In one of my questions (How to recover the integral group ring?), a statement is made for a commutative semisimple ring of characteristic $p^t, t\geq1$. Now I don't understand why there exist non-obvious examples of these. By Wedderburn-Artin, this ring is a direct product of fields $k_1 \times k_2 \times \ldots \times k_n$, and each of these factors has prime characteristic $p_i$. It seems to me that the characteristic of this product is equal to the product of all different (as in non-equal) characteristics appearing, which would imply that for a commutative semisimple ring of characteristic $p^t$, necessarily $t=1$, and it is thus a direct product of fields of characteristic $p$.


Well, the characteristic of a finite product of rings is the LCM of the characteristics of the rings involved.

Since a commutative semisimple ring is a product of fields, the characteristics are each prime or possibly 0. That makes the characteristic of the whole ring either 0, if one of them is 0, or else the LCM of all the prime characteristics, which will wind up being a squarefree number.

So, you're right: it's not possible to get characteristic $p^t$, $t>1$ for a commutative semisimple ring.

Let's discuss commutative rings with nonzero characteristic for a moment.

  1. In general, if $\{0\}$ is a prime ideal, the characteristic is prime.

  2. If $\{0\}$ is a semiprime ideal, the characteristic is a squarefree number.

  3. If $\{0\}$ is a primary ideal, then the characteristic is a prime power.

The zero ideal of semisimple rings is semiprime, so that's why we are in that case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.