Let $R$ be a commutative unital ring. Let $n\in\mathbb{N}_+$. Consider a $n\times n$-matrix $A=(a_{ij})_{i,j=1}^n$ with entries in $R$. A diagonal matrix is defined obviously. We can define several notions about diagonalizability:

1. $A$ is diagonalizable in $R$: if there exists an $n\times n$-matrix $P$ with entires in $R$ such that $\det(P)\in R^\times$ and $PAP^{-1}$ is diagonal;
2. $A$ is diagonalizable in $R_{\mathfrak{p}}$: consider $A$ as a matrix over $R_{\mathfrak{p}}$ and defined in an obvious manner;
3. $A$ is diagonalizable in $\kappa(\mathfrak{p})$: similarly defined.

Obviously $(1)\implies(2)\implies(3)$. Moreover $(3)$ is over a field and then we have the minimal polynomial criterion. I do not expect that $(3)$ implies $(2)$. My question is about $(2)\implies(1)$.

For the title I mean the following.

If $A$ is diagonalizable in $R_{\mathfrak{p}}$ for all primes $\mathfrak{p}$, can we claim that $A$ is diagonalizable?