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Let $R$ be a commutative ring (with $1$) and $R^{n \times n}$ be the ring of $n \times n$ matrices with entries in $R$.

In addition, suppose that $R$ is a ring in which every non-zero element is either a zero divisor or a unit [For example: take any finite ring or any field.] My question:

Is every non-zero element of $R^{n \times n}$ a zero divisor or a unit as well?

We know that if $A \in R^{n \times n}$, then $AC=CA=\mathrm{det}(A)I_n$ where $C$ is the classical adjoint of $A$ and $I_n$ is the identity matrix.

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.

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.

Things to consider:

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 $AB=0$.

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:

$$A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \qquad \mathrm{implies} \qquad \mathrm{classical\;adjoint} = 0 $$ (All $2 \times 2$ sub-determinants are zero.)

3) This is true when $R$ is finite (since $R^{n \times n}$ would be finite as well).

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).

Edit: Not totally unrelated... http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor

Edit: One more thing to consider...

5) 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):

$$ A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \qquad \Longrightarrow \qquad \mathrm{classical\;adjoint} = C = \begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{22} \end{bmatrix} $$

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.

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I am going to try reposting this on Mathoverflow. –  Bill Cook Oct 11 '11 at 13:36
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up vote 4 down vote accepted

As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $\det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $\det A$ is a zero divisor, and thus $R^{n\times n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.

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Thanks @joriki ! For anyone who cares, the question was answered on Mathoverflow just now as well: mathoverflow.net/questions/77816/… –  Bill Cook Oct 11 '11 at 17:26
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