# Annihilators in Matrix Rings

I think this is not a research question, but in stackExchange remained unanswered.

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?

Let $k$ be a finite field, and let $R := k[X,Y] / (X^2,XY,Y^2)$. Then $R = k \oplus kx \oplus ky$ is a finite ring of order $|k|^3$ with maximal ideal $\mathfrak{m} := kx \oplus ky$ of square zero.

Now let $A = \begin{pmatrix} 0 & x \newline 0 & y \end{pmatrix} \in M_2(R)$. Then

$\begin{pmatrix} 0 & x \newline 0 & y \end{pmatrix} \begin{pmatrix} a & b \newline c & d \end{pmatrix} = \begin{pmatrix} cx & dx \newline cy & dy\end{pmatrix}$ whereas $\begin{pmatrix} a & b \newline c & d \end{pmatrix}\begin{pmatrix} 0 & x \newline 0 & y \end{pmatrix} = \begin{pmatrix} 0 & ax +by \newline 0 & cx + dy\end{pmatrix}$.

The equation $ax + by = 0$ with $a,b \in R$ implies $a, b \in \mathfrak{m}$ as can be seen by writing $a = \lambda + u, b = \mu + v$ for some $\lambda,\mu \in k$ and $u,v \in \mathfrak{m}$.

Hence rann$(A) = \begin{pmatrix} R & R \newline \mathfrak{m} & \mathfrak{m}\end{pmatrix}$ has size $|k|^{10}$, whereas lann$(A) = \begin{pmatrix} \mathfrak{m} & \mathfrak{m} \newline \mathfrak{m} & \mathfrak{m}\end{pmatrix}$ has size $|k|^8$.

• Thank you Konstantin. This is a nice counterexample. I think that the answer is yes in the case in which the finite commutative ring has identity (the proof of Ralph assumes that the ring has identity ). Jul 14 '12 at 13:47
• The ring $R$ in my example also has an identity. Ralph proved that the answer is yes when $R$ is a product of local principal ideal rings (with identity), but not every finite commutative local ring has to have the property that every ideal is principal, such as my $R$. Jul 14 '12 at 15:07

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

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

• Just out of curiosity, does Kaplansky's result also hold for nonsquare matrices ? Jul 13 '12 at 21:38
• Yes, it does. .. Jul 13 '12 at 22:01