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.

Answer 1

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".

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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.