Existence of irreducible polynomials with certain criteria

Question

Let $\mathbb{F}_{q}$ be the finite field with $q$ elements, where $q$ is an odd prime power. The question is as follows:

Does there exists $a\in \mathbb{F}_{q}^*\setminus (\mathbb{F}_{q}^*)^2$, such that $x^4-a$ is irreducible in $\mathbb{F}_{q}[x]$?

I was looking at the following theorem on Serg Lang's book "Algebra". (Pg 297 3rd ed)

Theorem: Let $k$ be a field and $n$ an integer $\geq$ 2. Let $a\in k, a\neq 0$. Assume that for all primes $p$ such that $p\mid n$, we have $a\notin k^p$, and if $4\mid n$, then $a\notin -4k^4$. Then $X^n-a$ is irreducible in $k[x]$.

I think this theorem might be of some help. I have been thinking as follows:

Let $4\mid q-1$. Let, $M=-4(\mathbb{F}_{q}^*)^4$. Then $|M|=|(\mathbb{F}_{q}^*)^4|$. Now, $|(\mathbb{F}_{q}^*)^4|=\frac{q-1}{(4, q-1)}=\frac{q-1}{4}$. But, number of non-squares is $\frac{q-1}{2}$. So, there must exist a $a\in \mathbb{F}_{q}^*\setminus (\mathbb{F}_{q}^*)^2$, such that $a\notin M$, and hence by the theorem $X^4-a$ is irreducible. I hope this is true.

I have no clue about the case $4\nmid q-1$. So, I need assistance for this.

There can be easier methods to do this, without using this theorem maybe, which I am not aware of. Thank you in advance for any kind of help!

Answer 1

Your hunch is correct. The key is whether $q\equiv1\pmod 4$ or not. Remember that the multiplicative group of the field $\Bbb{F}_{q^n}$ is cyclic of order $q^n-1$.

Assume first that $q\equiv1\pmod4$.

We can write $q-1=2^\ell r$ with $r$ odd and $\ell\ge2$. Cyclicity of $\Bbb{F}_q$ implies that we can locate an element $a\in\Bbb{F}_q^*$ of order $2^\ell$. I claim that with this specific $a$ the polynomial $f(x)=x^4-a$ is irreducible over $\Bbb{F}_q$. This is seen as follows.

Let $\alpha$ be a zero of $f(x)$ in some extension field $K$. Without loss of generality we can assume that $K=\Bbb{F}_q(\alpha)$. We know that $\alpha^4=a$. From the first course on cyclic groups, we also recall the formula for the order of a power of an element of a known order: $$ 2^\ell=\operatorname{ord}(a)= \operatorname{ord}(\alpha^4)=\frac{\operatorname{ord}(\alpha)}{\gcd(4, \operatorname{ord}(\alpha))}. $$ This formula leaves as the only possibility the result $$ \operatorname{ord}(\alpha)=2^{\ell+2}. $$

• Because $2^{\ell+2}\nmid q-1$ we can deduce that $\alpha\notin\Bbb{F}_q$. This means that $f(x)$ has no zeros in the field $\Bbb{F}_q$, and therefore no linear factors either.
• But, $q\equiv1\pmod4$ implies that $q+1\equiv2\pmod4$. Therefore $$q^2-1=(q-1)(q+1)=2^{\ell+1}\cdot r\cdot\frac{q+1}2.$$ This is not divisible by $2^{\ell+2}$, so we can conclude that there are no elements of order $2^{\ell+2}$ in the field $\Bbb{F}_{q^2}$ either. Therefore $\alpha\notin\Bbb{F}_{q^2}$. This means that $f(x)$ cannot have any quadratic factors either. This proves that $f(x)$ is irreducible over $\Bbb{F}_q$.

Assume then that $q\equiv-1\pmod4$.

I claim that in this case $x^4-a$ is never irreducible over $\Bbb{F}_q$. You may be able to guess how it goes. Let's fix $a\in\Bbb{F}_q^*$ (the case $a=0$ is not interesting). The order of $a$ is then some factor $m$ of $q-1$. Again, let $\alpha$ be a zero of $f(x)=x^4-a$ in some extension field of $\Bbb{F}_q$. Again, $$m=\operatorname{ord}(a)= \operatorname{ord}(\alpha^4)=\frac{\operatorname{ord}(\alpha)}{\gcd(4, \operatorname{ord}(\alpha))}.$$ However, because we don't know whether $m$ is divisible by two, we cannot deduce the order of $\alpha$ from this. Nevertheless, it is obvious that $\alpha^{4m}=a^m=1$, so the order of $\alpha$ is a factor of $4m$.

The upshot is that this time $4\mid q+1$. Therefore $$ 4m\mid (q+1)(q-1)=q^2-1. $$ All the solutions of $x^{q^2-1}=1$ (in, say, an algebraic closure of $\Bbb{F}_q$ or some other big umbrella field) belong to the field $\Bbb{F}_{q^2}$. In particular, $\alpha\in\Bbb{F}_{q^2}$.

We have shown that the minimal polynomial $m(x)$ of $\alpha$ over $\Bbb{F}_q$ has degree two at most. Therefore $m(x)$ is a proper factor of $f(x)$ proving the claim.