**Problem**. Let be a finite ring with and suppose that satisfies the following properties

i)

ii)

Show that either or

**Solution**. First notice that both and satisfy i), ii). Since is the only ring with two elements, we may assume that Let First we show that is even. Suppose, to the contrary, that is odd. Then the group would have no element of order i.e. for all and thus

contradicting the property i).

Now let We show that for some prime and To see that, let be the prime factorization of If then

contradicting the property ii), because are non-zero distinct elements of So i.e. for some prime number and positive integer Now if then we can write

and again, by the minimal property of the elements and are non-zero and distinct and that contradicts the property ii). So

We now show that in fact We just proved that for some prime and Suppose that is a prime divisor of Then there exists that has order But then which gives the false result because are coprime. So is a power of But we have already showed that is even. So and thus either or If then would be a vector space over i.e. for some integer because and But then

contradicting the property ii). So and hence contains a subring

Finally, we show that Suppose, to the contrary, that and let We have because and so, since we must have either or by property ii). If then which gives by property ii), and that’s a contradiction. So i.e. we have shown that for any element But if then too, because and so which gives contradiction. So