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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.

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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

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

and

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

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).

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  • $\begingroup$ 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
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    $\begingroup$ @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

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