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

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- Feb 11th 2009, 11:48 AMdori1123polynomial rings
$\displaystyle \mathbb{Z}_3[x]/(x^2+1)$ is a field of order 9. How do I find a generator for the cyclic multiplicative group of nonzero elements?

- Feb 11th 2009, 02:22 PMThePerfectHacker
Let $\displaystyle \alpha = x + (x^2+1)$. All elements in this field are $\displaystyle a+b\alpha$ where $\displaystyle a,b\in \mathbb{Z}_3$. Therefore, the elements are:

$\displaystyle 0$, $\displaystyle \alpha$, $\displaystyle 2\alpha$,$\displaystyle 1$,$\displaystyle 1+\alpha$, $\displaystyle 1+2\alpha$,$\displaystyle 2$, $\displaystyle 2+\alpha$, $\displaystyle 2+2\alpha$.

You also know that $\displaystyle \alpha^2 + 1 = 0$ so $\displaystyle \alpha^2 = 2$.

From this list find an element $\displaystyle y$ so that $\displaystyle y^4\not = 1$.

And that would be your generator. (Nod) - Feb 11th 2009, 02:55 PMdori1123
Why are you letting $\displaystyle \alpha = x + (x^2+1)$, not $\displaystyle \alpha=x^2+1$.

- Feb 11th 2009, 03:20 PMThePerfectHacker
First because $\displaystyle \alpha = x^2+1$ makes no sense since $\displaystyle \alpha$ needs to be an element of $\displaystyle \mathbb{Z}_3[x]/(x^2+1)$ and $\displaystyle \mathbb{Z}_3[x]/(x^2+1)$ consists of cosets of $\displaystyle (x^2+1)$. Second, $\displaystyle \{1 ,\alpha\}$ is a basis for this field over $\displaystyle \mathbb{Z}_3$. Therefore, everything has the form $\displaystyle a+b\alpha$.