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.
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 $k \equiv i j \mod 10$.
Note that if $1+1=0$ then $\mathrm{char}R =2$ and the additive group if it will have order $10$ then it should have a subgroup of order $5$ which is cyclic and generated by $x$. But $x^2=x+x=0$ contradiction.