Show that a polynomial is irreducible in $\mathbb{F}_5$ I need to show that $x^2+x+1$ is irreducible in $\mathbb{F}_5[x]$ and in $\mathbb{F}_{29}[x]$.
I know that I can start by finding the roots of the polynomial, but I'm not sure how to follow.
 A: We can show that $x^2+x+1$ is irreducible in both $\mathbb{F}_5[x]$ and $\mathbb{F}_{29}[x]$ with a few observations.
Observation 1: Let $K$ be a field and let $f(x)\in K[x]$. If the degree of $f(x)$ is $2$ or $3$, then $f(x)$ is irreducible iff $f(x)$ has no roots.
Observation 2: $(x^2+x+1)(x-1)=x^3-1$. So if $\alpha$ is a root of $x^2+x+1$, then $\alpha$ is also a root of $x^3-1$. Hence $\alpha^3=1$. If the characteristic of the field is not $3$, then $1$ is not a root of $x^2+x+1$, so the roots of $x^2+x+1$ are elements $\alpha$ with $\alpha^3=1$ and $\alpha\ne1$.
Observation 3: Let $K=\mathbb{F}_p$ for some prime $p$. Then $K^{\times}=K-\{0\}$ is a group under multiplication. An element $\alpha\in K$ with $\alpha^3=1$ and $\alpha\ne1$ would be element of order $3$ in $K^{\times}$. Note that the order of $K^{\times}$ is $p-1$. Hence if $K^{\times}$ has elements of order $3$, then $3$ divides $p-1$.
Comment: In fact, if $K$ is a field of order $p$, then $K^{\times}$ is cyclic. So $K^{\times}$ will have elements of order $3$ iff $3$ divides $p-1$, but we won't need this to show that $x^2+x+1$ is irreducible over $\mathbb{F}_5[x]$ and $\mathbb{F}_{29}[x]$.
Conclusion: Since the order of $\mathbb{F}_5$ is $5$ and the order of $\mathbb{F}_{29}$ is $29$, and both $5,29\equiv2$ (mod $3$), it follows that $\mathbb{F}_5^{\times}$ and $\mathbb{F}_{29}^{\times}$ don't have elements of order $3$. So $x^2+x+1$ is irreducible in both $\mathbb{F}_5[x]$ and $\mathbb{F}_{29}[x]$.
A: In $\mathbf F_5[x]$, you can easily check there's not root, writing the elements of $\mathbf F_5$ as $0, \pm 1,\pm 2$ to speed up a bit the computations.
As to $\mathbf F_{29}[x]$, simply compute the discriminant:
$\Delta=-3$, and use the law of quadratic reciprocity:
$$ \biggl(\frac{-3}{29}\biggr)= \biggl(\frac{-1}{29}\biggr)\biggl(\frac{3}{29}\biggr)=
\bigl(-1\bigr)^{\tfrac{28}2}\biggl(\frac{29}{3}\biggr)\bigl(-1\bigr)^{\tfrac{28}2\cdot\tfrac22}=\biggl(\frac{29}{3}\biggr)=\biggl(\frac{2}{3}\biggr)=-1, $$
so $\Delta$ has no root in $\mathbf F_{29}$, and the polynomial has no root in this field.
