Galois group of $f(x)=x^5+2x+1\in\mathbb{Z}_3[x]$

Question

consider $f(x)=x^5+2x+1\in\mathbb{Z}_3[x]$,what is the splitting field of $f$ and its Galois group?

I know it is a Galois extension. But i do not know the degree of the extension for i have no idea of the root of $f$, the answer is the cyclic group of order 5.

any idea will be helpful!

Answer 1

If $f$ is irreducible then $\Bbb{F}_3[x]/(f)$ is a field of $3^5$ elements. Since all fields of $3^5$ elements are isomorphic, adjoining any root of $f$ yields the same field (up to isomorphism) and hence this field is a splitting field of $f$, and so $$|\operatorname{Gal}(f)|=[\Bbb{F}_3[x]/(f):\Bbb{F}_3]=5.$$ Every group of order $5$ is cyclic.

Answer 2

Hint: if $f$ is irriducible then you can consider the field exstension $F'=\Bbb Z_3[x]/I$, where $I:=(f(x))$, the ideal generated by $f$. Now for the theorem of Kronecker exist a root $\alpha$ of $f$ ( $\alpha:= x+I$). The elements of $F'$ are $$F'=\{a+b\alpha+c\alpha ^2+d\alpha ^3+e\alpha ^4 : a,b,c,d,e \in \Bbb Z_3[x]\}.$$ Using the euclidean division we obtain: $$x^5+2x+1=(x^4+\alpha x^3+\alpha ^2x^2+\alpha ^3x+(2+\alpha ^4))(x-\alpha ).$$ Now yuo can prove that the other roots belong to $F'$. This show that $F'$ is the splitting field of $f$ and that $G(F'/F)=Z_5$.