Finding a cubic polynomial whose splitting field over $\mathbb{Q}$ equals $\mathbb{Q}(a)$ if $a$ is any of its roots

Question

Question: Let $\alpha$, $\beta$ and $\gamma$ be the roots of a rational cubic polynomial $q$. Can we find a (non-trivial) example where the splitting field of $q$ over $\mathbb{Q}$ equals $\mathbb{Q}(\alpha), \mathbb{Q}(\beta)$ and $\mathbb{Q}(\gamma)$?

Comments: This is from an old algebra exam. Clearly one solution (the one given by the professor) is $q(x)=(x-1)^3$ where the splitting field is $\mathbb{Q}(1) = \mathbb{Q}$, but I was wondering: does there exist a $q$ as described above so that $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$ = $\mathbb{Q}(\gamma)$ but still so that they are different from $\mathbb{Q}$? If not, can we prove that it does not exist?

Answer 1

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Yes. The basic construction is actually quite simple. You want a Galois extension to be generated by your root, so that--by normality--any separable polynomial with a single root in the field splits completely.

Since the degree is $3$, this means we want an extension with Galois group $\Bbb Z/3\Bbb Z$. By general theory we know that all abelian extensions are contained within cyclotomic extensions (or at least that all cyclotomic extensions are abelian, that's enough to find an example.)

So let us look at numbers $n$ such that $3|\varphi(n)=[\Bbb Q(\zeta_n):\Bbb Q]$, where here $\zeta_n$ is a primitive $n^{th}$ root of unity and $\varphi$ is Euler's totient function.

Such a number is $n=7$. Then within $\Bbb Q(\zeta_7)$ we have a Galois subfield of degree $3$, which is fixed, therefore, by any element of order $2$ because $\varphi(7)=6=2\cdot 3$. Since Complex conjugation is one such (indeed the only one), we see that $\Bbb Q\left(\cos\left({2\pi\over 7}\right)\right)=\Bbb Q\left(\zeta_7+\zeta_7^{-1}\right)=\Bbb Q(\zeta_7)\cap\Bbb R$ is a Galois extension of $\Bbb Q$ of degree $6/2=3$. Hence the polynomial $q$ can be taken to be the minimal polynomial of $\zeta_7+\zeta_7^{-1}$ over $\Bbb Q$.

Furthermore, if you want to compute the polynomial explicitly, you can always do the trick where you compute the characteristic polynomial of multiplication by $\zeta_7+\zeta_7^{-1}$ on the basis $\{1,\zeta_7+\zeta_7^{-1}, \zeta_7^2+\zeta_7^{-2}\}$ for $\Bbb Q\left(\zeta_7+\zeta_7^{-1}\right)$, which gives the matrix

$$\begin{pmatrix} \cdot & |& 1 & \zeta_7+\zeta_7^{-1} & \zeta_7^2+\zeta_7^{-2} \\ --- &\cdot & --- & --- & ---\\ 1 & | & 0 & 1 & 0 \\ \zeta_7+\zeta_7^{-1} & | & 2 & 0 & 1 \\ \zeta_7^2+\zeta_7^{-2} & | &-1 & 0 & -1 \end{pmatrix}$$

with characteristic polynomial

$$q(x) = \begin{vmatrix} x & -1 & 0 \\ -2 & x & -1 \\ 1 & 0 & x+1 \end{vmatrix} = x^3 +x^2-2x-1$$

which is an explicit example.

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Nitty gritty in case you haven't done many of those computations:

The idea is to take a basis for $\Bbb Q\left(\zeta_7+\zeta_7^{-1}\right)$ and look at the matrix which comes out of multiplying this basis by $\zeta_7+\zeta_7^{-1}$. I chose the basis $\{1, \zeta_7+\zeta_7^{-1}, \zeta_7^2+\zeta_7^{-2}\}$. How does multiplication by $\zeta_7+\zeta_7^{-1}$ affect this?

It does this

$$\begin{cases} 1\mapsto \zeta_7+\zeta_7^{-1} \\ \zeta_7+\zeta_7^{-1}\mapsto (\zeta_7+\zeta_7^{-1})^2=2\cdot 1+1\cdot(\zeta_7^2+\zeta_7^{-2}) \\ \zeta_7^2+\zeta_7^{-2}\mapsto \zeta_7^3+\zeta_7^{-3}+\zeta_7+\zeta_7^{-1} \end{cases}$$

Now we know that the sum of all $n^{th}$ roots of unity is $0$, so we do write that here as

$$1+\zeta_7+\zeta_7^2+\zeta_7^3+\zeta_7^{-3}+\zeta_7^{-2}+\zeta_7^{-1}=0$$

so that we have $\zeta_7^3+\zeta_7^{-3}=-1-(\zeta_7+\zeta_7^{-1})-(\zeta_7^2+\zeta_7^{-2})$ which makes the last case

$$\zeta_7^2+\zeta_7^{-2}\mapsto -1\cdot (1)\cancel{-1\cdot (\zeta_7+\zeta_7^{-1})}+\cancel{1\cdot (\zeta_7+\zeta_7^{-1})}-1\cdot(\zeta_7^2+\zeta_7^{-2})$$ $$=-1\cdot(1)-1\cdot(\zeta_7^2+\zeta_7^{-2})$$

Each of these give us the rows of the matrix.

Finally, one just computes the characteristic polynomial and that produces the $q$ I showed above.