# Non-Examples for Krull-Schmidt-Azumaya

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 $R = \mathbb{Z}_{(2)}$?

• (1) Maybe you want to say $I\oplus I^{-1}$. Dec 8, 2014 at 9:38
• @user26857: I am quite sure it was $I \oplus I$ but I would also be satisfied with an example for $I \oplus I^{-1} = A \oplus A$.
– Dune
Dec 8, 2014 at 12:38
• 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. Dec 8, 2014 at 15:21
• @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?
– Dune
Dec 8, 2014 at 15:46
• Look for "Steinitz Isomorphism Theorem". Dec 8, 2014 at 16:01

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.
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.