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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 11 down vote accepted

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 $|k|^8$.

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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 ). –  zacarias Jul 14 '12 at 13:47
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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 $R$. –  Konstantin Ardakov Jul 14 '12 at 15:07
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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

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

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