arithmetic $\!\bmod 3.\,$ The coefficient arithmetic will be slightly simpler if we use $\,-1\,$ vs. $\,2\,$ as our rep for $\,2+3\Bbb Z,\,$ which also serves to clarify innate algebraic structure, e.g. $\,(x+1)(x-1) = \color{#c00}{x^2}-\color{#c00}1\equiv 0\,$ vs. $\,(x+1)(x+2)\equiv \color{#c00}{x^2}+3x+\color{#c00}2\equiv 0$