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I know what a non-degenerate bi-linear form is, but what does it mean for say a left $R$-module $M$ to be non-degenerate? (Here $R$ is a ring without unit$)

I came across a module being called non-degenerate studying representation theory.

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    @MTurgeon: Actually, you right: the ring doesn't have a unit.2012-08-30

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Here is the definition which I have seen in Cartier's article on the Representation theory of p-adic groups (it appears in the Corvallis proceedings):

A module $M$ over a ring $R$ without unit is called non-degenerate if any element can be written as $a_1m_1+\cdots+a_nm_n$, with $a_i\in R$ and $m_i\in M$.

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    @MTurgeon Thanks... I did not anticipate that could happen!2012-08-31
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In Bourbaki's Commutative Algebra, one has the following definition.

Suppose $A$ is a (commutative) ring and $S$ is a multiplicative subset containing no zero divisors. Let $B=S^{-1}A$, so that there is an injection $A\to B$. An $A$-submodule $M$ of $B$ is said to be non-degenrate (non-dégénéré) if $BM=B$.

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This is proving amazingly resistant to an internet search.

A few of the first hits left me with the impression it might just mean that the map $M\times R\rightarrow M$ is a nondegenerate bilinear map.

Edit: The link in the comments for this solution say something of this sort. They say "if $xR=0$ for an $x$ in the module, then $x=0$" I think it is mainly meant to guarantee the the annihilator of an element isn't the whole ring (having an identity normally precludes that.)

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    @MTurgeon [This](http://arxiv.org/pdf/0806.2089.pdf) is what gave me the impression, I think.2012-08-30