## Tensor product of division algebras (1)

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