## Tensor product of division algebras (3)

Posted: October 29, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

Here you can see part (2). We are now ready to prove a nice result.

Theorem. Let $k$ be an algebraically closed field. Let $A$ be a commutative $k$-domain and let $B$ be a $k$-domain. Then $A \otimes_k B$ is a $k$-domain.

Proof. Suppose that $A \otimes_k B$ is not a domain. So there exist non-zero elements $u, v \in A \otimes_k B$ such that $uv=0.$ Let $u = \sum_{i=1}^m a_i \otimes_k b_i$ and $v = \sum_{i=1}^n a_i' \otimes_k b_i',$ where $a_i,a_i' \in A, \ b_i,b_i' \in B$ and both sets $\{b_1, \ldots , b_m\}$ and $\{b_1', \ldots , b_n'\}$ are $k$-linearly independent. Let $C=k[a_1, \ldots , a_m, a_1', \ldots , a_n'],$ which is a commutative domain because $C \subseteq A.$ Also, note that $u,v \in C \otimes_k B \subseteq A \otimes_k B.$ Now,  since $u, v \neq0,$ there exist integers $r,s$ such that $a_r \neq 0$ and $a_s' \neq 0.$ Therefore, by the third part of the corollary in part (2), there exists a ring homomorphism $\varphi : C \longrightarrow k$ such that $\varphi(a_r) \neq 0$ and $\varphi(a_s') \neq 0.$ Let $\psi = \varphi \otimes \text{id}_B.$ Then $\psi : C \otimes_k B \longrightarrow k \otimes_k B \cong B$ is a ring homomorphism and hence $\psi(u)\psi(v)=\psi(uv)=0.$ Therefore either $\psi(u)=0$ or $\psi(v)=0,$ because $B$ is a domain. But $\psi(u)=\sum_{i=1}^m \varphi(a_i) \otimes_k b_i \neq 0,$ because $\{b_1, \ldots , b_m \}$ is $k$-linearly independent and $\varphi(a_r) \neq 0.$ Also, $\psi(v)=\sum_{i=1}^n \varphi(a_i') \otimes_k b_i' \neq 0,$ because $\{b_1', \ldots , b_n' \}$ is $k$-linearly independent and $\varphi(a_s') \neq 0.$ This contradiction proves that $A \otimes_k B$ is  a domain. $\Box$

Let $k$ be an algebraically closed field. A trivial corollary of the theorem is a well-known result in field theory: if $F_1,F_2$ are two fields which contain $k,$ then $F_1 \otimes_k F_2$ is a commutative domain. Another trivial result is this: if $k$ is contained in both a field $F$ and the center of a division algebra $D,$ then $F \otimes_k D$ is a domain.

Question. Let $k$ be an algebraically closed field and let $D_1,D_2$ be two finite dimensional division $k$-algebras. Will $D_1 \otimes_k D_2$ always be a domain?

Answer. No! See the recent paper of Louis Rowen and David Saltman for an example.

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Comments
1. Daniel says:

In their paper, the two division k-algebras were finite dimensional over their centers. If k is algebraically closed then we can’t even have any finite-dimensional division k-algebras over k, other than k itself.

• galoistron says:

$\mathbb{H}$ is not a division algebra over $\mathbb{C}$. It is over $\mathbb{R}$. There are no division algebras over an algebraically closed field.

• Yaghoub Sharifi says:

Sure, if k is an algebraically closed field, then there exists no finite dimensional “central” division k-algebra (except k itself of course).
But here (and in Rowen’s paper) k is not the center, it’s inside the center and division algebras are finite dimensional over their center not over k.