## Tensor product of simple algebras (3)

Posted: January 23, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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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! $\Box$