Do all finite rings which are not fields have a non-trivial zero divisor?

Question

I know that any commutative ring without zero divisors must be a field, but if we have a finite ring, which isn't necessarily commutative or unital, must there exist a zero divisor? I guess $\mathbb{Z}/n\mathbb{Z}$ for even $n$ will always have a zero divisor, but what about other finite rings?

Answer 1

A finite ring without zero divisors is a field by Weddurburn's little theorem. Note that the top answer to this question shows that a finite ring that contains a nonzero divisor has a unit.

Answer 2

Let $R$ be a finite ring, not necessarily unital, not necesarrily commutative, but with no zero divisors. Pick any $r \in R$, $r \ne 0$, and consider the map

$f_r:R \to R, \; f_r(a) = ra, \; a \in R; \tag 1$

we note that the condition that $R$ be without zero divisors implies $f_r$ is injective:

$f_r(a_1) = f_r(a_2) \Longleftrightarrow ra_1 = ra_2 \Longleftrightarrow r(a_1 - a_2) = 0 \Longleftrightarrow a_1 - a_2 = 0 \Longleftrightarrow a_1 = a_2; \tag 2$

since $R$ is finite and $f_r$ is injective, it is also surjective; thus there exists $e \in R$ with

$re = f_r(e) = r; \tag 3$

next, consider the map

$g_r: R \to R, \; g(a) = ar, \; a \in R; \tag 4$

an argument parallel to that just given for $f_r$ shows that $g_r$ is also injective and surjective; thus for any $s \in R$ there is $t \in R$ with

$tr = g_r(t) = s; \tag 5$

it follows that

$se = (tr)e = t(re) = tr = s; \tag 6$

this shows that $e$ is a right multiplicative identity for $R$. Now, the above method and logic can easily be applied to show that there exists a left multiplicative identity $e' \in R$; that is,

$e'r = r, \; \forall r \in R\; \tag 7$

then

$e' = e'e = e, \tag 8$

which shows the left and right multiplicative identities are equal; we thus denote this element by $1_R$:

$1_R r = r = r 1_R, \; \forall r \in R; \tag 9$

next, we observe that the surjectivity of $f_r$ implies the existence of $r' \in R$ with

$rr' = f_r(r') = 1_R; \tag{10}$

likewise, the surjectivity of $g_r$ forces the existence of $r'' \in R$ with

$r''r = 1_R; \tag{11}$

also,

$r'' = r'' 1_R = r''(rr') = (r''r)r' = 1_R r' = r'; \tag{12}$

we thus see that each $r \in R$ has a unique two-sided multiplicative inverse, which we deonte by $r^{-1}$:

$r' = r'' = r^{-1}. \tag{13}$

We have thus established than any finite ring with no zero divisors has a multiplicative identity and an inverse for each non-zero element; thus it is a division ring, and we are now in position to invoke Wedderburn's little theorem to conclude that $R$ must be commutative, i.e. $R$ is a field.

Working backwards via contraposition, we see that any finite ring which is not a field must be possessed of zero divisors.

Note: When I learned Wedderburn's Little Theorem, it was stated as A finite division ring is a field. Based on this memory, I established that $R$ is a division ring before invoking Wedderburn. But I see from the linked citing that the theorem is often stated as A finite domain is a field, which I guess means I've done a little extra (and possibly redundant) work here. End of Note.

Answer 3

A finite ring $R$ (not necessarily commutative or unital) is certainly (left and right) artinian.

Suppose $R$ has no left zero divisor, that is, for $x\in R$, $x\ne0$, $rx=0$ implies $r=0$; in other words the left annihilator of $x$ is $\{0\}$. If $I$ is a minimal left ideal and $x\in I$, $x\ne0$, we have $Rx=I$; however, $Rx$ is isomorphic to $R$ modulo the left annihilator of $x$. Therefore $R\cong I$ and so $R$ is simple artinian (in particular semisimple).

By Wedderburn “big” theorem, which also applies to nonunital rings, $R$ is isomorphic to the full $n\times n$ matrix ring over a (finite) division ring. No zero divisors implies $n=1$. So $R$ is a finite division ring which, by Wedderburn's “little” theorem it is commutative.