Galois group of $x^6+3$ over $\mathbb{F}_5$
问题内容
One question from a past qualifying exam is to find the Galois group of $x^6+3$ over the finite field $\mathbb{F}_5$. With some creativity, we can write: $$x^6+3=x^6+8=(x^2)^3+2^3=(x^2+2)(x^4-2x^2+4)=(x^2+2)(x^4+4x^2+4-16x^2)\\ =(x^2+2)((x^2+2)^2-(4x)^2)=(x^2+2)(x^2-2x+2)(x^2+2x+2).$$
Since $x^6+3$ does not have any root in $\mathbb{F}_5$, each degree-2 polynomial is irreducible. Each term has the same splitting field, which is $\mathbb{F}_5(\alpha); \alpha^2=-2.$ Consequently, the Galois group is $\mathbb{Z}/2\mathbb{Z}$.
Now, I am wondering if we are not able to do the factorization, what would be the standard approach to solve this problem? I was hopeful to take advantage of $x^6+3=x^2+3 \mod5$, but it seems $x^2+3$ does not even divide $x^6+3$ in $\mod5$. So the only general way that comes to my mind is to check whether there exist $(a,b,c,d,e,f)$ satisfying $$(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)=x^6+3$$ or $$(x^2+ax+b)(x^4+cx^3+dx^2+ex+f)=x^6+3.$$
If not, then the Galois group would be $\mathbb{Z}/6\mathbb{Z}$. If the first case is satisfied, then the Galois group would be $\mathbb{Z}/3\mathbb{Z}$, and if the last case is satisfied, then the Galois group would be $\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}/4\mathbb{Z}$ depending on the irreducibility of the degree 4 term. This approach seems not appropriate for the exam, so what do you suggest? Only creativity, or is there something to take advantage of?
回答 (3)
For $f(x) = x^6 +3$ over $\mathbf{F}_5$, let $\alpha$ be any root of $f$. If $\alpha \in \mathbf{F}_5$, this would imply $\alpha^4 = 1$ and hence $\alpha^2 + 3 =0$, which is not possible in $\mathbf{F}_5$. Thus $f$ does not have any root in $\mathbf{F}_5$. Moreover, since $\alpha^{24} = (-3)^4 =1$, this implies $\alpha \in \mathbf{F}_{25}$. Thus the splitting field of $f$ is $\mathbf{F}_{25}$, and its Galois group is of order $2$ and equal to $\mathbf{Z}/2\mathbf{Z}$.
The Galois group is cyclic, generated by the Frobenius map $F\colon\alpha\mapsto\alpha^5$. If $K$ is the splitting field of $x^6+3$ and $\alpha\in K$ is a root of the polynomial then $F(\alpha) = \alpha^6/\alpha = -3/\alpha$, and so $F(F(\alpha))=\alpha$. This means that $F\in\operatorname{Gal}(K/\mathbb{F}_5)$ satisfies $F^2=1$ and so the order of $\operatorname{Gal}(K/\mathbb{F}_5)$ is at most $2$. Since $K$ is strictly larger than $\mathbb{F}_5$ (the polynomial has no root in $\mathbb{F}_5$, since such a root would satisfy also $\alpha^2+3=0$, which has no solution in $\mathbb{F}_5$), the order must be $2$.
In $F_5$ all non-zero $x$ satisfy that $x^4=1$. This shows immediately that the polynomial does not split in $F_5$. Alternatively just note that a primitive 24th root of unity in $F_{25}$ satisfies either $x^6+3=0$ or $x^{18}+3=0$ which means that you have an element of order $8$ or $24$ in the splitting field which is not possible in $F_5$. (In fact you can show that there exists a primitive 24th root of unity which satisfies that $x^6+3=0$.)