It is well known that for a matrix $A$ in $\mathfrak{sl}_n(\mathbb{C})$, we have the following equivalence: $$\dim Z(A) \text{ is minimal} \leftrightarrow A \text{ is cyclic}$$ where $Z(A)$ is the centralizer of $A$ (elements commuting with $A$). The minimal dimension is the rank of the Lie algebra, so $n-1$ in this case. Cyclic means that there is a vector generating the whole vector space under the action of $A$.

My question is whether there is a "doubled" version of this: Is it true that for a pair of *commuting* matrices $(A,B)$ in $\mathfrak{sl}_n(\mathbb{C})$ we have
$$\dim Z(A,B) \text{ is minimal} \leftrightarrow (A,B) \text{ is cyclic}$$
where $Z(A,B)=Z(A)\cap Z(B)$ is the common centralizer (elements commuting with $A$ and $B$)? Here the centralizer is meant in $\mathfrak{sl}_n$, and minimal means minimal *among commuting pairs*.
Remark that the minimal dimension of the centralizer of a commuting pair is also the rank ($n-1$ here) of the Lie algebra.

It is not so difficult to show that $(A,B)$ cyclic implies that the centralizer is of minimal dimension. For the direct implication, in the case of one matrix, one can use the Frobenius decomposition, itself based on the invariant factor decomposition which only works over euclidean rings. For one matrix, we have the ring $\mathbb{C}[x]$ of polynomials in one variable, which is euclidean. But for the doubled version, we need $\mathbb{C}[x,y]$ which is not euclidean (two commuting matrices gives the vector space the structure of a $\mathbb{C}[x,y]$-module). Is there another argument for the direct implication?

Edit: $(A,B)$ is cyclic means that $\mathbb{C}^n$, as module over the subalgebra generated by $\{A,B\}$, is generated by a single element. For commuting $A,B$, this means here that there exists $v\in\mathbb{C}^n$ such that $\mathbb{C}[A,B]v=\mathbb{C}^n$.