0
$\begingroup$

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 $ax+b+A$.

$\endgroup$
1
$\begingroup$
  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 $\mathbb{Z}_3[x]$).
$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.