# If R is a finite ring without unity, then every element of R must be zero divisor?

I know two results- "$$R$$ is a finite ring with unity, each element of $$R$$ is a unit or a zero divisor"

"If $$R$$ is a finite ring without zero divisors, $$R$$ must have a unity"

The commutativity does not require in both case as we can consider both $$aR$$ and $$Ra$$

Now, my question is, if $$R$$ is a finite ring without unity, then every element of $$R$$ must be zero divisor?

I am talking about finite non-trivial rings only.

In $$2\mathbb{Z}$$, we find no elements as unit, nor a zero divisor

This is simply not possible in finite ring.

In there is no zero divisor, there comes unity, and therefore every element must be a unit

If there is atleast one zero divisor, then we already find one zero divisor. Now the most doubtful question arises when it is a ring with zero divisors, and without unity, can we find a ring where some elements are zero divisors and some are nothing (since, no unity, they can't be unit, they are also not zero divisors as per my statement)

Just like in $$M_2(2\mathbb{Z})$$, we find zero divisors, and some elements who are not unit nor a zero divisor

But failed to find such an example in case of finite ring

Every element is becoming zero divisor in this case (left or right side)

## 2 Answers

The original proof (when $$R$$ has an unity) goes like this:

Consider for $$r \in R$$ the map $$\begin{array}{rrcl} \tau_r & R & \to & R\\ &s&\mapsto & rs \end{array}$$ It is injective iff it is surjective since $$R$$ is finite.

Case 1: it is bijective, then $$r$$ is an unit (just consider $$r^{-1} = \tau_r^{-1} (1)$$.

Case 2: Since $$\tau_r$$ is an endomomorphism of right $$R$$-modules ( $$\tau_r \in \operatorname{End}_{\operatorname{Mod} R} (R_R)$$). Not being injective is equivalent to have a non trivial kernel (monomorphism are injective morphism in this category), in this case you have $$\ker \tau_r \neq \{0\}$$ thus there exists $$r' \in R\setminus \{0\}$$ such that $$rr' = \tau_r(r') = 0$$.

The converse is true for right multiplication (reasoning with left modules).

NB: $$R_R$$ is the right module which underlying set is $$R$$ and the multiplication is given by $$r \cdot s = rs$$. Symmetrically, $${}_R \! R$$ is $$R$$ endowed with the natural left $$R$$-module structure. You can probably avoid using modules by considering $$R$$ as an additive group instead of module.

For your question, take $$R$$ a finite ring without unity, this is equivalent to have $$\tau_r \neq \operatorname{id}_{R_R}$$ for every $$r$$. So the ring homomorphism: $$\begin{array}{rcl} R & \to & \operatorname{End}_{\operatorname{Mod} R} (R_R)\\ r&\mapsto & \tau_r \end{array}$$ Does not have $$\operatorname{id}_{R_R}$$ in its image. Now suppose it has an automorphism in its image, $$\phi = \tau_r$$ for some $$r$$. Since $$\operatorname{End}_{\operatorname{Mod} R} (R_R)$$ has cardinality at most $$\sharp R^{\sharp R}$$, it is finite, so $$\langle \phi \rangle \subset \operatorname{Aut}_{\operatorname{Mod}-R} (R_R)$$ is cyclic, thus there exists $$n >0$$ such that $$\phi^n = \operatorname{id}_{R_R}$$. This means that $$\tau_{r}^n = \tau_{r^n} = \operatorname{id}_{R_R}$$ which is absurd.

Thus $$\forall r \in R , \, \tau_R \notin \operatorname{Aut}_{\operatorname{Mod}-R} (R_R)$$, which means that it is not bijective, thus not injective, thus its kernel is non-trivial, thus $$r$$ is a divisor of $$0$$.

In conclusion, every element of your non-unital ring is a divisor of $$0$$.

EDIT: I think it might not be clear for everyone that in this setting, an unit on the left is an unit on the right (or equivalently for $$I \in R, \, \tau_i = \operatorname{id} R_R \iff \forall r \in R, \, ir=r=ri$$.
The proof is very similar: let us consider the ring morphism: $$\begin{array}{rcl} R^\text{op} &\to& \operatorname{End}_{R-\operatorname{Mod}} ({}_R \! R)\\ r &\mapsto& (\tau_r': s \mapsto sr) \end{array}$$ If we suppose that $$\tau_i = \operatorname{id}_{R_R}$$ (which means $$\forall r, \, ir = r$$) then since $$\tau_i'$$ belongs to the finite group $$\operatorname{Aut}_{R-\operatorname{Mod}} ({}_R \! R)$$, it has finite order, hence $$\exists k>0$$ such that $${\tau_i '}^k = \operatorname{id}_{{}_R \! R}$$.
But since $$i^k =i$$, $${\tau_i '}^k = \tau_{i^k}' = \tau_i'$$, thus $$\tau_i ' = \operatorname{id}_{{}_R} \! R$$ hence $$\forall r \in R, \, ri = r$$.

If $$A$$ is a finite ring without an identity, suppose $$r \neq 0$$ is not a zero-divisor. Consider $$\{ r^n \colon n \geq 1 \}$$. Since this is infinite, there must be a minimal $$k$$ such that $$r^k = r^l$$ for some $$l. Then $$r^{l-1}(r^{k-l+1}-r)=0$$ so as $$r$$ is not a zero-divisor we must have $$r^d=r$$ where $$d=k-l+1\geq 2$$.

Now if $$\rho(a) = ra$$ is the operation of right-multiplication by $$r \in A$$ then $$\rho$$ is a left A-module homomorphism of $$A$$. Thus it is injective if and only if $$\ker(\rho)=0$$, which is if and only if $$r$$ is not a right zero-divisor. Since we are assuming $$r$$ is not a zero-divisor, it follows that $$\rho$$ is injective and hence, as $$A$$ is finite, bijective. Thus $$A=A \cdot a$$. Similarly $$\lambda(a)= ra$$ is injective and hence bijective, so that $$A=a \cdot A$$.

But now it is clear that $$r_1 = r^{d-1}$$ is an identity for $$A$$: if $$a \in A$$ then as as the property that $$r_1 \cdot a = a$$ for all $$a \in A$$. Indeed if $$a \in A$$ we may write $$a=r \cdot a_1=a_2 \cdot r$$, and thus $$r_1 \cdot a=r^da_1 =ra_1 =a$$, and similarly $$a \cdot r_1 = a_2 r^d = a_2 r = a$$.

Thus if $$A$$ is a finite ring without an identity, every element must be a zero-divisor.