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How does one construct a polynomial with Galois Group $D_{2n}$? A general method would be preferable or if that's impractical then an example of it being done for any n would be appreciated.


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Over $\mathbb{Q}$, I assume? –  Qiaochu Yuan Mar 18 '11 at 15:52

2 Answers 2

One general method proceeds by making use of invariant polynomials. Let $G$ be a candidate Galois group for an irreducible polynomial of degree $n$ over a field $F$ so that in particular $G$ has a transitive permutation action on $n$ objects $r_{i}$ which we identify with the roots of the polynomial. Basic Galois theory then tells us that any polynomial in the $r_{i}$ which is invariant under the action of $G$ must then lie in $F$. Thus $G$ stabilizes the invariant polynomial ring $F[r_{1}, \cdots, r_{n}]^{G}$.

Now, since $G$ is a subgroup of $S_{n}$ the invariant ring includes the elementary symmetric polynomials $\sigma_{i}$ defined as

$ \sigma_{1}=r_{1}+r_{2}+\cdots +r_{n}, $

$ \sigma_{2}=r_{1}r_{2}+r_{1}r_{3}+\cdots +r_{n-1}r_{n}, $

$\cdots = \cdots ,$

$\sigma_{n} = r_{1}r_{2}\cdots r_{n}. $

On the other hand in the splitting field, a polynomial with roots $r_{i}$ can be completely factored as

$\prod_{i}(z-r_{i}) = z^{n}-\sigma_{1}z^{n-1}+\sigma_{2}z^{n-2}+\cdots + (-1)^{n}\sigma_{n}.$

To construct a polynomial with Galois group $G$ we can then simply choose the $\sigma_{i}$ to be consistent with whatever relations occur in the invariant ring and write a polynomial as above. In general this leads to a polynomial whose Galois group is a subgroup of $G$, but provided we choose the invariants sufficiently generically the Galois group will in fact be $G$ itself. In general this method works well for groups of small order where the invariant rings are managable.

Now let us apply this to produce an example of a degree four polynomial over $\mathbb{Q}$ with Galois group $D_{4}$. Up to a shift in the indeterminate we may assume that this polynomial takes the form

$ p(z)=z^{4}+\sigma_{2}z^{2}-\sigma_{3}z+\sigma_{4}. $

In other words without loss of generality we may assume that the sum of the roots of $p(z)$ vanishes. An elementary problem in the theory of finite group representations shows that the invariant ring of $D_{4}$ acting as permutation on four objects $r_{i}$ subject to the constraint that $\sum_{i}r_{i}=0$ is generated by four objects $\alpha, \beta, \chi, \lambda$ subject to the single relation $\alpha \lambda =\chi^{2}$. Thus the relevant invariant polynomial ring is simply

$ \mathbb{Q}[r_{1}, r_{2}, r_{3}, -r_{1}-r_{2}-r_{3}]^{D_{4}}\cong \mathbb{Q}[\alpha, \beta, \chi, \lambda]/\langle \alpha \lambda=\chi^{2}\rangle. $

Then the symmetric polynomials are expressed in terms of the generators of the invariant ring as

$ \sigma_{2}=-\frac{1}{8}(\beta+\alpha), $

$\sigma_{3}= \frac{\chi}{16},$

$\sigma_{4}= \frac{1}{256}((\alpha-\beta)^{2}-\lambda).$

Thus any polynomial with Galois group $D_{4}$ can be written by choosing arbitrary $\alpha, \beta, \chi, \lambda$ subject to the single constraint in the invariant ring and plugging into the above. For example, one solution with integer coefficients is given by

$ p(z)=z^{4}+z^{2}+2z+1. $

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This is done in:

MR1697454 (2000e:12013) Ledet, Arne(3-QEN) Dihedral extensions in characteristic 0. (English, French summary) C. R. Math. Acad. Sci. Soc. R. Can. 21 (1999), no. 2, 46–52. 12F12 (11R20)

I am having trouble finding the actual paper, but if you write to the author, I am sure he can help out.

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I have found this one, which may be interesting : library.msri.org/books/Book45/files/book45.pdf –  Auguste Hoang Duc Mar 18 '11 at 21:22

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