# For every ring R, is there a block-diagonal canonical form for a square matrix over R?

This question asks whether there exists an analogue of the Jordan decomposition for an arbitrary ring $$R$$. This analogue is not necessarily the Jordan-Chevalley decomposition, which is unnecessarily strong. This follows from this question, but you don't need to read it.

Given a ring $$R$$, let $$J(R)$$ be the monoid whose elements are all square matrices over $$R$$, and where the monoid operation is $$\oplus$$ denoting direct sum of matrices. Let $$A { \sim_\text{S}} B$$ mean that there exists an invertible matrix $$P$$ such that $$PAP^{-1} = B$$. Is the monoid $$J(R)/{ \sim_\text{S}}$$ always a free abelian monoid?

For perfect fields, the answer is yes by the Jordan-Chevalley decomposition. What about for every ring? Is there a counterexample?

I've referenced this question in this paper: Towards a Singular Value Decomposition and spectral theory for all rings.

• This will be true if R is Artinian Nov 23, 2021 at 14:20
• I'm assuming in my comment R is commutative but I'm not sure it is necessary. If R is commutative Artinian you are looking at indecomposable R[x] modules with a bounded length and can use Krull-Schmidt Nov 23, 2021 at 14:28
• @BenjaminSteinberg Thanks. That's very helpful. I'm not assuming that $R$ is commutative because I know that $J(\mathbb H)/{\sim_{\text S}}$ is free abelian where $\mathbb H$ is the quaternions.
Nov 23, 2021 at 14:44
• It should be true for R Artinian in general. Nov 23, 2021 at 14:48
• But I am not immediately sure how to phrase things in a module theory way. You basically want to make a category with objects pairs (V,A) with A is an nxn matrix over R and V an A-invariant submodule and show this is a krull schmidt category when R is artinian, which seems ok to me. Nov 23, 2021 at 14:51

Your question is equivalent to whether the category $$\mathcal{E}$$ of pairs $$(V,f)$$ consisting of a finitely generated free (right) $$R$$-module and an endomorphism $$f$$ of $$V$$ is a Krull-Schmidt category, i.e., an additive category where every object decomposes as a direct sum of finitely many indecomposable objects and the decomposition is unique up to isomorphism and reordering. (The category $$\cal E$$ is also equivalent to the category of right $$R[t]$$-modules which are f.g. free $$R$$-modules, and it is rarely Krull-Schmidt, but I won't use this point of view.)

Here is one possible counterexample, phrased using the category $$\mathcal{E}$$ of pairs $$(V,f)$$ above: Take $$R$$ to be a Dedekind domain admitting a non-free rank-$$1$$ projective module $$L$$ such that $$L\oplus L\cong R\oplus R$$ (equivalently, $$L$$ represents an element of order $$2$$ in the Picard group of $$R$$). Let $$f_L : L^2\to L^2$$ be defined by $$f_L(x,y)=(0,x)$$, and define $$f_R:R^2\to R^2$$ similarly. The isomorphism $$L\oplus L\cong R\oplus R$$ gives rise to an isomorphism $$(L^2,f_L)\oplus (L^2,f_L) \cong (R^2,f_R)\oplus (R^2,f_R)$$ in $$\cal E$$. One readily checks that $${\rm End}_{\cal E}(L^2,f_L)\cong R[\epsilon|\epsilon^2=0]$$ and $${\rm End}_{\cal E}(R^2,f_R)\cong R[\epsilon|\epsilon^2=0]$$, so the endomorphism rings of $$(L^2,f_L)$$ and $$(R^2,f_R)$$ contain no nontrivial idempotents. This means that these objects are indecomposable in $$\cal E$$. On the other hand, $$(L^2,f_L)\ncong (R^2,f_R)$$ because $$\ker f_L\cong L\ncong R\cong \ker f_R$$. Consequently, the monoid $$(\cal E/\cong, \oplus)$$ (which is isomorphic to $$(J(R)/\sim, \oplus)$$ in your question) is not a free abelian monoid (because $$2x=2y$$ implies $$x=y$$ in a free abelian monoid).

One the other hand, a sufficient (but not necessary) condition for the category $$\cal E$$ to be Krull-Schmidt is that the endomorphism ring of every object $$(V,f)$$, i.e., the centralizer of $$f$$ in $${\rm End}_R(V)\cong {\rm M}_n(R)$$, is a semiperfect ring.

For example, if $$R$$ is commutative and aritinian as in Benjamin Steinberg's comment, then $${\rm End}_{\cal E}(V,f)$$ will be an $$R$$-subalgebra of $${\rm End}_R(V)$$, hence artinian, and in particular semiperfect.

The semiperfectness $${\rm End}_{\cal E}(V,f)$$ for all f.g. free $$V$$ is actually true even if $$R$$ is non-commutative one-sided artinian, and even if $$R$$ is just semiprimary. This appears implicitly in a paper of mine (page 20 & Thm. 8.3(iii) & Remark 2.9). When $$R$$ is commutative noetherian and local, one can use a theorem of Azumaya (Theorem 22) and a little work to show that this property is equivalent to $$R$$ being henselian.

• Could you please provide an example of such an $L$ and $R$? Thanks
• @ogogmad Take $R=\mathbb{Z}[\sqrt{-5}]$ and $L=(2,1+\sqrt{-5})$, for instance. One has $L\cdot L=2R$. This is the simplest example I know. Nov 23, 2021 at 16:55