Degree of splitting field for $x^3-3x-1$ over $\Bbb Q$ and $\Bbb F_5$

Question

I want to find the degree of the splitting field for $x^3-3x-1$ over $\Bbb Q$ and $\Bbb F_5$.

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My attempt is contained below.

Answer 1

Since $f(x)= x^3-3x-1$ is cubic, if it can be reduced, it will have a linear factor, however by the rational root test, $\pm1$ are not roots, hence $f$ is irreducible over $\Bbb Q$.

$\text{disc}(f(x))=-4(-3)^3-27(-1)^2=4(27)-27=3^4=9^2$ Since this cubic has a square discriminant, we have $\text{Gal}(L/K)=A_3$ and hence $[L:K]=3$

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We see that over $\Bbb F_5$, $f(x)=x^3-3x-1$ has $f(0)=4,f(1)=2 ,f(2)=1,f(3)=2,f(4)=1$ so that $f$ is irreducible over $\Bbb F_5$.

Now $\Bbb F_5$ is a finite field, so is perfect and any algebraic extensions shall be perfect.

(WE CAN EMBED $\text{Gal}(L_2/\Bbb F_5)$ INTO $\text{Gal}(L/\Bbb Q)$ not sure how to write this rigorously)

Answer 2

I will only discuss the splitting field of $f$ over $\Bbb{F}_5$ since you have already done the splitting field over $\Bbb{Q}$.

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Let $L,L'$ be the splitting fields of $f$ over $\Bbb{F}_5$ and $\Bbb{Q}$ respectively. Your method for the splitting field of $f$ over $\Bbb{F}_5$ uses Dedekind's Theorem: if $\phi\in \text{Gal}(L/\Bbb{F}_5)$, then $\text{Gal}(L'/\Bbb{Q})$ contains an element with the same cycle type as $\phi$, assuming $f$ is separable over $\Bbb{F}_5$. In particular, $\text{Gal}(L/\Bbb{F}_5)$ cannot contain a $2$-cycle as $\text{Gal}(L'/\Bbb{Q})$ does not, and so $\text{Gal}(L/\Bbb{F}_5)\cong A_3$.

There is a much simpler approach. Let $f\in\Bbb{F}_q[x]$ be an irreducible polynomial of degree $n$ and let $L$ be the splitting field of $f$ over $\Bbb{F}_q$ (here $\Bbb{F}_q$ is the finite field with $q$ elements). Then, $L\cong \Bbb{F}_{q^n}$.

To see this, recall that $\text{Gal}(\Bbb{F}_{q^n}/\Bbb{F_q})$ is cyclic and generated by the Frobenius $\text{Fr}:x\mapsto x^q$. If $\alpha$ is a root of $f$ in $L$, then $\Bbb{F}_{q^n}=\Bbb{F}_q(\alpha)$, and so $\alpha,\alpha^q,\cdots,\alpha^{q^{n-1}}$ are $n$ distinct roots of $f$ in $\Bbb{F}_q(\alpha)$. This shows that $L=\Bbb{F}_q(\alpha)=\Bbb{F}_{q^n}$.