## 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 $xy=0. \ \Box$

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.