bilinearity and commutative rings
Let be a ring and and be -modules. If is -bilinear then is also -middle linear.
Given , and then and so . ∎
We may assume that is left non-degenerate. Let . Then for all and it follows that
Therefore , where . This makes an element of the left radical of as it is true for all . However is non-degenerate so the radical is trivial and so for all . Since is a faithful -module this makes for all . That is, is commutative. ∎
Alternatively we can interpret the result in a weaker fashion as:
Let be a ring and and be -modules. If is -bilinear with then every element acts trivially on one of the three modules , or .
Suppose , and . Then we have shown for all and . As it follows that . ∎
Whenever a non-commutative ring is required for a biadditive map it is therefore often preferable to use a scalar map instead.
|Title||bilinearity and commutative rings|
|Date of creation||2013-03-22 17:24:19|
|Last modified on||2013-03-22 17:24:19|
|Last modified by||Algeboy (12884)|