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Let $R$ be a non-trivial commutative ring, hence $R$ has IBN property.

Let $M$,$N$ be free $R$-modules.

Then the tensor product $M\otimes_R N$ is free and $rnk(M\otimes_R N)=rnk(M)rnk(N)$.

Let $A,B$ be bases for $M,N$ respectively.

Then how do I prove that $a\otimes b=a'\otimes b'$ iff $a=a'$ and $b=b'$, ($a\in A, b\in B$)using only above facts?

I know that it can be shown by applying a finite sequence of natural isomorphisms to have $M\otimes_R N\cong \otimes_{(i,j)\in A\times B} R$ and this proves the question.

However, I'm curious whether it can be proven not using isomorphisms but just with the rank.

Rubertos
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1 Answers1

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Clearly, the set $\{a\otimes b|a\in A,b\in B\}$ generates $M\otimes_RN$. If you know that the latter is free and of rank $|A|\cdot|B|$, it follows that the elements in the above set have to be distinct.

Amitai Yuval
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  • I'm asking about infinite cases – Rubertos Mar 17 '15 at 08:54
  • @Rubertos In that case, the rank equation you write seems to be less useful. On the other hand, you are asking for a proof based only on this equation... I don't know if there is one. – Amitai Yuval Mar 17 '15 at 08:59
  • It is maybe off the topic, but I'm not even sure that if a set $A$ generates $M\otimes_R N$, then $|A|$ should be not smaller than the rank (your argument uses this) When $R$ is a division ring, this is true, but when it is a commutative ring, is it still true? – Rubertos Mar 17 '15 at 09:06
  • What is your definition for rank? – Amitai Yuval Mar 17 '15 at 09:25
  • Cardinality of a basis, which is uniquely defined for IBN rings such as commutative ring and division ring etc. – Rubertos Mar 17 '15 at 09:26