A criteria for Jordan decomposition of a matrix over a general ring

Question

When I look at different proofs of the Jordan-Chevalley decomposition of a matrix, the minimal hypothesis I usually found is about the perfection of the field over which such decomposition occurs (e.g. Wikipedia article). But it seems to me that the gist of the proofs is not about the field but about the type of characteristic polynomial of the matrix.

More precisely, by using the two following definitions for a general ring $A$

Definition 1 A $n\times n$-matrix $M$ with coefficients in the ring A is semisimple if $A^n$ is a semi-simple $A[M]$-module (here I am using the definition provided in the answer of @ಠ_ಠ below)

Definition 2 A separable polynomial $P$ of $A[X]$ is a polynomial whose discriminant is invertible in $A$

Then is the following statement true?

A matrix $M$ over $A$ have a unique Jordan-Chevalley decomposition if its characteristic polynomial divides a power of a separable polynomial.

Edit By Jordan decomposition I mean here the additive one, i.e. there exist $M_s$ and $M_n$, two $n\times n$ $A$-matrices respectively semi-simple and nilpotent, such that $$M=M_s+M_n$$

If it is not, what kind of general property should the ring $A$ have to make it true (like $A$ is a domain, etc.)

Edit 2 I believe my definition 1 is equivalent in a domain to the fact that the matrix is diagonalizable in an algebraic extension of the fraction field which is equivalent to the fact that the matrix is cancelled by a separable polynomial (definition 2) in $A[X]$

Answer 1

Let $A\in M_n(R)$ where $R$ is some commutative ring, $p$ be the "square free" polynomial that divides $\chi_A$ (the characteristic polynomial of $A$) and $p'$ be the derivative of $p$.

We consider the pseudo Newton algorithm

$X_0=A , X_{k+1}=X_k-p(X_k)(p'(X_k))^{-1}$. When $R$ is a perfect field, after a finite number of iterations, $X_k=D$, the semi simple part of $A$.

Of course, we have to give conditions on $R$, for example, as assumed by Minseon Shin,

"Let $R$ be an integral domain whose fraction field $K$ is perfect".

We can chain the iterations iff $p'(X_k)$ is invertible over $M_n(R)$, that is, if $\det(p'(X_k))$ is an invertible of $R$. In fact, at each iteration, $\det(p'(X_k))=\det(p'(A))$ and the invertibility of $\det(p'(A))$ suffices. It is not difficult to see that $\det(p'(A))$ is a divisor of a power of $discrim(p)$.

Finally, if $discrim(p)$ is an invertible (that is equivalent to the condition considered by the OP), then we can chain the iterations.

When $R=\mathbb{Z}/m\mathbb{Z}$, the entire algorithm seems to work; I did not find any counter-examples.

EDIT. Answer to the OP.

Assume that $R$ is an integral domain. If $p\in R[x]$, then $discrim(p)\not= 0$ implies that $p,p'$ have no common roots over an algebraically closed field containing $R$, that implies that $p$ has only simple roots. If moreover, $discrim(p)$ is invertible and $p$ divides $\chi_A$, then I think that the above algorithm formally works and that we obtain $D$ s.t. $p(D)=0$ and $A-D$ nilpotent.

However, practically, how to find such a $p$ ? A priori, $R[x]$ is not an UFD; on the other hand,it seems difficult to work over a field that contains $R$ and the roots of $\chi_A$; how to prove that $p$ divides $\chi_A$ if $R$ is not euclidean or not pincipal?

Anyway, in general, $p=\dfrac{\chi_A}{gcd(\chi_A,\chi'_A)}$. Note that if $R$ is an UFD, then the gcd always exists in $R[x]$; yet, in general, there is no algorithm to do that.

Finally, if $R$ is an Euclidean domain, I think that we can explicitly calculate $D,N$. However, in theory, perhaps the hypothesis UFD suffices...

Answer 2

If $K$ is a commutative ring and $V$ is a $K$-module, then a choice of endomorphism $f \in \text{End}_K(V)$ is the same as a representation of the $K$-algebra $K[f]$ on $V$. This is proved in Aluffi's Algebra text in section VI.7.1. Then we say $f$ is simple (resp. semisimple) if $V$ is simple (resp. semisimple) as a $K[f]$-module.

Sorry I can't help with the rest of your questions.

Answer 3

(This is only a partial answer and I suspect it only confirms what you already knew but it was an interesting exercise for me and I thought I'd write it up.)

I will use the following interpretation of your question:

Let $A$ be an integral domain whose fraction field $K$ is perfect. For a matrix $M \in \mathrm{Mat}_{n \times n}(A)$, let $$ M = M_{\mathrm{ss}} + M_{\mathrm{n}} $$ be its Jordan decomposition over $K$ (so $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(K)$). We may ask whether $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$.

