<s>[INST] $x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
 [/INST] Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.
The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$. 
Therefore, the smallest field containing $\mathbb{F}_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t. 
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