Given R is a field, $A\in M_n(R)$ is not a zero divisor implies A is invertible. Counterexample to R commutative? I am attempting to prove that given a field R, a matrix $A \in M_n(R)$ is invertible if and only if $A$ is not a zero divisor in $M_n(R)$. Furthermore I am trying to figure out whether or not this iff holds when R is only assumed commutative.
I have seen other posts asking similar questions, but they use methods (e.g. fraction fields) that are at least three lectures ahead of where I am currently at in my algebra class. I have proven the "invertible implies not a zero divisor" direction, can someone please help me with the converse and in finding this counterexample to the commutative R version - if there even exists one, but with how the proof  of the R is a field case is going it seems like we do need R to be a field.
I have only just learned about matrix rings as well as types of rings like fields and domains and subrings.
 A: [Disclaimer: I have to assume that you know a little about finite-dimensional vector spaces over a field. I don't see how you could approach this question otherwise.]
If $R$ is a field, then $M_n(R)$ is a vector space over $R$ under matrix addition and  scalar multiplication. For each $A \in M_n(R)$, the mapping $L_A : M_n(R) \to M_n(R)$ such that $L_A(B) = AB$ for all $B$ is a linear transformation from $M_n(R)$ to itself. If $A$ is not a zero divisor, then $L_A$ is injective (because $L_A(B) = AB = 0$ iff only $B = 0$, implying that $\ker(L_A) = \{0\}$). But $M_n(R)$ is finite dimensional (with dimension $n^2)$, so if $L_A$ is injective, it is also surjective. This implies that there is some $A' \in M_n(R)$ with $AA' = L_A(A') = I$ (where $I$ is the unit matrix). So we have that (i):
$$AA' = I$$
and that (ii):
$$L_A(A'A) = A(A'A) = (AA')A = IA = A = AI = L_A(I)$$
and so as $L_A$ is injective, (ii) gives us that (iii):
$$A'A = I$$
(i) and (iii) give us that $A$ is invertible with inverse $A'$.
The comments have dealt with things that can go wrong if $R$ is not assumed to be a field. E.g., if $R$ is only assumed to be an integral domain, $A$ could represent scalar multiplication by a non-zero scalar with no inverse.
A: Here’s a trick to see that it does not only hold for fields.
Let’s say you believe that $M_4(F)$ has that property, that all elements are either a zero divisor or a unit.  Then you already know that $M_2(M_2(F))$ (it’s the same ring) also has that property, and $M_2(F)$ is certainly not a field.
There is a very large class of rings which has this property: they are called right Artinian rings.  This post had a lot of information on that in the solutions.
It turns out that square matrix rings over right Artinian rings are right Artinian too, so it gives you lots of noncommutative examples.
