Why is $X^4 + \overline{2}$ irreducible in $\mathbb{F}_{125}[X]$?

I want to prove that $f = X^4 + \overline{2}$ is irreducible in $\mathbb{F}_{125}[X]$.

I know that $\mathbb{F}_{125}$ is the splitting field of $X^{125} - X$ over $\mathbb{F}_5$, and that this is a field extension of $\mathbb{F}_5$ of degree 3, and that $\mathbb{F}_{125} = \mathbb{F}_5[X]/(g)$ where $g = X^3 + X - \overline{1}$.

If we put $\alpha = X + (g)$ then $\{\overline{1}, \alpha, \alpha^2\}$ is a basis for $\mathbb{F}_{125}$ considered as a vector space over $\mathbb{F}_5$. Hence every element $x \in \mathbb{F}_{125}$ can be written $x = a_0 + a_1\alpha + a_2\alpha^2$, with $a_i \in \mathbb{F}_5$.

How do I proceed?

Solution 1:

The roots of the polynomial $x^4+2=x^4-3$ are sixteenth primitive roots of unity because $3$ is a fourth primitive root of unity. This is not immediate, but it does follow. For if $\alpha$ is such a root, then $\alpha^{16}=1$. Hence the order of $\alpha$ is a factor of $16$. But $\alpha^8=3^2\neq1$, so no proper factor of $16$ will work.

The smallest extension field of $K=\mathbb{F}_5$ that contains sixteenth roots of unity is the field $E=\mathbb{F}_{625}$. This is because $16$ does not divide either $24=5^2-1$ nor $124=5^3-1$, but it does divide $624=5^4-1$. Consequently $$ f(x)=(x-\alpha)(x-\alpha^5)(x-\alpha^{25})(x-\alpha^{125}) $$ for some $\alpha\in E=K[\alpha]$, and $f$ is irreducible over $K$ as the minimal polynomial of $\alpha$.

So any factor of $f$ must have its coefficients in $E$, as the said coefficients are sums of products of conjugates of $\alpha$. Let $L=\mathbb{F}_{125}$. As $E\cap L=K$, any factor with coefficients in $L$ must have its coefficients in the prime field $K$. But we just saw that over $K$ this polynomial is irreducible, so no such factor can exist.

Solution 2:

Let me justify my comment above. Actually it was a little off in the sense that it is not necessary to check or roots in the extension field. What I want to show is the following:

Proposition 1: Let $K$ be a field, $f \in K[t]$ an irreducible polynomial of degree $d$, and let $L/K$ be a field extension of degree $n$. Then if $d$ and $n$ are relatively prime, then $f \in L[t]$ remains irreducible.

Proof: Let $\alpha$ be a root of $f$ in an algebraic closure of $K$ which contains $L$. I claim that the natural map $\Phi: K[\alpha] \otimes_K L \rightarrow K[\alpha] L$ is an isomorphism. Indeed it is always surjective, and since $K[\alpha]L$ is a field extension of $K$ containing both $K$ and $L$, its degree must be divisible by both $d$ and $n$. Since $d$ and $n$ are relatively prime, we have $[K[\alpha]L:K] = dn = \dim_K K[\alpha] \otimes_K L$. So $\Phi$ is a surjection between two $F$-algebras of the same finite dimension, hence it is an isomorphism.

Thus, since $K[\alpha] \cong K[x]/(f)$, $K[\alpha] \otimes_K L \cong L[x]/(f)$. Because this is a field, $(f)$ is a maximal ideal of $L[x]$, hence a prime ideal, so $f$ is irreducible in $L[t]$.

Added later: here is a more elementary proof of the above result. Suppose that $f$ becomes reducible in $L$, thus over $L$ there is an irreducible factor $g$ of $f$ of degree $k \leq \frac{d}{2} < d$. Let $M = L[x]/(g)$. Then $f$ has a root $\alpha$ in $M$ and $[M:K] = kn$. On the other hand, since $f$ is irreducible over $K$, the map $K[t] \rightarrow M$ which sends $t$ to $\alpha$ has kernel $(f)$ and thus we get a field homomorphism $K[t]/(f) \rightarrow M$, which shows that $d \mid kn$. Since $\operatorname{gcd}(d,n) = 1$, we get $d \mid k$, and since $k < d$, this is a contradiction.

Solution 3:

I know it's an old question, but I thought someone may benefit from seeing yet another approach, so here we go.

Let $\alpha\in\overline{\mathbb{F}_5}$ be a root of $f$. Then $\alpha^4=-2$ and so $\alpha^{16}=1$. You can check that $\alpha^m\neq 1$ for all $m<16, m\mid 16$ so the order of $\alpha$ is 16. Thus $$\alpha\in\mathbb{F}_{5^n}\iff 16\mid 5^n-1\iff 5^n\equiv 1\mod 16$$ and since the order of $5\in\mathbb{Z}/16\mathbb{Z}$ is 4, it follows that $\alpha\in\mathbb{F}_{5^n}\iff 4\mid n$. So we have that $\deg f_{\mathbb{F}_5}^\alpha=4$ and since $f_{\mathbb{F}_5}^\alpha\mid f$ it follows that $f=f_{\mathbb{F}_5}^\alpha$ is irreducible over $\mathbb{F}_5[X]$, but we don't know yet if it splits in $\mathbb F_{5^3}[X]$. As $\gcd(4,3)=1$, $$[\mathbb F_{5^3}(\alpha):\mathbb F_{5}]= [\mathbb F_{5^3}:\mathbb F_{5}]\cdot[\mathbb F_{5}(\alpha):\mathbb F_{5}]=12$$ and so $[\mathbb F_{5^3}(\alpha):\mathbb F_{5^3}]=4$. Thus $\deg f_{\mathbb F_{5^3}}^\alpha=4$ and $f=f_{\mathbb F_{5^3}}^\alpha$ is irreducible.

The nice thing about this approach is that it generalizes nicely to prove that $f$ is irreducible in $\mathbb{F}_{5^n}[X]$ for all odd $n$.