By Corollary 1, if are simple -algebras and the center of or is then is simple. The condition that is the center of or cannot be omited, as the following examples show.

**Example 1**. Let be a prime number, and , where is the field of rational functions in over Then is not simple.

*Proof*. Since a commutative simple ring is a field, we just need to show that is not a field. We will do that by proving that has a non-zero nilpotent element and thus is not even a domain! Let . The set is linearly independent over because otherwise we would have for some which is absurd. So by Lemma 1. We now show that which will complete the proof: note that since we have for all Thus

**Example 2**. is not simple.

*Proof*. Again, if was simple, then it would be a field because it is commutative. We will show that is not even a domain. Let and Then, both are non-zero elements of by Lemma 1 and the fact that is linearly independent over It is quickly seen that

The first example actually provides an example of a simple, finite dimensional algebra whose base extension along a purely inseparable extension becomes even not semi simple.

Is it true that if is a finite dimensional algebra, and a separable extension, then is semisimple?

No. An easy example:

Let be a field of characteristic zero and consider the 3-dimensional -algebra Then and so is not semisimple.

Thanks for the quick response, but, sorry,I mistyped the condition : I meant the following — if A is a *semisimple* algebra, say finite dimensional over a field k, and k’ a finite separable extension of k, is it true that A \otimes_k k’ remains semisimple? I think this should at least be true when k has char 0, right? I might be wrong… Best regard,

Yes, if is a field of characteristic zero and if is a finite dimensional semisimple -algebra and if is a field extension, then is certainly semisimple (see page 108 of Anthony Knapp’s book “Advanced Algebra”).

Neat post! Thank you for your effort.