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.

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