Degree of splitting field of $x^3-5$ over $\mathbb{F}_7$

Find the degree of the splitting field of $$f(x):=x^3-5$$ over $$F:=\mathbb{F}_7$$.

Attempt:

$$f$$ is irreduicible in $$F[x]$$ (suppose in contradiction it is reducible, thus it splits to at least one linear normalized polynomial element but this is contradiction beacuse $$f$$ has no roots in $$F$$). Thus, if $$E$$ is the splitting field of $$f$$ over $$F$$, we get $$F\subset F(5^{1\over3})\subseteq E$$. So I get that the degree is at least $$2$$.

If $$[E:F]=2$$ and $$\alpha^3=5$$ then there exists some polynomial $$g$$ of degree $$2$$ such that $$g(\alpha)=0$$. But then $$f(x)=g(x)q(x)+r(x)$$ for some polynomials $$q$$ and $$r$$, and the degree of $$r$$ is at most $$1$$. Therefore $$0=f(\alpha)=q(\alpha)q(\alpha)+r(\alpha)=r(\alpha)$$ So $$r=0$$. This is a contradiction because $$f$$ is irreducible.

• Why does $|E:F|=2$ imply the existence of $g$? – J. Doe May 2 at 15:17
• It is the very definition of the degree of an extension. – ajotatxe May 2 at 17:48

Hint: $$f(x)=x^3-5$$ is irreducible over $$\Bbb F_7$$ since it has no root.

Further reference:

Splitting field of $x^3 - 2$ over $\mathbb{F}_5$

• I noticed that but not sure how it helps. – J. Doe May 2 at 14:54
• And if $\alpha$ is one root then $\alpha^7$ is another... – Jyrki Lahtonen May 2 at 14:55

Hint:

If $$\omega$$ is a root of $$x^3-5$$ in some extension (it has no root in $$\mathbf F_7$$), search for cube roots in $$\mathbf F_7$$. Deduce the splitting field is $$\mathbf F_7[\omega]$$.