# when upper triangular matrix modulo prime ideals implies upper triangular?

Let $E/ \mathbb{Q}_p$ be a finite extension, let $\mathcal{O}$ be the ring of integers of $E$. Let $A$ be a reduced noetherian local complete $\mathcal{O}$-algebra with the maximal ideal $\mathfrak{m}$. (I guess this particular setting does not really matter and the statement should be true for more general $A$'s)

Let $M \in GL_n(A)$ be a $n \times n$-matrix. Suppose that for every height $1$ prime ideal $\mathfrak{p}$ of $A$, the reduced mod $\mathfrak{p}$ matrix $M / \mathfrak{p}$ is conjugate to an upper triangular matrix with coefficients in $A / \mathfrak{p}$. Under what conditions (on $M$ or even better: just on $M / \mathfrak{m}$) the matrix $M$ is conjugate to an upper triangular matrix with coefficients in $A$?

I'll be grateful for your help.

-
I think you need to be more precise about what kind of conditions you are looking for. Otherwise, all the entries below the diagonal are zero would seem to be a correct answer to your question. –  Simon Wadsley May 31 '13 at 11:28
Could you spell out what you mean by "congruent"? Thanks! –  David Speyer May 31 '13 at 12:10
@Simon Wadsley: you are right - I'm looking for the conditions which would only involve $M / \mathfrak{m}$ in fact (for example, $M / \mathfrak{m}$ is conjugate to an upper triangular matrix will the pair-wise distinct terms on the diagonal - I think that might work, but I don't know how to prove it). @David Speyer: I meant to write "conjugate" (in the respective $GL_n(A/\mathfrak{p})$, sorry! –  Przemyslaw Chojecki May 31 '13 at 12:51