Zero divisor or unit

Question

Let $R$ be a finite ring with unity. Prove that every $x \neq 0$ is a zero divisor or a unit.

I made an attempt, but got stuck. I took a non-zero $x$ and defined the mapping $f: R \to R, y \mapsto xy$. This mapping may not be injective, but if it is, since it maps from $R$ to $R$ and is finite, $f$ is also surjective, and since $1$ is an element of $R$, there exists $y$ such that $xy = 1$. The ring is not necessarily commutative, however, so I also need to find a $y$ such that $yx = 1$. My thought was to define an analagous map $g: R \to R, y \mapsto yx$, which may or may not be injective. But then I run into problems where the first map is injective, but the second isn't.

Any help would be appreciated.

EDIT: The linked question does not answer this because it assumes that the ring is commutative. This ring is not commutative.

The linked question does not assume the ring is commutative, and several answers there (including the one by the user who marked this a duplicate) work just fine in the case of a noncommutative ring. — KReiser, Nov 01 '20 at 01:54

score 9 · Answer 1 · answered Oct 31 '20 at 23:45

What you have shown is:

If $x \neq 0$ is not a left (resp. right) zero divisor, then there exists $y$ such that $xy = 1$ (resp. $yx = 1$).

If $x$ is a left zero divisor but not a right zero divisor, then we will have $xz = 0$ for some $z \neq 0$ and $yx = 1$ for some $y$.

We then get $$0 = y\cdot 0 = yxz = 1 \cdot z = z,$$ which contradicts $z \neq 0$.

Therefore this situation cannot happen. Every left zero divisor is automatically a right zero divisor, and vice versa, so we may say unambiguously "zero divisor".

Thus when $x$ is not a zero divisor, we will have $xy = 1$ and $zx = 1$ for some $y, z$. This then gives $$y = 1 \cdot y = zxy = z \cdot 1 = z,$$ and we see that $y = z$ is the multiplicative inverse of $x$, hence $x$ is a unit.

Zero divisor or unit

1 Answers1

Linked