Showing $x^4 + 2$ is irreducible in $F_5[x]$ There is a similar question like this here
but I don't understand the solution. Since this is a fourth degree and since it has no root in $F_5$, it can only have a quadratic reduction. But how do we rule out the possibility that $x^4 + 2 = (x^2 + ax + b)(x^2 + cx + d)?$ Where the quadratic is irreducible in $F_5$?
There are a lot of similar questions like this on my practice exams.
Thanks
 A: If we defined the Frobenius map by $F(a)=a^5$, then $a$ lies in $\Bbb F_{5^k}$
iff $F^k(a)=a$.
Here let $a$ be a root of $x^4+2=0$. Then $a^4=-2$ and $F(a)=a^5=-2a\ne a$.
So $a\notin\Bbb F_5$.
Then $F^2(a)=F(-2a)=(-2a)^5=4a=-a\ne a$. Therefore $a\notin\Bbb F_{25}$.
Then $F^3(a)=F(-a)=(-a)^5=2a\ne a$. Therefore $a\notin\Bbb F_{125}$.
Of course, $F^4(a)=F(2a)=-4a=a$, so $a\in\Bbb F_{625}$.
Therefore $a$ generates the field extension of degree $4$ over $\Bbb 5$, and
so $X^4+2$ is ireeducible.
A: Assume $x^4+2$ factors in $\mathbb{Z}_5[x]$ as
$$x^4+2=(x^2+ax+b)(x^2+cx+d)$$
If the $\text{RHS}$ is expanded,

*

*The coefficient of the $x^3$ term is $a+c$, hence we must have $c=-a$.$\\[4pt]$

*The constant term is $bd$, hence we must have $d={\large{\frac{2}{b}}}$.  

hence the factorization can be rewritten as
$$x^4+2=(x^2+ax+b)(x^2-ax+\frac{2}{b})$$
Expanding, we get
$$x^4+2=x^4+\left(b+\frac{2}{b}-a^2\right)x^2+\left(\frac{2a}{b}-ab\right)x+2$$
hence we must have
$$
\left\lbrace
\begin{align*}
&\frac{2a}{b}-ab=0&&(\text{eq}1)\\[4pt]
&b+\frac{2}{b}-a^2=0&&(\text{eq}2)\\[4pt]
\end{align*}
\right.
$$
If $a\ne 0$, $(\text{eq}1)$ yields $b^2=2$, and if $a=0$, $(\text{eq}2)$ yields $b^2=-2$.

Either way we have a contradiction since $2$ and $-2$ are not squares in $\mathbb{Z}_5$.
A: Here's another proof . . .

Assume $x^4+2$ factors in $\mathbb{Z}_5[x]$ as
$$x^4+2=(x^2+ax+b)(x^2+cx+d)$$
Let $K$ be an algebraic closure of $\mathbb{Z}_5$, and let $r$ be a root in $K$ of the polynomial $x^4+2$.

Then identically we have
$$r^4=(-r)^4=(2r)^4=(-2r)^4$$
hence $\pm r,\pm 2r$ are roots in $K$ of $x^4+2$, and are distinct since $r\ne 0$.

Since $b$ is the product of the roots in $K$ of $x^2+ax+b$, it follows that $b$ is the product of two distinct elements of the set $\{\pm r,\pm 2r\}$, hence $b\in\{-r^2,\pm 2r^2,-4r^2\}$.

In any case, since $b\in\mathbb{Z}_5$, it follows that $r^2\in\mathbb{Z}_5$.

But $r^2\in\mathbb{Z}_5$ implies $r^4$ is a square in $\mathbb{Z}_5$, contradiction, since $r^4=-2$ which is not a square in $\mathbb{Z}_5$.
