# Polynomial in $\Bbb Q[X]$ with splitting field having Galois group $\Bbb Z_6$ over $\Bbb Q$

I would like to find a polynomial in $\Bbb Q[X]$ whose splitting field has Galois group $\Bbb Z_6$.

Since $\Bbb Z_6$ can be embedded in $S_n$ only when $n \ge 5$, the degree of the polynomial must be $\ge 5$.

Since $\Bbb Z_6 \cong \Bbb Z_2 \times \Bbb Z_3$, our polynomial is reducible.

Aiming for a minimal example, my polynomial would have degree exactly $5$ and would reduce into two irreducible polynomials, one with degree $2$ and one with degree $3$.

To prevent the one with degree $3$ from contributing $S_3$ instead of $\Bbb Z_3$, its discriminant must be a rational square, so it may for example be $x^3-3x+1$. It is irreducible by rational root theorem.

Our factor of degree $2$ could be $x^2-2$.

Therefore, our polynomial is $(x^2-2)(x^3-3x+1) = x^5-5x^3+x^2+6x-2$.

Am I correct?

• Yes. If $L/\mathbb{Q}$ is an abelian extension $\text{Gal}(L/\mathbb{Q})\simeq \mathbb{Z}_6$ then let $\beta \in L, [\mathbb{Q}(\beta):\mathbb{Q}] = 2$ and $\alpha \in L, [\mathbb{Q}(\alpha):\mathbb{Q}] = 3$ both abelian extensions so that $L = \mathbb{Q}(\alpha,\beta)= \mathbb{Q}(\alpha,\sqrt{d})$ with $d \in \mathbb{Q}\setminus \mathbb{Q}^2$ and $\alpha$ the root of an irreducible rational cubic with square discriminant. – reuns Oct 10 '17 at 7:40
• @reuns Try to make it an answer :) – Kenny Lau Oct 10 '17 at 7:45
• @reuns Like $x^7-1$? – Kenny Lau Oct 10 '17 at 7:46
• Then you can use this to write what looks like a primitive element and its minimal polynomial. I'd say the splitting field of $x^7-1$ contains $\sqrt{-7}$ and $\zeta_7+\zeta_7^{-1}$ – reuns Oct 10 '17 at 7:49
• You say, “Since $\Bbb Z_6\simeq\Bbb Z_2\times\Bbb Z_3$, our polynomial is reducible.” This does not follow. The cyclotomic polynomial $\Phi_7(X)$ is irreducible but the Galois group is $\Bbb Z_6$. – Lubin Oct 13 '17 at 4:55