Tensor product of simple algebras (3)

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

Note. To remind the reader again that if A and B are k-algebras and k is a field, then in A \otimes_k B we can identify A \otimes_k 1=\{a \otimes_k 1: \ a \in A\} and 1 \otimes_k B=\{1 \otimes_k b: \ b \in B\} with A and B respectively.

Theorem. Let A be a central simple k-algebra and let B be an arbitrary k-algebra. If J \neq (0) is a two-sided ideal of A \otimes_k B, then J \cap B \neq (0).

Proof. Among all non-zero elements of J choose

c = \sum_{i=1}^n a_i \otimes_k b_i,

where a_i \in A, \ b_i \in B and such that n is minimal. Then the set \{a_1, \cdots , a_n \} is linearly independent over k because otherwise one of a_i, say a_n, would be a linear combination of a_1, \cdots , a_{n-1} and that would give us b_i' \in B such that c = \sum_{i=1}^{n-1}a_i \otimes_k b_i' contradicting the minimality of n. Similarly, the set \{b_1, \cdots , b_n \} is linearly independent over k. Now, since A is simple and Aa_1A is a two-sided ideal of A and it contains the non-zero element a_1, we have Aa_1A=A. Thus there exist x_j, y_j \in A and a positive integer m such that

\sum_{j=1}^m x_j a_1y_j=1. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

Let z_i = \sum_{j=1}^m x_j a_i y_j, \ i=1, \cdots , n. Then z_1=1 by (1). Let

c' = \sum_{i=1}^n z_i \otimes_k b_i. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

Claim 1. c' \in J. Because J is a two-sided ideal and c \in J. Thus c' = \sum_{j=1}^m (x_j \otimes_k 1)c(y_j \otimes_k 1) \in J.

Claim 2. c' \in B. To see this, let a \in A. Then (2) gives us

(a \otimes_k 1)c' - c'(a \otimes_k 1) =\sum_{i=1}^n(az_i - z_ia) \otimes_k b_i = \sum_{i=2}^n(az_i - z_i a) \otimes_k b_i, \ \ \ \ \ \ \ (3)

because z_1=1. But, since (a \otimes_k 1) c' - c'(a \otimes_k 1) \in J, we will get (a \otimes_k)c' - c'(a \otimes_k 1)=0 by the minimality of n. Thus, by (3) and Lemma 1, we must have az_i-z_ia=0 for all i. That means z_i \in Z(A)=k. Hence by (2)

c'=1 \otimes \sum_{i=1}^n z_ib_i \in 1 \otimes_k B = B.

Claim 3. c' \neq 0.  Suppose to the contrary  that c' = 0. Then, by Lemma 1, \sum_{i=1}^n z_ib_i=0 and so b_i=0 for all i, because \{b_1, \cdots , b_n \} is k-linearly independent and z_i \in k. So c = \sum_{i=1}^n a_i \otimes_k b_i = 0. Contradiction! \Box

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Comments
  1. jaime says:

    The link to the lemma 1 are unavailable. Thanks

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