Tensor product of simple algebras (4)

Posted: January 23, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: ,

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 2S \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

  1. Reza says:

    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.

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