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

Question

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 $2$.

Answer 1

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 $\mathbb{F}_5$

Answer 2

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 $\mathbf F_7[\omega]$.

Answer 3

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.

Answer 4

From "Lehrbuch der Algebra (Gerd Fischer)":

Let $p$ be prime, $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ the finite field of characteristic $p, n \in \mathbb{N}\setminus\{0\}$ and $q := p^n$

• If $f \in \mathbb{F}_p[X]$ is irreducible with $deg f = n$, then $\mathbb{F}_q \cong \mathbb{F}_p[X] / (f)$
• $\forall \alpha, f(\alpha) = 0: \alpha$ is a primitive element of $\mathbb{F}_q / \mathbb{F}_p \Longrightarrow \mathbb{F}_q = \mathbb{F}_p(\alpha)$

Applying this to your problem yields:

• $x^3 - 5$ is irreducible over $\mathbb{F}_7 \iff \mathbb{F}_7[X] / (x^3 - 5) \cong \mathbb{F}_{p^n} = \mathbb{F}_{7^3} = \mathbb{F}_{343}$ $$\iff [\mathbb{F}_7[X] : (x^3 - 5)] = 343$$

Since in this case $\mathbb{F}_7[X] / (x^3 - 5)$ is the splitting field of $f$ over $\mathbb{F}_7[X]$, 343 is the solution.