Perhaps one can somehow apply Eisenstein's sign here by considering $f(x+1)$, but by default it is formulated for the expansion over $\mathbb{Q}$ of a polynomial from $\mathbb{Z}[x]$. Here we have $GF(5)$ and it seems that the sign becomes inapplicable.
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J. W. Tanner
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mackenzie
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1You cannot use Eisenstein because there are no irreducibles in $\mathrm{GF}(5)$. – Arturo Magidin May 13 '24 at 01:16
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@ArturoMagidin: what do you mean there are no irreducibles in GF($5$)? isn’t $x^2-2$ irreducible in GF($5$)? – J. W. Tanner May 13 '24 at 01:28
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1@J.W.Tanner It's irreducible in $\mathrm{GF}(5)[x]$ (or over $\mathrm{GF}(5)$). Eisenstein requires prime elements in the ring of coefficients, and $\mathrm{GF}(5)$ has no irreducibles, hence no primes. – Arturo Magidin May 13 '24 at 01:32
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since GF($5$) is a field and every nonzero element is a unit? Thanks for explaining – J. W. Tanner May 13 '24 at 01:34
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1@J.W.Tanner Indeed. And there is no domain $D$ that is not a field that has $\mathrm{GF}(5)$ as its field of fractions. – Arturo Magidin May 13 '24 at 01:35
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2This is a special case of the following observation. The line of reasoning in that old answer is essentially the one adopted by the answerers. The techniques recurs - I have used it on this site more often than I care to admit :-) – Jyrki Lahtonen May 13 '24 at 02:30
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This is true, because $f(x)=\frac{1-x^7}{1-x}$, that is the roots of $f$ are all $7$-th roots of $1$. Let $\alpha$ be a root of $f$, consider the Galois orbit $\alpha, \alpha^5, \alpha^{5^2}, \cdots$. As $5$ is a primitive root mod $7$, the Galois orbit of $\alpha$ has precisely $6$ elements. Therefore all the roots of $f$ are Galois conjugate to $\alpha$, so $f$ must be irreducible.
Another way to finish without Galois theory: $\alpha$ generates a multiplicative subgroup of order $7$ in $F[\alpha]^{\times}$. If $|F[\alpha]| = 5^m$, then by Lagrange $7 | 5^m - 1$, and again since $5$ is a primitive root mod $7$, $m\ge 6$.
J. W. Tanner
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Just a user
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If $f(x)$ is not irreducible, then it has an irreducible factor of degree $2$ or $3$ (it has no linear factors).
Suppose $\alpha$ is a root of $f(x)$ in some extension $K$ of $\mathbb{F}_5$. Then $\alpha$ satisfies $$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$ So the multiplicative order of $\alpha$ is either $1$ (which it cannot be since $\alpha=1$ is not a root of $f(x)$), or $7$. So the order of $\mathbb{F}_5(\alpha)$ is of the form $7k+1$, as well as a power of $5$.
Since neither $5^2$ nor $5^3$ are congruent to $1$ modulo $7$, $f(x)$ does not have an irreducible factor of degree $2$ or $3$. So $f(x)$ must be irreducible over $\mathrm{GF}(5)$.
(In fact, $5^6$ is the smallest power of $5$ that is $1$ mod $7$, so $\mathrm{GF}(5^6)$ is the smallest extension of $\mathrm{GF}(5)$ that has a primitive $7$th root of unity.)
Arturo Magidin
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