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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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In what context have you seen this terminology? – M Turgeon Aug 30 '12 at 20:08
@MTurgeon: I updated. – Thomas Aug 30 '12 at 20:10
Can you give a link to a book/paper where it is used, or maybe you could add a quote to your question? – M Turgeon Aug 30 '12 at 20:11
I have seen this terminology once in the context of modules over a ring without unit, but I don't remember where exactly. That is why I'm asking for more context. – M Turgeon Aug 30 '12 at 20:13
@MTurgeon: Actually, you right: the ring doesn't have a unit. – Thomas Aug 30 '12 at 20:14
up vote 2 down vote accepted

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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Excellent. I actually happened to have that on my bookshelf. – Thomas Aug 30 '12 at 20:28
So are the $m_i$ fixed or something? Right now, to me, it looks like any module with a generating set has that property. – rschwieb Aug 30 '12 at 20:44
@rschwieb They're not fixed; in particular, they depend on $m$. The point is, if $R$ has a unit, every element is of this form. But it may not be the case if $R$ has no $1$. But you are right: if the module has a generating set, it is non-degenerate. – M Turgeon Aug 30 '12 at 21:08
@rschwieb I guess the most important point is, in a ring without unit, the generating elements may not be in the module they generate. – M Turgeon Aug 30 '12 at 21:09
In Cartier , the real point is rings with "sufficiently many idempotents": given any finite set of elements $x_1,\ldots,x_n$ in the ring, there is an idempotent $e$ s.t. $ex_i=x_i$ for all $i$. The idempotents have that property as well, so, given any finite set of them, there is a "lcm/gcd" idempotent which acts trivially on all of them. The modules of interest are "smooth" in the sense that every finite set of elements is fixed by some idempotent. My point is that the "non-degenerate" notion may be entirely fine in that continuation, but have complications otherwise. I don't know. – paul garrett Aug 30 '12 at 23:46

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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Could I ask what hit(s) left you with this impression? It is an interesting definition. – M Turgeon Aug 30 '12 at 20:29
@MTurgeon I glanced at probably half a dozen links, and only one or two bothered to talk about definitions of nondegeneracy. I gleaned that they were talking about the multiplication operation being nondegenerate in an algebra, and likewise in a module. I was surprised that almost nobody seemed to define it in the hits. – rschwieb Aug 30 '12 at 20:35
@MTurgeon This is what gave me the impression, I think. – rschwieb Aug 30 '12 at 20:39

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