Tensor product of simple algebras need not be simple

Posted: January 29, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: ,

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

Example 1. Let p be a prime number, F :=\mathbb{Z}/p\mathbb{Z} and R:=F(x) \otimes_{F(x^p)} F(x), where  F(x) is the field of rational functions in x over F. Then R is not simple.

Proof. Since a commutative simple ring is a field, we just need to show that R is not a field. We will do that by proving that R has a non-zero nilpotent element and thus R is not even a domain! Let y=x \otimes 1 - 1 \otimes x. The set \{1,x \} is linearly independent over F(x^p) because otherwise we would have f(x^p)=xg(x^p), for  some f,g \in F[x], which is absurd. So y \neq 0, by Lemma 1. We now show that y^p=0, which will complete the proof: note that since x^p \in F(x^p), we have x^p \otimes u = 1 \otimes x^pu for all u \in F(x). Thus

y^p = x^p \otimes 1 - 1 \otimes x^p = 1 \otimes x^p - 1 \otimes x^p=0. \ \Box

Example 2. R := \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} is not simple.

Proof. Again, if R was simple, then it would be a field because it is commutative. We will show that R is not even a domain. Let x = 1 \otimes i - i \otimes 1 and y = 1 \otimes i + i \otimes 1. Then, both x,y are non-zero elements of R by Lemma 1 and the fact that \{1,i\} is linearly independent over \mathbb{R}. It is quickly seen that xy=0. \ \Box

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Comments
  1. Kyon S says:

    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 A is a finite dimensional k algebra, and K/k a separable extension, then A \otimes_k K is semisimple?

    • Yaghoub Sharifi says:

      No. An easy example:
      Let K=k be a field of characteristic zero and consider the 3-dimensional k-algebra A: = \begin{pmatrix} k & k \\ 0 & k \end{pmatrix}. Then J(A)=\begin{pmatrix} 0 & k \\ 0 & 0 \end{pmatrix} \neq 0 and so A \otimes_k K \cong A is not semisimple.

      • Kyon S says:

        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,

      • Yaghoub Sharifi says:

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

  2. JoseBrox says:

    Neat post! Thank you for your effort.

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