What (permutation) groups can occur as galois groups of irreducible polynomials of degree n - MathOverflow

What (permutation) groups can occur as galois groups of irreducible polynomials of degree n

2010-11-19T21:03:14Z

I think the answers for the first few degrees ($n$) are:

$n=2$, $S_2$

$n=3$, $S_3,A_3$

$n=4$, $S_4,A_4,D_4,\mathbb{Z}_4,K_4$ ($K_4$ is the Klein four group)

$n=5$, $S_5,A_5,D_5,\mathbb{Z}_5,Fr_5$ ($Fr_5$ is a Frobenius group)

What results do we have for higher orders and are there any results for a general $n$?

Edit: Sorry, I have aptly demonstrated that I am still rough on some of my concepts here. I believe I was thinking about the case where $f(x)\in \mathbb{Z}[x]$ and the base field is a finite extension of the rational numbers. I hope I have gotten it right this time and the above answers are right for this situation.

Answer by Joël for What (permutation) groups can occur as galois groups of irreducible polynomials of degree n

2010-11-19T23:43:51Z

Let $G$ be a transitive subgroup of $S_n$. Let $L=\mathbb{C}(X_1,\dots,X_n)$. The group $G$ acts on $L$ by permutations of the variables. Let $K=L^{G}$. Then $L/K$ is Galois of Galois group $G$. Let $P$ be the minimal polynomial of $X_1$ over $G$. It is irreducible. It has degree at least $n$ because all the $X_i$ are Galois conjugate of $X_1$ over $K$, by transitivity of $G$. On the other hand, it divides $(X-X_1)\dots(X-X_n)$ which has coefficients in $L^{S_n} \subset K$. Hence $P$ has degree $n$. The splitting field of $P$ over $K$ is $L$ obviously, so its Galois group is $G$.

Conclusion: the subgroups of $S_n$ that are the Galois group of an irreducible polynomial are the transitive ones.

Answer by Alex Bartel for What (permutation) groups can occur as galois groups of irreducible polynomials of degree n

2010-11-20T03:09:03Z

This question is ambiguous and can be understood in various ways, since no mention is made of the base field. If you are asking "for what subgroups $G$ of $S_n$ do there exist fields $L,K$ such that $L/K$ is a Galois with Galois group $G$", then Joël has given a complete answer. If you fix the base field $K$ and ask what groups can arise as Galois groups over $K$, then the question can be anything between trivial and open, depending on $K$. Here are are some examples of the spectrum of hardness:

If the base field is finite, then of course finite $G$ can be realised as a Galois group if and only if $G$ is cyclic.
If $K$ is a local field with finite residue field, then a necessary condition for $G$ to be realisable as a Galois group over $K$ is that $G$ be solvable. There are other restrictions as well, coming from the filtration on $G$ by higher ramification groups. However, not all finite solvable groups can be realised over a given local field. A fun exercise (if you know about the structure of higher ramification groups) is: classify all primes $p$ and $q$ such that the dihedral group $D_{2q}$ is realisable as a Galois group over the field $\mathbb{Q_p}$. Actually, the absolute Galois groups of local fields are fairly well understood. They are certain pro-solvable groups explicitly given by generators and relations (see Neukirch's book on Cohomology of Number fields). So in principle, given any finite group, you could try checking whether it's a quotient of the absolute Galois group (which might still be very difficult).
If $K$ is a number field, then the question you are posing is a famous open problem. Shafarevich proved that all solvable finite groups are Galois groups over $\mathbb{Q}$ and people are chipping away at special cases of insoluble groups, but so far, no uniform picture is apparent.

Note, that a generalisation of the Inverse Galois Problem over a given field, which should help answering the inverse Galois problem over all finite extensions of your given field, is the so-called Embedding problem.