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Thread: polynomial rings

  1. #1
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    polynomial 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?
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  2. #2
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    Quote Originally Posted by dori1123 View Post
    $\displaystyle \mathbb{Z}_3[x]/(x^2+1)$ is a field of order 9. How do I find a generator for the cyclic multiplicative group of nonzero elements?
    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.
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  3. #3
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    Why are you letting $\displaystyle \alpha = x + (x^2+1)$, not $\displaystyle \alpha=x^2+1$.
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  4. #4
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    Quote Originally Posted by dori1123 View Post
    Why are you letting $\displaystyle \alpha = x + (x^2+1)$, not $\displaystyle \alpha=x^2+1$.
    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 $\displaystyle a+b\alpha$.
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