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

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 $$ax+b+A$$.

## 2 Answers

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 $$\mathbb{Z}_3[x]$$).

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. $$\,(x+1)(x+2)\equiv \color{#c00}{x^2}+3x+\color{#c00}2\equiv 0$$