How to find a ring with a given automorphism group

Question

I'm hoping to find an example of a ring whose automorphism group is D8, i.e. the order 8 dihedral group. This raises the question of how, given a group $G$ to find a ring $R$ whose automorphism group is $G$?

Any ideas on how to approach this problem in general or for the specific example would be appreciated.

So far I can see that the automorphism group of $\bf{Z} \times ... \times \bf{Z}$ $n$-times is $S_n$. Also I noticed that the group ring ${\bf Z}G$ contains $G$ in its automorphism group, but it is often larger.

Answer 1

By Shafarevich's theorem every finite solvable group (including $D_8$) is the automorphism group of some Galois extension of the field of rational numbers. See also this paper: K. Hashimoto & K. Miyake, Inverse Galois problem for dihedral groups, Developments in Mathematics 2, Kluwer Academic Publishers, 1999, 165– 181. Or this paper, page 3. Or this book, table on p.11.

Answer 2

In this special case you can in fact take $R$ to be a number field $\mathbb{Q}(\sqrt[4]{2}, i)$, the splitting field of $x^4 - 2$, so $D_8$ is its Galois group; this is a nice exercise. It's a famous open problem, the inverse Galois problem over $\mathbb{Q}$, whether every finite group occurs as the Galois group of a Galois extension of $\mathbb{Q}$ in this way.

In general the case of infinite groups seems hard but I think the case of finite groups is doable. The strategy I have in mind is to mimic the construction in this math.SE answer of mine: given a finite group $G$ pick some action of it on an affine space $\mathbb{A}^n$ (say coming from a faithful linear representation with no trivial components) over $\mathbb{Q}$, then find a subvariety $V \subsetneq \mathbb{A}^n$ which admits no nontrivial automorphisms. Then we can hope that the subvariety of all translates $gV, g \in G$ should have automorphism group exactly $G$, and take $R$ to be its ring of functions.