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

**Theorem**. Let be an algebraically closed field. Let be a commutative -domain and let be a -domain. Then is a -domain.

*Proof*. Suppose that is not a domain. So there exist non-zero elements such that Let and where and both sets and are -linearly independent. Let which is a commutative domain because Also, note that Now, since there exist integers such that and Therefore, by the third part of the corollary in part (2), there exists a ring homomorphism such that and Let Then is a ring homomorphism and hence Therefore either or because is a domain. But because is -linearly independent and Also, because is -linearly independent and This contradiction proves that is a domain.

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

**Question**. Let be an algebraically closed field and let be two finite dimensional division -algebras. Will always be a domain?

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

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.

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

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.