Let R be a commutative ring (with 1) and Rn×n be the ring of n×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 Rn×n a zero divisor or a unit as well?

We know that if A∈Rn×n, then AC=CA=det(A)In where C is the classical adjoint of A and In is the identity matrix.

This means that if det(A) is a unit of R, then A is a unit of Rn×n (since A−1=(det(A))−1C). Also, the converse holds, if A is a unit of Rn×n, then det(A) is a unit.

I would like to know if one can show 0≠A∈Rn×n is a zero divisor if det(A) is zero or a zero divisor.

Things to consider:

1) This is true when R=F a field. Since over a field (no zero divisors) and if det(A)=0 then Ax=0 has a non-trivial solution and so B=[x|0|⋯|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=[111000000]impliesclassicaladjoint=0

(All 2×2 sub-determinants are zero.)3) This is true when R is finite (since Rn×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=diag(r,1,…,1) (this is non-zero, not a zero divisor, and is not a unit).

Edit:Not totally unrelated…

https://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=[a11a12a21a22]⟹classicaladjoint=C=[a22−a12−a21a11]

Thus if det(A)b=0 for some b≠0, then either bC=0 so that all of the entries of both A and C are annihilated by b so that A(bI2)=0 or bC≠0 and so A(Cb)=det(A)bI2=0I2=0. Thus A is a zero divisor.

**Answer**

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 detA 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 detA is a zero divisor, and thus Rn×n inherits from R the property that all non-zero elements are either units or zero divisors.

**Attribution***Source : Link , Question Author : Bill Cook , Answer Author : Community*