Annihilators in Matrix Rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:22:32Z http://mathoverflow.net/feeds/question/102164 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102164/annihilators-in-matrix-rings Annihilators in Matrix Rings zacarias 2012-07-13T18:40:19Z 2012-07-13T22:33:56Z <p>I think this is not a research question, but in stackExchange remained unanswered.</p> <p>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?</p> http://mathoverflow.net/questions/102164/annihilators-in-matrix-rings/102182#102182 Answer by Ralph for Annihilators in Matrix Rings Ralph 2012-07-13T20:54:55Z 2012-07-13T21:16:01Z <p>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. </p> <blockquote> <p>If each $R_i$ is a local principal ideal ring (or more generally an elementary divisor ring), then $$|\text{Ann}^l(A)|=|\text{Ann}^r(A)|.$$ </p> </blockquote> <p>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. </p> <p><strong>Remark:</strong> If $R$ is a (not necessarily finite) product of commutative principal ideal rings, then the proof above shows </p> <blockquote> <p>$\qquad\qquad \text{Ann}^r(A) = V^{-1}U^T \text{Ann}^l(A)^T$</p> </blockquote> http://mathoverflow.net/questions/102164/annihilators-in-matrix-rings/102189#102189 Answer by Konstantin Ardakov for Annihilators in Matrix Rings Konstantin Ardakov 2012-07-13T22:33:56Z 2012-07-13T22:33:56Z <p>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.</p> <p>Now let $A = \begin{pmatrix} 0 &amp; x \newline 0 &amp; y \end{pmatrix} \in M_2(R)$. Then</p> <p>$\begin{pmatrix} 0 &amp; x \newline 0 &amp; y \end{pmatrix} \begin{pmatrix} a &amp; b \newline c &amp; d \end{pmatrix} = \begin{pmatrix} cx &amp; dx \newline cy &amp; dy\end{pmatrix}$ whereas $\begin{pmatrix} a &amp; b \newline c &amp; d \end{pmatrix}\begin{pmatrix} 0 &amp; x \newline 0 &amp; y \end{pmatrix} = \begin{pmatrix} 0 &amp; ax +by \newline 0 &amp; cx + dy\end{pmatrix}$.</p> <p>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 $u,v \in \mathfrak{m}$. </p> <p>Hence rann$(A) = \begin{pmatrix} R &amp; R \newline \mathfrak{m} &amp; \mathfrak{m}\end{pmatrix}$ has size $|k|^{10}$, whereas lann$(A) = \begin{pmatrix} \mathfrak{m} &amp; \mathfrak{m} \newline \mathfrak{m} &amp; \mathfrak{m}\end{pmatrix}$ has size $|k|^8$.</p>