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Let $F:= \mathbb{F}_7[x]/(x^2+3x+1)$

  1. Is it a field?
  2. Find all the roots in F of the polynom $f (Y) := Y^2+[3]_{F}Y +[1]_{F} \in F[Y]$.

Attempt:

  1. It is a field, because $x^2+3x+1$ is irreducible $\in \mathbb{F}_7[x]$. In fact it has no roots $\in \mathbb{F}_7$.
  2. I suppose I can't just replace numbers from $0$ to $6$ in the place of the $Y$. What should you do to solve this problem?
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  • $\begingroup$ Since $Y$ is an element of the field $F$, $Y$ is a polynomial of degree at most one $1$. You want to find $Y=ax+b$ s.t. $f(Y)\equiv 0$. $\endgroup$ – Ragnar Jan 25 '14 at 15:42
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Well, $f$ has $[x]$ as a root, the other root has to be $-3-[x]$ (Vieta).

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  • $\begingroup$ Could you please explayin why those are the roots? If I use the quadratic formula $D=b^2-4ac$ the result is $5$, then I need to calculate $\sqrt{5}$ and I can't find an element $x \in Z_{7}$ such that $x^2=5$ $\endgroup$ – Angelo Tricarico Jan 25 '14 at 17:48
  • $\begingroup$ As I've said, use Vieta. By the way (but this is really irrelevant here), $\sqrt{5} \notin \mathbb{F}_7$ (but $\sqrt{5} \in \mathbb{F}_{7^2}$). $\endgroup$ – Martin Brandenburg Jan 25 '14 at 18:41
  • $\begingroup$ I read the page but I didn't understand how to get to the result in this case. Would you please explain? $\endgroup$ – Angelo Tricarico Jan 29 '14 at 7:20
  • $\begingroup$ The sum of the roots is $-3$ by Vieta. $\endgroup$ – Martin Brandenburg Jan 29 '14 at 8:51

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