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.


  • 2
    $\begingroup$ Over $\mathbb{Q}$, I assume? $\endgroup$ Mar 18 '11 at 15:52

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. $


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.


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.