Ring with 10 elements is isomorphic to $\mathbb{Z}/10 \mathbb{Z} $

How do I prove that a finite ring $R$ of order 10 is isomorphic to the ring $\mathbb{Z}/10 \mathbb{Z}$?

I know that as a group under addition, $(R,+)$ is isomorphic to the group $(\mathbb{Z}/10 \mathbb{Z}, + )$, but the multiplication is rather mysterious to me.

Solution 1:

I assume your definition of "ring" requires a unit element, which I'll write as $1$ (without that requirement, the statement is false: you could make all products be $0$). Now if $1+1=0$, we'd have $r+r = (1+1)\cdot r = 0$ for all $r \in R$, but then the order of the additive group of $R$ couldn't be $10$. Similarly, $1+1+1+1+1$ can't be $0$. So $1$ must have order $10$ in the additive group of $R$, and all members of $R$ are $0, 1, 2=1+1, \ldots, 9=1+1+1+1+1+1+1+1+1$. Expanding it out, any $i \cdot j$ is the sum of $i j$ $1$'s, and this is $k$ where $k \equiv i j \mod 10$.