Finite ring ideal

Question

Does a finite ring $R$ may have any other ideal $I$ that satisfies the condition $I \subset R$ except $\langle 0 \rangle$

I could prove that it is the only one ideal $\subset R$ by saying: let $I$ be left ideal, than $\exists x,y \in R (x \neq y), \forall a \in I, (a \neq 0) \rightarrow (ax = bx) \rightarrow x = y $ But I'm not sure especially the assumption that $(ax = ay) \rightarrow x = y$ when $a \neq 0$

Answer 1

Hint : Consider the ring $\mathbb{Z}/6\mathbb{Z}$. Let $I = \langle 3 \rangle$. Then does $I$ satisfy the condition?

Let me add the following: Here we can see that $3 \cdot 2 = 0 = 3 \cdot 4$, but $2 \neq 4$ in $\mathbb{Z}/6\mathbb{Z}$.

Answer 2

Suppose that, for $a\ne0$, $ax=ay$ implies $x=y$. Then the map $x\mapsto ax$ is injective and therefore surjective. This implies every $a\ne0$ has a right inverse.

In particular, $R$ is a division ring, because if every nonzero element has a right inverse, then every nonzero element has an inverse: let $a\ne0$; then there exists $b$ so that $ab=1$; since $b\ne0$, there is $c$ so that $bc=1$. Hence $$ a=a1=a(bc)=(ab)c=1c=c $$ so also $ba=1$.

A finite division ring is commutative (Wedderburn), so it is a field.

Can you think to a finite ring that's not a field? Hint: if $n>0$, $\mathbb{Z}/n\mathbb{Z}$ is a field if and only if $n$ is prime.