We know from linear algebra that if an $n \times n$ matrix $A$ over a field $k$ is diagonalizable (that is, there exists $P \in GL_n(k)$ such that $PAP^{-1}$ is a diagonal matrix), then this diagonal matrix is unique up to permutation of diagonal entries.

Are there any analogous results in the literature if we replace the field $k$ with the ring $\mathbb{Z}_{p^k}$ where $p$ is a prime and $k \geq 2$?

  • $\begingroup$ Your first sentence is missing a "then", and the parenthetical doesn't end. Did you mean to include something there? $\endgroup$ – user44191 Jun 26 '18 at 1:39
  • $\begingroup$ The ring has zero divisors, so I would not expect as much uniqueness. Gerhard "Will Think About Four Elements" Paseman, 2018.06.25. $\endgroup$ – Gerhard Paseman Jun 26 '18 at 1:44
  • $\begingroup$ I have edited the question so that it reads as a complete sentence. $\endgroup$ – BDS Jun 26 '18 at 1:52
  • $\begingroup$ To Gerhard Paseman: That is an understandable assumption; indeed the JCF (with at least one nontrivial Jordan block) over such rings doesn't have 'uniqueness' in the sense that my question asks concerning diagonal matrices. $\endgroup$ – BDS Jul 14 '18 at 2:46

The answer is yes, that you can get an analogous theorem, because $\mathbb{Z}_{p^k}$ is a local ring.

The equation $PAP^{-1} = B$ can be rewritten $PA = BP$. Write $A = \text{diag}(a), B = \text{diag}(b), a = \{a_i\}, b = \{b_j\}$. Then this implies that for any $i, j$, we have that $P_{i, j} a_i = P_{i, j} b_j$.

As $P \in GL_n(\mathbb{Z}_{p^k})$, we know that $\text{det}(P)$ must be a unit. We can write $\text{det}(P) = \sum_{\sigma \in S_n} (-1)^{\text{sgn}(\sigma)} \prod_i P_{i, \sigma(i)}$. As $\mathbb{Z}_{p^k}$ is a local ring, the set of nonunits is an additively closed, so at least one of the $\prod_i P_{i, \sigma(i)}$ must be a unit. Correspondingly, there must be some $\sigma$ such that $P_{i, \sigma(i)}$ is a unit for all $i$.

Then $P_{i, \sigma(i)} a_i = P_{i, \sigma(i)} b_{\sigma(i)}$. But $P_{i, \sigma(i)}$ is a unit for each $i$ - so we can cancel it to get $a_i = b_{\sigma(i)}$ for each $i$. Therefore, $\sigma$ is a permutation of diagonal entries that takes $A$ to $B$, and we are done.

  • $\begingroup$ Could you elaborate a little about about why $\sigma$ must exist as written? $\endgroup$ – BDS Jun 26 '18 at 2:57
  • 1
    $\begingroup$ @BDS I have done so. $\endgroup$ – user44191 Jun 26 '18 at 3:02

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