How to find all the roots in this ring?

Question

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

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

Attempt:

It is a field, because $x^2+3x+1$ is irreducible $\in \mathbb{F}_7[x]$. In fact it has no roots $\in \mathbb{F}_7$.
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?

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. $f(Y)\equiv 0$. — Ragnar, Jan 25 '14 at 15:42

score 3 · Answer 1 · answered Jan 25 '14 at 15:53

3

Well, $f$ has $[x]$ as a root, the other root has to be $-3-[x]$ (Vieta).

answered Jan 25 '14 at 15:53

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 $x^2=5$ – Angelo Tricarico Jan 25 '14 at 17:48
As I've said, use Vieta. By the way (but this is really irrelevant here), $\sqrt{5} \notin \mathbb{F}7$ (but $\sqrt{5} \in \mathbb{F}{7^2}$). – Martin Brandenburg Jan 25 '14 at 18:41
I read the page but I didn't understand how to get to the result in this case. Would you please explain? – Angelo Tricarico Jan 29 '14 at 07:20
The sum of the roots is $-3$ by Vieta. – Martin Brandenburg Jan 29 '14 at 08:51

1 Answers1