Jtdylktuy

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

Let $A =langle x^2+2rangle= {(x^2+2)f(x): f(x) in mathbb{Z}_3[x]}$ 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 $ax+b+A$.

1. Yes, that is the right derivation.


2. Now, divide $2x^2+x$ by $x^2+2$. That is easy:$$2x^2+x=2times(x^2+2).$$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 $,color{#c00}{x^2}equiv -2equivcolor{#c00}1,$ 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 $,rin {0,1}$

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 $,f - g notequiv 0pmod{!3}.,$ 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 $,a,bbmod 3$. 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 $,2+3Bbb Z,,$ which also serves to clarify innate algebraic structure, e.g. $,(x+1)(x-1) = color{#c00}{x^2}-color{#c00}1equiv 0,$ vs. $,(x+1)(x+2)equiv color{#c00}{x^2}+3x+color{#c00}2equiv 0$