Suppose $A$ is normal. Let $p_{M} \in A[t]$ be the characteristic polynomial of $M$. Let $\overline{K}$ be an algebraic closure of $K$, let $\overline{A}$ be the integral closure of $A$ in $\overline{K}$, and let $\lambda_{1},\dotsc,\lambda_{r} \in \overline{K}$ be the distinct roots of $p_{M}$. Since $\overline{A}$ is normal, we in fact have $\lambda_{i} \in \overline{A}$. If every pairwise difference $\lambda_{i} - \lambda_{j}$ (with $i \ne j$) is invertible in $\overline{A}$, then $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$.

Remark: Your condition that $p_{M}$ divides a power of a polynomial with invertible discriminant is nicer because it makes sense over any ring, but if $A$ is a normal domain, then I think it implies my condition that $\lambda_{i} - \lambda_{j}$ is invertible. It's not clear to me whether the converse is true if $A$ is a domain which is not a UFD.

Proof: Let $K'/K$ be a finite Galois extension over which $p_{M}$ splits, and let $A'$ be the integral closure of $K$ in $K'$. Since $A$ is normal, the inclusion $A \subseteq K \cap A'$ is an equality. Thus we may reduce to the case when $p_{M}$ splits into linear factors over $A$, say $p_{M}(t) = (t-\lambda_{1})^{n_{1}} \dotsb (t-\lambda_{r})^{n_{r}}$. The usual proof of Jordan-Chevalley decomposition for perfect fields (e.g. [2, Section 4.2]) says that if $p \in A[t]$ is any polynomial with $p-\lambda_{i} \in (t-\lambda_{i})^{n_{i}}$ for all $i$, then $M_{\mathrm{ss}} = p(M)$ and $M_{\mathrm{n}} = M-p(M)$. (Here often it is assumed that also $p \in (t)$, but if I understand correctly it is only used to show that Jordan decomposition is behaves well under passing to $M$-invariant subspaces, see part (c) of the Proposition in [2, Section 4.2].) Since by assumption the pairwise differences $\lambda_{i}-\lambda_{j}$ are invertible, the polynomials $(t-\lambda_{i})^{n_{i}}$ are comaximal in $A[t]$, so the Chinese remainder theorem applies and we can find such $p$.

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If the pairwise differences $\lambda_{i}-\lambda_{j}$ are not all invertible, it is not necessarily true that $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$. Here is an example following [1]. Let $a,b \in A$ and set $$ M_{(a,b)} := \begin{bmatrix} & & a^{2}b \\ 1 & & -(a^{2}+2ab) \\ & 1 & 2a+b \end{bmatrix} $$ which is the companion matrix associated to the polynomial $(x-a)^{2}(x-b)$. Then Example 2.3, Example 4.5 of [1] says that the matrices $M_{(a,b),\mathrm{ss}},M_{(a,b),\mathrm{n}}$ have coefficients in $A$ if and only if $a-b$ is a unit of $A$.

Let's work out the example with $A = \mathbb{Z}$ and $a = 0$ and $b = 2$. Then the eigenvalues of $$ M_{(0,2)} = \begin{bmatrix} & & \\ 1 & & \\ & 1 & 2 \end{bmatrix} $$ are $0,2$, the generalized eigenspace for $\lambda = 0$ is $\{v_{1} = [0,2,-1]^{T} , v_{2} = [2,-1,0]^{T}\}$ (with $M_{(0,2)}v_{2} = v_{1}$) and the eigenspace for $\lambda = 2$ is $\{v_{3} = [0,0,1]^{T}\}$. Set $U := [v_{1},v_{2},v_{3}]$; then $$ U^{-1}M_{(0,2)}U = \begin{bmatrix} 0 & 1 & \\ & 0 & \\ & & 2 \end{bmatrix} $$ is a Jordan normal form of $M_{(0,2)}$, and $$ M_{(0,2),\mathrm{ss}} = U \begin{bmatrix} 0 & & \\ & 0 & \\ & & 2 \end{bmatrix} U^{-1} = \begin{bmatrix} & & \\ & & \\ 1/2 & 1 & 2 \end{bmatrix} \qquad M_{(0,2),\mathrm{n}} = U \begin{bmatrix} & 1 & \\ & & \\ & & \end{bmatrix} U^{-1} = \begin{bmatrix} & & \\ 1 & & \\ -1/2 & & \end{bmatrix} $$ which are not contained in $\mathrm{Mat}_{3 \times 3}(\mathbb{Z})$.

References:

[1] Passi, Roggenkamp, Soriano, "Integral Jordan decomposition of matrices", Linear Algebra and its Applications, vol. 355 (2002), 241 -- 261 (link)

[2] Humphreys, Introduction to Lie Algebras and Representation Theory, Springer Graduate Texts in Mathematics, vol. 9 (1972)