How can I find all rings with 8 elements such that they do not have any nilpotent elements?

Question

I was thinking about Jacobson radical and all rings like $R/J(R)$. But I can not go further.

Answer 1

Such a ring has trivial Jacobson radical (for the elements of the radical of an artinian ring are nilpotent) and is therefore semisimple. Wedderburn's theorem tells us that the ring is a finite direct product of matrix rings over division rings. Now the division rings appearing in that description are finite, so by another of Wedderburn's theorems they are in fact fields. The fields have each a number of elements dividing 8, so they have 2, 4 or 8, and are, as all finite fields, determined uniquely up to isomorphism by their order. On the other hand, if anybody the factors were matrices of size larger than 1, we would have nilpotents. It follows that the ring is a direct product of fields of orders 2, 4 and 8

Now we have to consider cases.

• If there is just one factor, we have a field of order 8.

• If there are three factors or more, then, as each factor has at least two elements, there are exactly three, and each has two elements. We get another example, the direct product of three copies of the field with two elements.

• Finally, suppose there are two factors, of cardinals $a$ and $b$. Since $ab=8$, we may suppose that $a=2$ and $b=4$. In this case the ring is the direct product of two fields of order $2$ and $4$.