Hi all. This question is related to my previous one on tensor products:(adsbygoogle = window.adsbygoogle || []).push({});

Is there a way of "well-defining" a function on a tensor product M(x)N (where M,N are

both R-modules) ?

This is the motivating example for my question : Say we want to define a map f: M(x)M-->M by f(m(x)m')=m+m' . But then we have that (m(x)m')=(-m(x)-m') , so that f(m(x)m')=

f(-m(x)-m')=-m-m' . But in most cases we do not have m+m' =-m-m' , so this function is not well-defined, i.e., its value depends on the choice of representative .

Clearly we must find a way of defining a function that is constant in the classes (m(x)n) , but, how do we do that? One way would be to find a bilinear map defined on MxN and determining its image on M(x)N under the map (m,n)--> (m(x)n) , since the commutative triangle guarantees that the map is well-defined. Is there some other way of doing this, i.e., of "well-defining" maps on M(x)N ?

I suspect a necessary and sufficient condition is that a candidate function satisfies the tensor

relations:

i ) ((a+a')(x)b) = (a(x)b)+(a'(x)b)

ii) (a (x) (b+b'))=(a(x)b)+ a(x)b'

iii) r(a(x)b)=(ra(x)b)= (a(x)rb)

but I can't see a good way of proving this. Anyone?

Thanks.

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# Defining Functions on Tensor Products

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