bilinearity and commutative rings
We show that a bilinear map is almost always definable only for commutative rings. The exceptions lie only where non-trivial commutators act trivially on one of the three modules.
Lemma 1.
Let be a ring and and be -modules. If is -bilinear then is also -middle linear.
Proof.
Given , and then and so . ∎
Theorem 2.
Let be a ring and and be faithful -modules.
If is -bilinear and (left or right) non-degenerate,
then must be commutative
.
Proof.
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:
Corollary 3.
Let be a ring and and be -modules. If is -bilinear with then every element acts trivially on one of the three modules , or .
Proof.
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 |
---|---|
Canonical name | BilinearityAndCommutativeRings |
Date of creation | 2013-03-22 17:24:19 |
Last modified on | 2013-03-22 17:24:19 |
Owner | Algeboy (12884) |
Last modified by | Algeboy (12884) |
Numerical id | 5 |
Author | Algeboy (12884) |
Entry type | Theorem |
Classification | msc 13C99 |