Show some polynomial is irreducible over the field of 7 elements. I have to show that the polynomial $x^4+x^3+x^2+x+1$ is irreducible over the field $F_7$.
It doesn't have roots in $F_7$, but I can't show it does not have degree two irreducible factors in $F_7[x]$.  Please help me.  Thank you.
 A: Exercise 8 of Chapter 13.6 in Dummit & Foote's Abstract Algebra provides a step-by-step exercise (which is fairly easy) to determine the factorization of $\Phi_{\ell}(x)$ in $\mathbb F_p[x]$ where $p, \ell$ are primes and $\Phi_{\ell}$ is the $\ell^{\text{th}}$ cyclotomic polynomial. In your case you are trying to factor $\Phi_5$ over $\mathbb F_7$. The exercise determines the degrees of the irreducible factors of the factorization of $\Phi_{\ell}$ as soon as you know the order of $p$ in the multiplicative group of $\mathbb F_{\ell}$, i.e. the least positive integer $f$ such that $p^f \equiv 1 \pmod{\ell}$ (assuming $p \neq \ell$, of course ; the case $p=\ell$ is actually trivial, since $\Phi_p(x) = (x-1)^{p-1}$ because of the Frobenius automorphism). It requires some field theory though, so I don't know if you have the tools for this. I just thought this was of public interest. 
Hope that helps,
A: Try dividing your polynomial by the arbitrary quadratic (WLOG, monic) polynomial $x^2 + ax + b$.  The remainder will be a quadratic cubic polynomial in $a$ and $b$.  Check that it has no roots over $F_7$ by brute force.
A: The brute-force approaches of the earlier answers might be appropriate if the polynomial were essentially random, but it is not. It is the fifth cyclotomic polynomial! It will have a root in the smallest extension $\mathbb F_{7^d}$ of $\mathbb F_7$ that has a (necessarily cyclic) subgroup of order $5$. Since the multiplicative group of $\mathbb F_{7^d}$ is (cyclic) of order $7^d-1$, the question is what multiplicative order $7$ has mod $5$. Since $7=2 \mod 5$, and $2^2=4$, $2^3=8=3\mod 5$, and $2^4=2\cdot 3=6=1\mod 5$, the order of $7$ is $4$. That is, the smallest-degree extension containing a root is of degree $4$, and the $5$th cyclotomic polynomial is irreducible mod $7$.
