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 (for some prime $\mathfrak{p}\subseteq R$):

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 similarly defined;
3. $A$ is diagonalizable in $\kappa(\mathfrak{p})$: similarly defined;
4. $A$ is diagonalizable in $\overline{\kappa(\mathfrak{p})}$: similarly defined, where we consider an algebraic closure.

Obviously $(1)\implies(2)\implies(3)\implies(4)$. Moreover $(4)$ has the minimal polynomial criterion. I have two observations:

• (2) is an open condition on $\mathfrak{p}$: the primes $\mathfrak{p}$ such that $A$ is diagonalizable in $R_{\mathfrak{p}}$ form an open subset of $\mathrm{Spec}(R)$;
• (4) is a constructible condition on $\mathfrak{p}$: the primes $\mathfrak{p}$ such that $(4)$ holds form a constructible subset. This is seen from the morphism, whose image is where $(4)$ holds $$\mathrm{GL}_n\times\mathbb{A}^n\to \mathbb{A}^{n\times n},\quad P,\lambda_1,\cdots,\lambda_n\mapsto P\begin{pmatrix}\lambda_1\\&\ddots\\&&\lambda_n\end{pmatrix}P^{-1}. $$

I do not expect either $(3)\implies(2)$ or $(4)\implies(3)$, though my intuition of local diagonalizability is $(4)$. 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?