1. ## polynomial rings

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

2. 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.

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

4. 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 $a+b\alpha$.