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Let $R$ be a finite commutative ring. For $n>1$ consider the full matrix ring $M_n(R)$. For a matrix $A\in M_n(R)$ is true that the cardinality of the left annihilator (in $M_n(R)$) of $A$ equals the cardinality of the right annhilator?

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up vote 2 down vote accepted

No, this isn't true in general, but it's true in principal ideal rings. Consider $R=\mathbb{F}_2[x,y]/(x,y)^2$ (a finite ring that isn't a principal ideal ring, which I found in this answer by Zev). The matrix


annihilates the $16$ vectors that have no constant term, both on the left and on the right, but on the right it also annihilates the $16$ vectors that have a constant term in both components, for a total of $32$. Thus the left annihilator in $M_n(R)$ has $256$ elements, whereas the right annihilator has $1024$.

However, if $R$ is an elementary divisor ring, we can bring $A$ into Smith normal form using invertible matrices. Since this is diagonal, its left and right annihilators are the same up to transposition, and the invertible matrices don't change their cardinality. (See this MO question.)

According to Theorem 15.9 in Matrices over commutative rings, all principal ideal rings are elementary divisor rings. (Wikipedia only claims this for principal ideal domains.) According to this paper, a necessary condition for a ring to be an elementary divisor ring is that all finitely generated ideals are principal. Since in a finite ring all ideals are finitely generated, a finite elementary divisor ring must be a principal ideal ring. Thus a finite ring is an elementary divisor ring if and only if it is a principal ideal ring.

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