I'm trying to determine which subgroups of $S_5$ occur as the Galois group of an irreducible quintic $f\in\Bbb{Z}[X]$. I know such a subgroup of $S_5$ should be transitive, leaving only five possibilities (up to isomorphism/automorphism), which are determined by their size. It is not hard to find quintic $f\in\Bbb{Z}[X]$ with Galois group of order $120$ or $20$, and with some effort I've found quintic $f\in\Bbb{Z}[X]$ with Galois group of order $60$. That leaves the problem of finding irreducible $f\in\Bbb{Z}[X]$ with Galois groups of order $5$ and $10$, or showing that no such $f\in\Bbb{Z}[X]$ exists.

If such $f\in\Bbb{Z}[X]$ exist then their Galois group is contained in $A_5\subset S_5$, hence $\Delta f$ is a square in $\Bbb{Z}$. However, I have no clue where to start looking for such $f$; the discriminant of a quintic is a huge and nasty expression to work with, and I haven't had any succes restricting to polynomials of the form $f=x^5+ax+b$, though I haven't tried much as I don't know what to try.

In general (this is perhaps a different question), given a (small) finite group $G$ that is a transitive subgroup of $S_n$, how would one go about determining whether there exists an irreducible $f\in\Bbb{Z}[X]$ with Galois group isomorphic to $G$? And how to find one if it exists? I'm not expecting any algorithmic answer as I think that would solve the inverse Galois problem, though this would be welcome, but some heuristic method would be appreciated.


For abelian subgroups, in particular the cyclic group $C_5$ of order $5$ in this case, the Kronecker-Weber theorem tells you that the splitting field of the polynomial is a subfield of a cyclotomic field $\mathbb{Q}(\zeta_n)$ for some $n$. Galois theory can be used to complete classify the subfields of $\mathbb{Q}(\zeta_n)$ and their Galois groups: for every subgroup $H \subseteq G$ where $G = \text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong \mathbb{Z}_n^{\times}$ is the Galois group, the fixed field of $H$ is a subfield of $\mathbb{Q}(\zeta_n)$ with Galois group $G/H$.

Hence we can find an irreducible polynomial $f$ of degree $5$ with Galois group $C_5$ as follows. The smallest $n$ for which $\mathbb{Z}_n^{\times}$ has a quotient isomorphic to $C_5$ is $n = 11$, where $\mathbb{Z}_{11}^{\times} \cong C_{10}$. Hence we can take the splitting field of $f$ to be the fixed field of a subgroup of $C_{10}$ of order $2$. There is a unique such subgroup generated by complex conjugation $\zeta_{11} \mapsto \zeta_{11}^{-1}$, so we can take the splitting field of $f$ to be the real subfield

$$\mathbb{Q}(\zeta_{11} + \zeta_{11}^{-1}) = \mathbb{Q} \left(2 \cos \frac{2\pi}{11} \right).$$

In particular, we can take $f$ itself to be the minimal polynomial of $\alpha = 2 \cos \frac{2\pi}{11} = \zeta_{11} + \zeta_{11}^{-1}$, which can be computed in terms of Chebyshev polynomials as follows. Let $T_n(x)$ denote the unique polynomial of degree $n$ such that

$$T_n(2 \cos \theta) = 2 \cos n \theta$$

for all $\theta \in \mathbb{R}$, or equivalently such that

$$T_n(z + z^{-1}) = z^n + z^{-n}$$

for all $z \in \mathbb{C}$. For example, $T_2(x) = x^2 - 2$. (This is a different normalization than Wikipedia is using, but I prefer it because it makes $T_n(x)$ monic.) Then we have

$$T_5 \left( 2 \cos \frac{2\pi}{11} \right) = \cos \frac{10\pi}{11} = -\cos \frac{\pi}{11}$$


$$T_6 \left( 2 \cos \frac{2\pi}{11} \right) = \cos \frac{12\pi}{11} = -\cos \frac{\pi}{11}$$

from which it follows that $T_5(\alpha) = T_6(\alpha)$. This gives a polynomial of degree $6$ satisfied by $\alpha$ which has an additional root of $2$ (since $T_5(2) = T_6(2) = 2$), so dividing out by that factor gives a polynomial of degree $5$ satisfied by $\alpha$ which must be its minimal polynomial. The resulting polynomial $f$ must have Galois group $C_5$ (and incidentally this computation makes no use of the Kronecker-Weber theorem and is just basic Galois theory, but the Kronecker-Weber theorem told us where to look).

  • $\begingroup$ My (belated) thanks for your very clear answer, it has been a great help. However, the case of a quintic $f\in\Bbb{Z}[X]$ with Galois group of order $10$ is left unanswered. I have found that $$f=X^5+X^4-5X^3-4X^2+3X+1,$$ with discriminant $401^2$ does the trick, but I can only show it by means of 'magic'. Is there a method to find such polynomials like the method you describe for abelian groups? $\endgroup$ – Servaes Jul 1 '15 at 12:03
  • 1
    $\begingroup$ @Servaes: not sure off the top of my head. Probably it's known how to get Galois group the dihedral group in general. Again I would start by first trying to find a Galois extension with that Galois group and only then finding a polynomial (by finding a primitive element). A dihedral extension is a cyclic extension of a quadratic extension, so probably Kronecker's Jugendtraum is relevant. But maybe that's working too hard. You could try asking this as a separate question. $\endgroup$ – Qiaochu Yuan Jul 2 '15 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.