Tensor product of division algebras (1)

Posted: September 1, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , , ,

In this post we showed that \mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \cong M_4(\mathbb{R}). So the tensor product of two (finite dimensional) division algebras may not even be a reduced algebra let alone a division algebra. There is however the following simple result.

Theorem. Let D_1, D_2 be finite dimensional central division k-algebras. If \gcd(\dim_k D_1, \dim_k D_2)=1, then D_1 \otimes_k D_2 is a division algebra.

Proof. By the corollary in this post, D_1 \otimes_k D_2 is a finite dimensional central simple k-algebra. So

D_1 \otimes_k D_2 \cong M_n(D) \ \ \ \ \ \ \ \ \ \ (1)

for some finite dimensional central division k-algebra D and some integer n \geq 1.  We need to prove that n=1. Let \dim_k D_i=m_i, \ i=1,2. We are given that

\gcd(m_1,m_2)=1. \ \ \ \ \ \ \ \ \ \ (2)

Now, we have

D_i^{op} \otimes_k D \cong M_{r_i}(D_i'), \ \ i=1,2, \ \ \ \ \ \ \ \ \ (3)

for some finite dimensional central division k-algebras D_i' and integers r_i \geq 1. Also, by the theorem in this post,

D_i^{op} \otimes_k D_i \cong M_{m_i}(k), \ i=1,2. \ \ \ \ \ \ \ \ \ \ (4)

Thus, using (1), (3) and (4), we have

M_{m_1}(D_2) \cong M_{m_1}(k) \otimes_k D_2 \cong D_1^{op} \otimes_k D_1 \otimes_k D_2 \cong D_1^{op} \otimes_k M_n(D) \cong

M_n(D_1^{op} \otimes_k D) \cong M_n(M_{r_1}(D_1')) \cong M_{nr_1}(D_1').

Hence m_1=nr_1. Similarly M_{m_2}(D_1) \cong M_{nr_2}(D_2') and so m_2=nr_2. So n divides both m_1 and m_2 and hence n=1, by (2). \ \Box

See part (2) here.

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