Irreducibility of $t ^ { 4 } + t - 1$ over $\mathbb{F}_{25}$

Question

Problem: Let $f ( t ) = t ^ { 4 } + t - 1 \in \mathbb { F } _ { 5 } [ t ]$, show that $f(t)$ has no roots in $\mathbb{F}_{25}$.

I tried the following: suppose $f(t)$ has a root in $\mathbb{F}_{25}$, then $f(t)$ can be written as the product of either two polynomials of degree 2, or the product of two polynomials of degree 1 and 3.

Suppose the first case, thus $t ^ { 4 } + t - 1 = (t^2+at+b)(t^2+ct+d)$ for some coefficients in $\mathbb{F}_{25}$.

$$\Rightarrow t ^ { 4 } + t - 1 = t^4 +t^3(c+a)+t^2(d+ac+b)+t(ad+bc)+bd$$

However I couldn't find a contradiction at this point

Suppose $f(t)$ has a root in $\mathbb{F}_{25}$, then $f(t)$ has a linear factor of degree $1$, namely $t-a$, where $a$ is the root. — Dietrich Burde, May 18 '19 at 10:20
Oh you're right, so the only way is to check every elements of $\mathbb{F}_{25}$? — NotAbelianGroup, May 18 '19 at 10:25
The title is a bit off. Your polynomial being irreducible over $\Bbb{F}5$ is equivalent to it not having any zeros in $\Bbb{F}{25}$. — Jyrki Lahtonen, May 18 '19 at 11:34
Anyway, the theory for checking irreducibility of a quartic is described here. — Jyrki Lahtonen, May 18 '19 at 19:28

lhf · Answer 1 · 2019-05-18T14:15:48.250

3

$t^{25}-t$ is the zero function in $\mathbb{F}_{25}$.

Then $\gcd(t^{25}-t,t^4+t-1)=1$ implies that $t^4+t-1$ has no zero in $\mathbb{F}_{25}$.

edited May 18 '19 at 14:15

answered May 18 '19 at 13:04

lhf

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Irreducibility of $t ^ { 4 } + t - 1$ over $\mathbb{F}_{25}$

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