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The cyclotomic polynomials $\Phi_n$ have Galois group $(\mathbb Z/n\mathbb Z)^\times$.

What other families of polynomials are there with known Galois groups like this?

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The Galois group of cyclotomic fields is not cyclic, in general. –  Mariano Suárez-Alvarez May 1 '11 at 21:20
    
@Mariano, thank you! That was a really silly mistake! –  quanta May 1 '11 at 21:25
    
Please edit the second line with good English. I am not able to understand it. –  Dinesh May 1 '11 at 22:08
    
@Dinesh, Is it okay now? –  quanta May 1 '11 at 22:09
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@quanta: "What are some other families of polynomials with known Galois groups?" (add a verb, and you can remove "like this" without changing the meaning) –  Jack Schmidt May 2 '11 at 0:07

4 Answers 4

up vote 7 down vote accepted

Your question is esssentially the starting point of the inverse Galois problem (over $\mathbb{Q}$): given a finite group $G$, construct an extension (or a family of extensions) of $\mathbb{Q}$ with this Galois group or prove that no such extension exists. In this generality, i.e. for arbitrary groups, this is still an open problem. There are many variants, e.g. changing the base field, or imposing conditions on the ramification in the extension. If you google for the inverse Galois problem, you find lots of literature on concrete families of groups.

Here is an extremely useful result in Inverse Galois theory:

Theorem (Hilbert): Let $\mathbb{Q}(T)$ be the function field in one variable over $\mathbb{Q}$ and let $f(T,X)\in\mathbb{Q}(T)[X]$ be an irreducible polynomial over this field. If the splitting field of $f$ has Galois group $G$, then there exist infinitely many $b\in \mathbb{Q}$ such that the splitting field of $f(b,X)$ over $\mathbb{Q}$ has Galois group $G$.

However, this result is non-constructive, i.e. it doesn't tell you where to find suitable $b$, so doesn't quite answer your question. But it does tell you that infinitely many polynomials with a certain Galois group exist, provided you can realise that Galois group over the function field.

For concrete families, here are some examples, that haven't been mentioned so far:

  • If you google for papers of Jensen, Yui and collaborators, you will see several papers on parametric families of polynomials with dihedral Galois groups, or more generally with certain Frobenius Galois groups.

  • Starting with elliptic curves over $\mathbb{Q}$ without complex multiplication and adjoining their $p$-torsion can give you Galois groups $\text{GL}_2(\mathbb{F}_p)$. In principle, you can write down parametric families of the corresponding polynomials for any fixed $p$ by writing down parametric families of elliptic curves with surjective image of Galois on $\text{Aut}(E[p])$, and by writing down their $p$-division polynomials, but it is going to be a lot of work.

  • Adjoining torsion points of higher-dimensional abelian varieties, you can get other interesting matrix groups. Again, googling something like "l-adic image Galois abelian variety" will give lots of examples.

There is also active research on the question: if $G$ is an extension of $N$ by $H$ and you managed to realise $N$ as a Galois extension of $\mathbb{Q}$ and $H$ as a Galois extension of a bigger field, can you realise $G$ as a Galois extension of $\mathbb{Q}$? Google "extension problem".

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Here's a general theorem, which might be what you're looking for: If $f$ is an irreducible rational polynomial of prime degree $p$ with exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_p$.

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Here is one family of polynomials, the Rikuna polynomials, I happen to have studied this past summer at an REU. The main result about Rikuna polynomials is:

Let $\ell$ be an odd prime. Let $K$ be a field whose characteristic does not divide $\ell$. Let $\zeta$ be a primitve $\ell$-th root of unity in some field containing $K$. We assume that $\zeta+\zeta^{-1}\in K$ but $\zeta\notin K$. Define $$p=\frac{\zeta^{-1}(x-\zeta)^\ell - \zeta(x-\zeta^{-1})^\ell}{\zeta^{-1}-\zeta}, \hskip0.5in q = \frac{(x-\zeta)^\ell -(x-\zeta^{-1})^\ell}{\zeta^{-1}-\zeta}.$$ Then $r=p-Tq\in K(T)[x]$ has Galois group $\mathbb{Z}/\ell\mathbb{Z}$ over $K(T)$.

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So the coefficients of $r$ involve $\zeta$, but $\zeta$ may not be in $K$, so $r$ isn't really in $K(T)[x]$, is it? –  Gerry Myerson May 2 '11 at 2:55
    
@Gerry: Whoops, I meant to assume $\zeta+\zeta^{-1}\in K$ but $\zeta\notin K$ (and now that that's corrected, it's clear $r\in K(T)[x]$ because it's invariant under $\zeta\mapsto\zeta^{-1}$). –  Zev Chonoles May 2 '11 at 3:51

You should be able to work out the group of $x^n-2$ for all $n$ (or, more generally, the group of $x^n-a$, with some mild restrictions on how the integers $n$ and $a$ relate).

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See Kummer theory. –  lhf May 2 '11 at 0:49
    
If I read that right, Kummer theory only applies when the ground field has $n$th roots of unity, so it's only part of the story for, say, $x^n-2$ over the rationals. I'm just trying to encourage quanta to look at this case. I think one can handle it by brute force, without appeal to notions beyond the basics. –  Gerry Myerson May 2 '11 at 5:07

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