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Show $x^4+x^2+x+1$ is irreducible in $\mathbb{F}_5[x]$.

I need to make use of Eisenstein's criterion, but I'm not sure how.

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    $\begingroup$ Eisenstein doesn't seem relevant here (it's a condition for irreducibility in $\mathbb Q[x]$). But it's only degree $4$! if it factors then either it has a root (easy to check) or it factors into two quadratics (just try to solve for the coefficients). $\endgroup$ – lulu Oct 29 '16 at 17:16
  • $\begingroup$ $(ax^2+bx+c)(dx^2+ex+f)$ so $ad=1, ae+bd=0 , be+cd+af=1, bf+ce=1, cf=1$ like this? $\endgroup$ – George Oct 29 '16 at 17:22
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    $\begingroup$ Well, yes. Only you can assume the coefficients of $x^2$ are both $1$ (divide through by $a$). $\endgroup$ – lulu Oct 29 '16 at 17:23
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    $\begingroup$ Eisenstein's criterion pertains to polynomials in $\mathbb Z[x]$. It is a condition for irreducibility over $\mathbb Q$. $\endgroup$ – lulu Oct 29 '16 at 17:38
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    $\begingroup$ Yes, that's the easiest way (I'd say). The posted solution by @JackD'Aurizio gives another approach; one that works a lot better for larger primes. After all, the approach I sketched relied on the fact that you could search the cases quickly. But perhaps the general method relies on more algebra than you have currently studied? $\endgroup$ – lulu Oct 29 '16 at 17:56
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Hint: Let $x^4+x^2+x+1=(x^2+ax+c)(x^2+bx+d)$ and you search for $a,b,c,d\in F_5[x]$.

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    $\begingroup$ That becomes even faster if we notice that we must have $a+b=0$ to ensure that the coefficient of $x^3$ is zero, and $cd=1$. $\endgroup$ – Jack D'Aurizio Oct 29 '16 at 22:46
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If $x^4+x^2+x+1$ were reducible over $\mathbb{F}_5$, it would have an irreducible factor with degree $\leq 2$. Since the polynomial $x^{25}-x$ is the product of all the irreducible monic polynomials over $\mathbb{F}_5$ with degree $1$ or $2$, it is enough to prove that over $\mathbb{F}_5$ we have $$\gcd(x^{25}-x,\,x^4+x^2+x+1)=1$$ to deduce that $x^4+x^2+x+1$ is an irreducible polynomial over $\mathbb{F}_5$. That is a completely algorithmic task.

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