Do these matrix rings have non-zero elements that are neither units nor zero divisors? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:57:14Z http://mathoverflow.net/feeds/question/77816 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77816/do-these-matrix-rings-have-non-zero-elements-that-are-neither-units-nor-zero-divi Do these matrix rings have non-zero elements that are neither units nor zero divisors? Bill Cook 2011-10-11T14:06:06Z 2011-10-11T16:21:05Z <p>First, a disclaimer: This is a repost of <a href="http://math.stackexchange.com/questions/71235/do-these-matrix-rings-have-non-zero-elements-that-are-neither-units-nor-zero-divi" rel="nofollow">a question I asked on stackexchange</a> (no answer there).</p> <p>Let $R$ be a commutative ring (with $1$) and $R^{n \times n}$ be the ring of $n \times n$ matrices with entries in $R$.</p> <p>In addition, suppose that $R$ is a ring in which <b>every non-zero element is either a zero divisor or a unit</b> [For example: take any finite ring or any field.] My question:</p> <blockquote> <p>Is every non-zero element of $R^{n \times n}$ a zero divisor or a unit as well?</p> </blockquote> <p>We know that if $A \in R^{n \times n}$, then $AC=CA=\mathrm{det}(A)I_n$ where $C$ is the <a href="http://en.wikipedia.org/wiki/Classical_adjoint" rel="nofollow">classical adjoint</a> of $A$ and $I_n$ is the identity matrix. </p> <p>This means that if $\mathrm{det}(A)$ is a unit of $R$, then $A$ is a unit of $R^{n \times n}$ (since $A^{-1}=(\mathrm{det}(A))^{-1}C$). Also, the converse holds, if $A$ is a unit of $R^{n \times n}$, then $\mathrm{det}(A)$ is a unit.</p> <p>I would like to know if one can show $0 \not= A \in R^{n \times n}$ is a zero divisor if $\mathrm{det}(A)$ is zero or a zero divisor.</p> <p>Things to consider:</p> <p>1) This is true when $R=\mathbb{F}$ a field. Since over a field (no zero divisors) and if $\mathrm{det}(A)=0$ then $Ax=0$ has a non-trivial solution and so $B=[x|0|\cdots|0]$ gives us a right zero divisor $AB=0$.</p> <p>2) You can't use the classical adjoint to construct a zero divisor since it can be zero even when $A$ is not zero. For example:</p> <p>$$A=\begin{pmatrix} 1 &amp; 1 &amp; 1 \cr 0 &amp; 0 &amp; 0 \cr 0 &amp; 0 &amp; 0 \end{pmatrix} \qquad \mathrm{implies} \qquad \mathrm{classical\;adjoint} = 0$$ (All $2 \times 2$ sub-determinants are zero.)</p> <p>3) This is true when $R$ is finite (since $R^{n \times n}$ would be finite as well).</p> <p>4) Of course the assumption that every non-zero element of $R$ is either a zero divisor or unit is necessary since otherwise take a non-zero, non-zero divisor, non-unit element $r$ and construct the diagonal matrix $D = \mathrm{diag}(r,1,\dots,1)$ (this is non-zero, not a zero divisor, and is not a unit).</p> <p>5) This is somewhat related to the question: <a href="http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor" rel="nofollow">http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor</a></p> <p>6) This is definitely true when $n=1$ and $n=2$. It is true for $n=1$ by assumption on $R$. To see that $n=2$ is true notice that the classical adjoint contains the same same elements as that of $A$ (or negations):</p> <p>$$A = \begin{pmatrix} a_{11} &amp; a_{12} \cr a_{21} &amp; a_{22} \end{pmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{pmatrix} a_{22} &amp; -a_{12} \ -a_{21} &amp; a_{22} \end{pmatrix}$$</p> <p>Thus if $\mathrm{det}(A)b=0$ for some $b \not=0$, then either $bC=0$ so that all of the entries of both $A$ and $C$ are annihilated by $b$ so that $A(bI_2)=0$ or $bC \not=0$ and so $A(Cb)=\mathrm{det}(A)bI_2 =0I_2=0$. Thus $A$ is a zero divisor. </p> <p>7) Apparently strange behavior can occur when $R$ is non-commutative (not surprising). Like a matrix can be both a <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.100.5549" rel="nofollow">left inverse and left zero divisor</a>. [The determinant keeps this from happening in the commutative case.]</p> http://mathoverflow.net/questions/77816/do-these-matrix-rings-have-non-zero-elements-that-are-neither-units-nor-zero-divi/77834#77834 Answer by Manny Reyes for Do these matrix rings have non-zero elements that are neither units nor zero divisors? Manny Reyes 2011-10-11T16:21:05Z 2011-10-11T16:21:05Z <p>The answer given by David Speyer can be strengthened as follows. If $A$ is a non-invertible $n\times n$ matrix with entries in $R$ as described in the problem, then the linear maps $R^n \to R^n$ defined by either left or right multiplication are non-injective. In particular, $A$ is both a left-zero-divisor and a right-zero-divisor.</p> <p>This is a consequence of McCoy's rank theorem. You can find a nice, brief account of this in Section 2 of <a href="http://math.berkeley.edu/~lam/amspfaff.pdf" rel="nofollow">this paper by Kodiyalam, Lam, and Swan</a>. One consequence of the theorem is that for any commutative ring $R$, a square matrix $A$ over $R$ has linearly independent columns if and only if its determinant is not a zero-divisor, if and only if its rows are linearly independent.</p> <p>So if every element of $R$ is either invertible or a zero-divisor, this means that every square matrix over $R$ defines a linear transformation that is either invertible or non-injective.</p>