polynomial rings

**dori1123** · February 11th 2009, 11:48 AM

$\mathbb{Z}_3[x]/(x^2+1)$ is a field of order 9. How do I find a generator for the cyclic multiplicative group of nonzero elements?

**ThePerfectHacker** · February 11th 2009, 02:22 PM

Originally Posted by dori1123

$\mathbb{Z}_3[x]/(x^2+1)$ is a field of order 9. How do I find a generator for the cyclic multiplicative group of nonzero elements?

Let $\alpha = x + (x^2+1)$ . All elements in this field are $a+b\alpha$ where $a,b\in \mathbb{Z}_3$ . Therefore, the elements are:
$0$ , $\alpha$ , $2\alpha$ , $1$ , $1+\alpha$ , $1+2\alpha$ , $2$ , $2+\alpha$ , $2+2\alpha$ .

You also know that $\alpha^2 + 1 = 0$ so $\alpha^2 = 2$ .
From this list find an element $y$ so that $y^4\not = 1$ .
And that would be your generator.

**dori1123** · February 11th 2009, 02:55 PM

Why are you letting $\alpha = x + (x^2+1)$ , not $\alpha=x^2+1$ .

**ThePerfectHacker** · February 11th 2009, 03:20 PM

Originally Posted by dori1123

Why are you letting $\alpha = x + (x^2+1)$ , not $\alpha=x^2+1$ .

First because $\alpha = x^2+1$ makes no sense since $\alpha$ needs to be an element of $\mathbb{Z}_3[x]/(x^2+1)$ and $\mathbb{Z}_3[x]/(x^2+1)$ consists of cosets of $(x^2+1)$ . Second, $\{1 ,\alpha\}$ is a basis for this field over $\mathbb{Z}_3$ . Therefore, everything has the form $a+b\alpha$ .

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