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Let $F(x)=x^4-7$. This is my attempt:

Note that $F(x)=x^4-2$ in $\mathbb{F}_5$. Then we can factor $F(x)=(x^2-a)(x^2+a)$ with $a^2=2, a\in \mathbb{F}_5$. But checking all elements of $\mathbb{F}_5$ we conclude that no such $a$ exists. None of the elements of $\mathbb{F}_5$ is a root for this polynomial so we cannot factor it into a linear and a cubic term. Therefore this polynomial is irreducible in $\mathbb{F}_5$. Is this enough?

mandella
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    Are you sure it couldn't have a factorisation $F(x)=(x^2+bx+c)(x^2-bx+c)$? – Angina Seng Nov 28 '19 at 17:22
  • So $c^2=-2$ which also has no solution? – mandella Nov 28 '19 at 17:26
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    I answered this as a part of this question by giving my algorithm for checking irreducibility of a binomial modulo a prime. $7=2$ has order four modulo five, so the zeros of $F(x)$ must be roots of unity order sixteen. The smallest extension field of $\Bbb{F}5$ containing roots of unity of order sixteen is $\Bbb{F}{5^4}$. Therefore the minimal polynomials of the zeros have degree four. Therefore $F(x)$ is irreducible. – Jyrki Lahtonen Nov 28 '19 at 18:52
  • Thank you for the information, I have not seen roots of unity yet but I am sure soon enough I will and this seems interesting to keep in mind. – mandella Nov 28 '19 at 18:54
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    Actually lhf used a similar technique in their answer. If the polynomial had a quadratic factor, the roots raised to power $24$ would have to be $1$. But $7^6\not\equiv1\pmod5$ so this is not the case. – Jyrki Lahtonen Nov 28 '19 at 18:56
  • yes, I can follow this, thank you for the comparison – mandella Nov 28 '19 at 18:58

3 Answers3

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In general, the factorization of a polynomial $F$ in $\mathbb F_p$ can be done by calculating the $\gcd$ of $F$ and $x^{p^d} - x$ for various $d$.

For this example: $\gcd(x^{5^2} - x, F) = 1$, so $F$ has no root in $\mathbb F_{5^2}$. Therefore all $4$ roots lie in $\mathbb F_{5^4}$ and the polynomial is irreducible.

The advantage of this method is that it gives an algorithm, thus works in general cases and can be performed by a computer.

WhatsUp
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    This is correct. I just want to point out that with a binomial of the form $x^n-a$ we can do it a bit faster by figuring out the multiplicative order of $a$, and then hopefully also of the zeros of the polynomial. – Jyrki Lahtonen Nov 28 '19 at 18:58
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$F$ has no roots in $\mathbb{F}_5$ and so no factors of degree $1$ or $3$.

If $F$ had a factor of degree $2$, then there would be $\alpha \in \mathbb{F}_{25}$ such that $\alpha^4=7=2$. Then $\alpha = \alpha^{25} = \alpha^{24} \alpha = (\alpha^4)^6 \alpha = 64 \alpha = 4 \alpha$, a contradiction since $\alpha\ne0$.

lhf
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Suppose that there is a solution $x$ in some field extension $K$ of $\mathbb{F}$, then since $x^4 =1 \mod 5$ for all elements in $\mathbb{F}_5$, $x,2x,3x,4x$ are solutions, in fact we have four automorphisms of the equation $F(x)$.

Since powers of $2$ generate $\mathbb{F}_5$, we have a automorphism of order $4$ given by $x \mapsto 2x$. If $K$ is the field $F =\mathbb{F}_5[x]$, then as $Gal(K/F)$ has at least 4 elements ($x\to 2x$ is in the Galois group and has order 4), the degree of this extension has to have at least degree 4, but if the polynomial is reducible it would have degree 2.

Note that you have to show that the polynomial $F(x)$ doesn't have any root to use a decomposition into degree 2 polynomials.

EDIT: the entire argument above is flawed. If $F$ decomposes into two monic factors $F = F_1 F_2$ and $\alpha$ is a solution of $F(\alpha) = 0$ without loss of generality we can assume that $F_1(\alpha) = 0$. By simple arguments one can show that both polynomials are of degree 2, or $F$ would be irreducible. Note that $\alpha, 2\alpha, 3\alpha, 4\alpha$ are also solutions of $F$ since $n^4 =1 \mod 5$ for $n$ in $\mathbb{F}_5^\times$, one of these must be the other solution of $F_1$ or else the degree of $F_2$ would not be $2$. Since it has solutions $\alpha$ and $n\cdot\alpha$, we have that $F_1(x) = F_1(n\cdot x)$, and a polynomial of degree 2 is invariant under this map if and only if $n^2 = 1\mod 5$, i.e. $n = 4$. On the other hand $F_1$ cant have a term of the form $a_1 x^1$ since these are not invariant unless $a_1 = 0$. i.e. $F_1(x) = x^2 + a_0$, and by doing the product $F_1 \cdot F_2$ we can see the same of $F_2 = x^2 + b_0$ or otherwise we would have a term of the form $Cx^3$ in $F$.

By checking all possible values for $a_0,b_0$ with the condition $a_0 b_0 = -2$ and $a_0 = -b_0$ we see that this can't be the case and $F$ is irreducible.

  • I have not heard about Galois groups yet. – mandella Nov 28 '19 at 17:56
  • See the edit for a simpler proof using the symmetries i wrote. – k76u4vkweek547v7 Nov 28 '19 at 18:26
  • Why is $n\cdot x$ the other solution? I am having a hard time following your answers, so I believe I do not have the background to follow your train of thought. Maybe when I learn more, I will get back to read them again. – mandella Nov 28 '19 at 18:56
  • I think your argument is not on a solid footing. Consider instead the polynomial $p(x)=x^4+1$ over $\Bbb{F}_5$. If $\alpha$ is one of the zeros (in some extension field), by your argument, the other zeros are again $2\alpha$, $3\alpha$ and $4\alpha$. If your argument would work, we could again conclude that $p(x)$ is irreducible. But $$x^4+1=x^4-4=(x^2-2)(x^2+2)$$ is not irreducible. The trouble with your argument is that $x\mapsto 2x$ does not necessarily come from an element of the Galois group. – Jyrki Lahtonen Nov 28 '19 at 20:54
  • You are absolute rigth, I just got home so I will review this. – k76u4vkweek547v7 Nov 28 '19 at 23:48
  • @mandella if $x$ is a solution, $n\cdot x$ is another solution since $F(x\cdot x) = n^4 x^4 -2 = x^4 -2 = F(x) = 0$. – k76u4vkweek547v7 Nov 29 '19 at 00:04