# Structure theorem for modules over Dedekind domains

I've come across the structure Theorem for fin. gen. Modules over a Dedekind domain several times now. It was formulated to us the following way:

Let $$R$$ be a Dedekind domain. For every element $$\alpha \in C(R)$$, let a representative $$I_{\alpha}$$ in the group of fractional ideals be chosen. Then, to a fin. gen. $$R$$-Mod $$M$$ there are unique natural numbers $$r$$ and $$s$$, $$\alpha \in C(R)$$ with $$\alpha = 0$$ if $$s = 0$$, and proper nonzero ideals $$I_r \subset ... \subset I_1$$ such that

$$M \cong R/I_1 \oplus... \oplus R/I_r, \text{if }s = 0$$

$$M \cong R/I_1 \oplus ... \oplus R/I_r \oplus R^{s-1} \oplus I_{\alpha}$$, if $$s > 0$$

Now I want to give a description of the finitely generated module over the Dedekind domain according to the structure theorem. In each case the elements of the class group are listed for you, each given by means of a representative ideal.

Dedekind domain $$\mathbb{Z}[\sqrt{79}]$$, class group of order 3, representatives $$(1), (3,\sqrt{79}+1), (3, \sqrt{79}-1)$$. The Module $$M = I \oplus I$$, where $$I = (3, \sqrt{79}+1)$$.

I actually don't know where to begin. I also have not found any examples on the web. Any of those would be very appreciated.

To do your exercise, you need one extra piece of information, a theorem of Steinitz:

Let $$r$$ and $$s$$ be non-negative integers, and let $$I_1,\ldots,I_r$$ and $$J_1,\ldots,J_s$$ be ideals in your Dedekind domain $$R$$. Then one has an isomorphism $$\bigoplus_{m=1}^r I_m\cong \bigoplus_{n=1}^sJ_n$$ of $$R$$-modules if and only if $$r=s$$ and the equality $$\prod_m [I_m] = \prod_n [J_n]$$ holds in the class group of $$R$$.

For a proof, see Curtis and Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley (1962), Section 22.

You should be able to solve your exercise with this extra information.

• @reuns I have added a reference to a proof of Steinitz's theorem. Commented May 20, 2020 at 21:44
• I'm asking about the simplest case, what is the idea to show that $I\oplus I^{-1}\cong R\oplus R$ ? Commented May 20, 2020 at 22:33
• @reuns I think the simplest case is actually when considering relatively prime ideals $I$ and $J$. Then $I+J = R$ and $I \cap J = IJ$, so the sequence $0 \to IJ \to I \oplus J \to R \to 0$ is short exact. Since $R$ is free (hence projective) then the sequence splits, so $I \oplus J \cong R \oplus IJ$. Commented May 21, 2020 at 2:04
• @Alex B. I may seem ignorant, but I'm still not able to solve it. I'm very unfamiliar with fractional ideals and the ideal class group $C(R)$, resp. we were only given the definition. Am I right in assuming that $r = 0$ i.e. there's no torsion-part (when looking at how I stated the theorem)? Then, I would need to show that $R^{s-1} \oplus I_{\alpha} \cong I \oplus I$ for some $\alpha$. Or have I already started the wrong way? Commented May 24, 2020 at 6:48
• @Tylwyth: that's right. You can also easily determine $s$ by noting that in order to have an isomorphism of modules, you should at least have an isomorphism of abelian groups. And lastly Steinitz's theorem will determine $\alpha$. Commented May 24, 2020 at 7:18