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Suppose $R$ is a commutative ring and $M$ is a $R$-module. Then we can define the tensor product $M\otimes_R M$ and more generally the $k$-fold tensor powers $\otimes_R^kM$ for any $k\in\mathbb{N}$, which are again $R$-modules.

In addition the tensor power $T_R(M):=R\oplus (\oplus_{k\in\mathbb{N}}\otimes_R^{k}M)$ is an $R$-algebra with respect to the concatenation product $\otimes: T_R(M)\times T_R(M)\to T_R(M)\;;\;(x,y)\mapsto x\otimes_R y$.

Now the question is:

Under what additional assumptions on $R$ or on the module $M$ can we say that any tensor $x\in T_R(M)$ is a $R$-linear combination of decomposable tensors, that is $x=\sum_{i\in I}a_i(x_{i,1}\otimes_R\cdots \otimes_R x_{i,n_i})$ for some index set $I$ and $x_{i,j}\in M$?

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1 Answer 1

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Why do you feel that additional assumptions are necessary? Any element of the tensor algebra is a sum of tensors, and any tensor is a sum of simple tensors.

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The tensor algebra is a direct sum, so any element is a sum of 'homogeneous' tensors $x\in \otimes^n_R M$.Ok. But the second point needs more explanation. Is it a linear combination of simple tensors, i.e a finite sum? And why? A proof directly from the definition would be nice. .. If this is obvious, a proof should be easy. –  Mark Neuhaus Oct 7 '13 at 7:41
    
The standard construction of the tensor product makes this obvious, since it's given as a quotient of a free module (or free abelian group) generated by (formal) simple tensors. And yes, that means that every element will be a finite sum of simple tensors. I don't know what definition you're using, but one way or another I think the proof will come down to that construction. Link here: en.wikipedia.org/wiki/Tensor_product_of_modules#Construction –  you-sir-33433 Oct 7 '13 at 7:46
    
ok. thank you for the answer. –  Mark Neuhaus Oct 7 '13 at 9:59

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