Showing $x^4 + 2$ is irreducible in $F_5[x]$

Question

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

Answer 1

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 $\mathbb{F}_5$, and so $X^4+2$ is ireeducible.

Answer 2

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 $\mathbb{Z}_5$.

Answer 3

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 $\mathbb{Z}_5$.