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In p.20 Stewart and Tall's algebraic number theory, it says that the polynomial $t^5-t+1$ is irreducible: consider $mod 5$, there is no linear factor since none of $0,1,2,3,4$ when substituted fir $t$, so the only possible way to factorize is $t^5-t+1=(t^2+bt+c)(t^3+dt^2+et+f)$, where $a,b,c,d,e,f$ take value $0,1,2,3,4 (mod 5)$. This gives a system of equations on comparing coefficients which are all easily eliminated and hence the polynomail is irreducible mod $5$, so irreducible over $\mathbb{Z}$.

My question is: can I just simply say that by fermat's little theorem, $t^5-t+1 \equiv t-t+1 =1 \pmod{5}$, which is clearly irreducible?

taylor
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1 Answers1

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Because $t^5-t+1\ne t-t+1$ in $\mathbb F_5[t]$. On of the key differences between finite and infinite fields is that two different polynomials over a finite field can agree at every point.

By the same logic we could argue that $t^5-1\equiv t-1$ is irreducible, but $t^5-1=(t-1)(t^4+t^3+t^2+t+1)$.

Alex Becker
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