I am looking for some rings $A$ and finitely generated $A$-modules $M$, where the conclusion of the Krull-Schmidt-Azumaya Theorem does not hold for $M$, i.e. where $M$ can be written as direct sums of indecomposable modules in multiple ways (the summands not being unique up to order and isomorphism). To be specific, I am looking for the following kinds of non-examples:

(1) I know that there is a Dedekind domain $A$ with a non-principal indecomposable ideal $I$ such that $I \oplus I \cong A \oplus A$, but I could not find a concrete example.

(2) Is there a non-example with $A = \mathbb{Z}G$ for some finite group $G$?

(3) Is there a non-example for $A = RG$ as in (2), where $R$ is a (non-complete) discrete valuation ring, for example $R = \mathbb{Z}_{(2)}$?

Thank you in advance!

  • $\begingroup$ (1) Maybe you want to say $I\oplus I^{-1}$. $\endgroup$
    – user26857
    Dec 8, 2014 at 9:38
  • $\begingroup$ @user26857: I am quite sure it was $I \oplus I$ but I would also be satisfied with an example for $I \oplus I^{-1} = A \oplus A$. $\endgroup$
    – Dune
    Dec 8, 2014 at 12:38
  • $\begingroup$ I've asked this for in a Dedekind domain $J\oplus J\simeq A\oplus IJ$, so if $J=I^{-1}$ we get what I've said. $\endgroup$
    – user26857
    Dec 8, 2014 at 15:21
  • 1
    $\begingroup$ @user26857: I see, so if $I^2$ is principal we also get the desired isomorphism, as in Georges' answer. But how does the isomorphism $I \oplus J \cong A \oplus IJ$ look like? Do you have a reference for this? $\endgroup$
    – Dune
    Dec 8, 2014 at 15:46
  • 1
    $\begingroup$ Look for "Steinitz Isomorphism Theorem". $\endgroup$
    – user26857
    Dec 8, 2014 at 16:01

3 Answers 3


Take $A=\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle=\mathbb R[x,y]$.
Then the ideal $I=\langle y,x-1\rangle\subset A$ is not principal but $I^2$ is principal .
Hence $I\oplus I\cong A\oplus I^2\cong A\oplus A$ as $A$-modules.



Jones, Alfredo, On representations of finite groups over valuation rings, Ill. J. Math. 9, 297-303 (1965). ZBL0132.27403,

Theorem 2 states that if $G$ is a finite abelian group of exponent $qp^n$, where $p$ is a prime not dividing $q$, then Krull-Schmidt holds for $\mathbb{Z}_{(p)}$-modules if and only if either $q=1$ or $p$ is a primitive root modulo $q$. So a cyclic group of order $14$ over $\mathbb{Z}_{(2)}$ will give a non-example. The paper also refers to a previous example of Berman and Gudivok.


For integral group rings of finite groups Krull-Schmidt holds in only a few cases, even when restricted to lattices which in this case is the better way to ask. Essentially small cyclic groups and small dihedral groups are the only positive answers. Check Theorem 1.6 in

Hindman, Peter; Klingler, Lee; Odenthal, Charles J. On the Krull-Schmidt-Azumaya theorem for integral group rings Comm. Algebra 26 (1998), no. 11, 3743–3758

for all the positive cases and the unique one remaining open (dihedral order 16).

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