# diagonalizing a matrix over the $\ell$-adics

Let $M$ be a $2 \times 2$ matrix with coefficients in $\mathbb{Z}_{\ell}$ whose characteristical polynomial is

$$P(T) = T^2- (a+d) T + (ad-bc).$$

I've encountered the following assertion: If $P(T)$ factors over $\mathbb{Z}/\ell \mathbb{Z}$ as

$$P(T) = (T-\lambda_1)(T-\lambda_2)$$ with $\lambda_1 \not\equiv \lambda_2 \pmod{\ell}$, then in fact $M$ is diagonalizable over $\mathbb{Z}_{\ell}$.

I was able to prove this using a straightforward argument by diagonalizing over $\mathbb{Z}/\ell \mathbb{Z}$ and then explicitly lifting. However, I was wondering if there was a more conceptual way to explain this in terms of Hensel's Lemma or something (i.e. a more conceptual way to package this lifting argument).

## 1 Answer

Let $R$ be a local henselian ring (e.g. complete local rings) with residue field $k$. Let $\phi$ be an endomorphism of $V=R^n$ such that its characteristic polynomial $P(T)\in R[T]$ is completely split and separable in $k[T]$. Then $\phi$ is diagonalizable over $R$.

First by Hensel property, $P(T)$ is completely split and separable in $R[T]$ (this is the only place we use henselian hypothesis). Denote by $a_1, \dots, a_n\in R$ the eigenvalues of $\phi$ and by $V_{a_i}$ the eigensubspace $\mathrm{ker}(\phi - a_i I_n)\subseteq V$. The point is to prove that $V=\oplus_{1\le i\le n} V_{a_i}$. As the submodules $V_{a_i}$ are in direct sum (note that $a_i-a_j$ is an unit in $R$ if $i\ne j$), it is enough to show that $V=\sum_{1\le i\le n} V_{a_i}$.

Consider the sub-$R$-algebra $A=R[\phi]$ of $\mathrm{End}(V)$ generated by $\phi$. For any $i\le n$, let $$e_i=\prod_{j\ne i} (\phi-a_j I_n)\in A.$$ We have $\phi e_i=(\phi-a_i I_n)e_i + a_i e_i=P(\phi)+a_ie_i=a_ie_i$. So $e_iV\subseteq V_{a_i}$ and it is enough to show that $I_n\in \sum_i e_iA$ or, equivalently, that $e_1, \dots, e_n$ generate the unit ideal of $A$. But all these properties are true over the residue field $k$ of $R$ by the diagonalizability condition over $k$. So $$A\otimes_R k=\sum_{1\le i\le n} \bar{e}_i A\otimes_R k.$$ (Of course this can also be checked directly over $k$.) Thus $$A/(\sum_i e_iA) \otimes_R k=0$$ and $A=\sum_i e_iA$ by Nakayama's lemma.

• Thanks! Exactly the kind of thing that I was looking for. – Tony Jan 6 '13 at 15:09