## Tensor product of simple algebras (4)

Posted: January 23, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Theorem. (Azumaya – Nakayama, 1947) Let $A$ be a central simple $k$-algebra and let $B$ be an arbitrary $k$-algebra. If $J$ is a two-sided ideal of $A \otimes_k B,$ then $J = A \otimes_k (J \cap B).$

Proof. It is obvious for $J = (0)$ and so we may assume that $J \neq (0).$ Let

$I = J \cap B.$

Consider the natural $k$-algebra homomorphism $f: A \otimes_k B \longrightarrow A \otimes_k (B/I).$

Claim 1. $f(J) \cap (B/I) = (0).$ To see this, let $y = 1 \otimes_k (b + I) \in f(J) \cap (B/I).$ Let $\{a_i \}$ be a $k$-basis of $A$ which contains 1. Let $x \in J$ be such that $f(x)=y.$ So $x = \sum_{i=1}^n a_i \otimes_k b_i,$ for some integer $n$ and $b_i \in B.$ Then

$y=1 \otimes_k (b+I)=f(x)=\sum_{i=1}^n a_i \otimes_k (b_i + I).$

Thus for some $i$ we have $a_i = 1, \ b_i + I = b+I$ and $b_j + I = 0,$ for all $j \neq i.$ But then $x = 1 \otimes_k b_i \in J \cap B = I$ and hence $y = 1 \otimes_k (b+I)=1 \otimes_k (b_i + I) = 0.$

Claim 2. $J \subseteq \ker f.$ This is easy to see: since $f$ is clearly onto, $f(J)$ is a two-sided ideal of $A \otimes_k (B/I).$ Thus, by the theorem we proved in part (3), we must have either $f(J) \cap (B/I) \neq (0),$ which is not true by cliam 1, or $f(J)=(0).$

Claim 3. $J \subseteq A \otimes_k I.$ This will be a result of claim 2 if we prove $\ker f = A \otimes_k I.$ But it is obvious that $A \otimes_k I \subseteq \ker f$ and so we only need to show that $\ker f \subseteq A \otimes_k I.$ Let $x \in \ker f.$ Let $\{a_i \}$ be a $k$-basis for $A.$ Then $x = \sum_{i=1}^n a_i \otimes_k b_i,$ for some positive integer $n$ and some $b_i \in B.$ Then

$0 = f(x) = \sum_{i=1}^n a_i \otimes_k (b_i + I)$

and thus, since $a_i$ are linearly independent, $b_i + I = 0$ for all $i.$ Hence $b_i \in I$ for all $i$ and so $x \in A \otimes_k I.$

Claim 4. $A \otimes_k I \subseteq J.$ The reason is that if $a \in A, \ b \in I,$ then $a \otimes_k b = (a \otimes_k 1)(1 \otimes_k b) \in J$ because $1 \otimes_k b \in I \subseteq J$ and $J$ is an ideal of $A \otimes_k B.$ $\Box$

Corollary. 1) If $A$ is a central simple $k$-algebra and $B$ is a simple $k$-algebra, then $A \otimes_k B$ is a simple $k$-algebra.

2) If both $A$ and $B$ are central simple $k$-algebras, then $A \otimes_k B$ is a central simple $k$-algebra.

Proof. Part 1) is a trivial result of the above theorem. Part 2) follows from part 1) and Lemma 2. $\Box$

Example. Let $\mathbb{H}$ be the division algebra of quaternions over $\mathbb{R}.$ Then $\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \cong M_2(\mathbb{C}).$

Proof. Let $S=\mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}.$ The first part of the above corollary implies that $S$ is simple. Moreover, by Lemma 2, $Z(S)=\mathbb{C}$ and so $S$ is a central simple $\mathbb{C}$-algebra. Hence, by Remark 2$S \cong M_n(\mathbb{C}),$ for some integer $n \geq 1.$ Since $\mathbb{C}$ is a field extension of $\mathbb{R},$ we have $\dim_{\mathbb{C}}S = \dim_{\mathbb{R}} \mathbb{H}=4$ and thus $4=\dim_{\mathbb{C}} M_n(\mathbb{C})=n^2.$ Hence $n=2. \Box$

I am working on algebras which are not necessarily unital. I would like to now that if there exists a similar results when at least one of algebras is not unital? In my special case $A$ is a central-simple algebra without $1$ and $B$ is a field extension of $k$. I will appreciate any clue.