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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.

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up vote 13 down vote accepted

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 $k \equiv i j \mod 10$.

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I am having trouble seeing why $1_R + 1_R = 0 $ implies that the order of the additive group of $R$ wouldn't be 10. All I am able to deduce is that $ r = - r $ for all $ r \in R$, and I don't see the collapsing that would occur. @RobertIsrael – Peter Feb 19 '12 at 3:55
@Peter A group with 10 elements must contain an element of order 5 ;) – N. S. Dec 10 '12 at 7:24
The additive group generated by $1$ has order $2$, $5$ or $10$. But it can't be $2$ or $5$ so it's $10$. And that doesn't leave room for anything else in $R$. – Robert Israel Feb 26 '15 at 3:32

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.

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