Can diagonalization mod p be generalized to diagonalization mod n?

Question

When you diagonalize a matrix $A$, your $D$ matrix will be the similar to if you diagonalized $A$ mod $p$ (but $D$ will also be mod $p$ in this scenario). I'm having a brainfart moment here. Does $p$ have to be prime for this to matter? Or can this be generalized to any integer $n$? For example, if I have $A$ mod $n$, will $D$ just be $D$ mod $n$? Or is something lost in translation if you don't have a prime modulus?

Answer 1

Maybe this is in the dirction you want: Assume you have $p,q$ with $\gcd(p,q)=1$, matrices $P,D,P'\in Mat_{n\times n}(\mathbb Z/p\mathbb Z)$ and $Q,E,Q'\in Mat_{n\times n}(\mathbb Z/q\mathbb Z)$, $A\in Mat_{n\times n}(\mathbb Z)$ such that $PP'=1$, $QQ'=1$, $PDP'\equiv A\pmod p$, $QEQ'\equiv A\pmod q$. Then you can find (using the Chinese remainder theorem), $R,F,R'\in Mat_{n\times n}(\mathbb Z/pq\mathbb Z)$ such that $R\equiv P\pmod p$, $R\equiv Q\pmod q$, $F\equiv D\pmod p$, $F\equiv E\pmod q$, $R'\equiv P'\pmod p$, $R'\equiv Q'\pmod q$. Then $RR'\equiv 1\pmod p$ and $RR'\equiv 1\pmod q$, hence $RR'\equiv 1\pmod{pq}$, and also $RFR'\equiv A\pmod p$ and $RFR'\equiv A\pmod q$, hence $RFR'\equiv A\pmod{pq}$.