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Is the polynomial $x^2+3$ irreducible in $\mathbb{Z}_7[x]$?

I am not sure how to even start the problem. I have been looking online trying to find help to teach myself how to do irreducibility and I am not having much luck. Can anyone offer any help as to where I should begin?

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2 Answers 2

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If $F$ is a field, and $P(x)$ is a polynomial of degree $2$ or $3$ over $F$, then $P(x)$ is irreducible over $F$ if and only if the equation $P(x)=0$ has no roots in $F$.

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  • $\begingroup$ So I am correct in thinking that it is reducible because it could be (x+a)(x+b)? $\endgroup$
    – girl
    Nov 18, 2013 at 2:09
  • $\begingroup$ We ask does the polynomial have a root in $\mathbb{Z}_7$? Try everything, that is, $0$ to $6$. We get that when $x=2$, the object $x^2+3$ is the zero element, so there is a root, the polynomial is not irreducible. It factors as $(x-2)(x-5)$, or equivalently $(x+5)(x+2)$. $\endgroup$
    – André Nicolas
    Nov 18, 2013 at 2:17
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Hint If $x^2+3$ is not irreducible then you can write

$$x^2+3=P(x)Q(x)$$ where $P(x)$ and $Q(x)$ are non-constant. What can the degree of $P,Q$ be? What can $P,Q$ be then?

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