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First note that $R$ as a finite ring is (like any Artinian ring) is a finite product of local rings: $R=\prod_i R_i$. Hence $M_n(R) =\prod_i M_n(R_i).$ So if we write $A=A_1 \times \cdots A_n$ with $A_i \in M_n(R_i)$, then $$\text{Ann}^l(A) = \text{Ann}^l(A_1) \times \cdots \text{Ann}^l(A_n)$$ and analogous for the right annulator.

If each $R_i$ is a local principal ideal ring (or more generally an elementary divisor ring), then $$|\text{Ann}^l(A)|=|\text{Ann}^r(A)|.$$

Proof: By the above we may assume wlog $R=R_i$ is a local PIR. By a theorem of Kaplansky, $A$ has a Smith normal form, i.e. there are invertible $U,V$ and a diagonal matrix $D=\text{diag}(d_,...,d_n)$ such that $A=UDV$. Thus $XA=0$ is equivalent to $(XU)D=0$, showing $\text{Ann}^l(A)=\text{Ann}^l(D)U^{-1}$. In particular both sets have the same cardinality. Write $X=(x_1,...,x_n)$. Then $XD=(d_1x_1,...,d_nx_n)$. Hence $$\text{Ann}^l(D)=\lbrace (x_1,...,x_n) \mid \forall i: x_i \in \text{Ann}_R(d_i)^n \rbrace.$$ Analogously, $\text{Ann}^r(A)=V^{-1}\text{Ann}^r(D)$ and $$\text{Ann}^r(D)=\lbrace (y_1^T,...,y_n^T) \mid \forall i: y_i \in \text{Ann}_R(d_i)^n \rbrace.$$ Hence $$|\text{Ann}^l(A)| = \prod_{i=1}^n |\text{Ann}_R(d_i)|^n = |\text{Ann}^r(A)|$$ and the assertion follows. q.e.d.

Remark: If $R$ is a (not necessarily finite) product of commutative principal ideal rings, then the proof above shows

$\qquad\qquad \text{Ann}^r(A) = V^{-1}U^T \text{Ann}^l(A)^T$

1

First note that $R$ as a finite ring is (like any Artinian ring) is a finite product of local rings: $R=\prod_i R_i$. Hence $M_n(R) =\prod_i M_n(R_i).$ So if we write $A=A_1 \times \cdots A_n$ with $A_i \in M_n(R_i)$, then $$\text{Ann}^l(A) = \text{Ann}^l(A_1) \times \cdots \text{Ann}^l(A_n)$$ and analogous for the right annulator.

If each $R_i$ is a local principal ideal ring (or more generally an elementary divisor ring), then $$|\text{Ann}^l(A)|=|\text{Ann}^r(A)|.$$

Proof: By the above we may assume wlog $R=R_i$ is a local PIR. By a theorem of Kaplansky, $A$ has a Smith normal form, i.e. there are invertible $U,V$ and a diagonal matrix $D=\text{diag}(d_,...,d_n)$ such that $A=UDV$. Thus $XA=0$ is equivalent to $(XU)D=0$, showing $\text{Ann}^l(A)=\text{Ann}^l(D)U^{-1}$. In particular both sets have the same cardinality. Write $X=(x_1,...,x_n)$. Then $XD=(d_1x_1,...,d_nx_n)$. Hence $$\text{Ann}^l(D)=\lbrace (x_1,...,x_n) \mid \forall i: x_i \in \text{Ann}_R(d_i)^n \rbrace.$$ Analogously, $\text{Ann}^r(A)=V^{-1}\text{Ann}^r(D)$ and $$\text{Ann}^r(D)=\lbrace (y_1^T,...,y_n^T) \mid \forall i: y_i \in \text{Ann}_R(d_i)^n \rbrace.$$ Hence $$|\text{Ann}^l(A)| = \prod_{i=1}^n |\text{Ann}_R(d_i)|^n = |\text{Ann}^r(A)|$$ and the assertion follows. q.e.d.