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**jakncoke** Ok. Let R be the ring. Since R has more than one element, take $\displaystyle \bar{R} = R - {0} $. Now since R is finite, so is $\displaystyle \bar{R}$. Since $\displaystyle \bar{R}$ is finite and closed under multiplication, take some element $\displaystyle x \in \bar{R} $ and $\displaystyle x * \bar{R} = \bar{R} = \bar{R} * x $ (This is because no zero divisors imply cancellation laws, so $\displaystyle x*x_0 = x*x_1 \text{ means } x_0 = x_1$. so There exists a $\displaystyle x_0 \in \bar{R} \text{ such that } x*x_0 = x $. Now for any element $\displaystyle y \in \bar{R} $ $\displaystyle x*x_0*y = x*y $ Since, no zero divisors imply cancellation laws are present, $\displaystyle x(x_0*y)=x(y) \text{ means } x_0*y = y $. Since $\displaystyle y * \bar{R} = \bar{R} $ say $\displaystyle y*x_1 = y $. So $\displaystyle x_0*y = y*x_1 = y $ Multiply both right side by y, you get $\displaystyle x_0*y*y = y*x_1*y = y^2 = y*(x_1*y) $ which means $\displaystyle y = x_1*y = y*x_1 $. Since y was any element from $\displaystyle \bar{R} $. the element $\displaystyle x_1 $ is the identity.

As for Inverses, since for any $\displaystyle x \in x*\bar{R} = \bar{R} =\bar{R}*x $ there exists some $\displaystyle x_0 \in R $ such that $\displaystyle x * x_0 = 1 = x_1 * x $. multiply left by$\displaystyle x_1$, you get $\displaystyle x_1*x*x_0 = x_1 $ and so since $\displaystyle x_1*x = 1 $, we get $\displaystyle 1*x_0 = x_1 $, so $\displaystyle x_0 * x = x * x_0 = 1 $ thus x has an inverse in $\displaystyle \bar{R} $.