Show that the polynomial $F(x)=x^4-7$ is irreducible in $\mathbb{F}_5$ 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? 
 A: 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.
A: $F$ 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 $\alpha\ne0$.
A: 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.
