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This question already has an answer here:

I am trying to find a right $R$-module $A$ and a left $R$-module $B$ such that the tensor product $A\otimes_R B$ has an element that is not of the form $a\otimes b$ for any $a,b$ in $A,B$ respectively. I could not find any. Is there any simple example? Thanks in advance

Edit: You may bring other examples, or consider my own example that taking $A,B$ to be $K^2$ for a field $K$, then $e_1\otimes e_1 + e_2\otimes e_2$ where $e_i$ are the canonical basis elements of $K$ is such an element but how can it be proved using elementary methods?

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marked as duplicate by Jendrik Stelzner, Namaste abstract-algebra Jun 22 '18 at 15:13

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Your example will do. Note in particular that every generator can be expanded as $$ (a_1 e_1 + a_2 e_2) \otimes (b_1 e_1 +b_2 e_2) = \\ a_1b_1\,e_1 \otimes e_1 + a_1 b_2 \,e_1\otimes e_2+ a_2b_1\,e_2 \otimes e_1 + a_2 b_2\,e_2 \otimes e_2 $$ From there, it suffices to prove that the system of equations $$ a_1b_1 = 1\\ a_1b_2 = 0\\ a_2 b_1 = 0\\ a_2b_2 = 1 $$ has no solutions (in any field). In particular, it suffices to observe because of the second equation, $a_1 = 0$ or $b_2 = 0$. Consequently, either the first or last equation must fail to be true.

An interesting observation is that the tensor $$ a_{11}\,e_1 \otimes e_1 + a_{12} \,e_1\otimes e_2+ a_{21}\,e_2 \otimes e_1 + a_{22}\,e_2 \otimes e_2 $$ can be written as a generator if and only if the matrix $$ A = \pmatrix{a_{11} & a_{12}\\a_{21} & a_{22}} $$ has rank $1$. In fact, this condition holds true for elements of $K^m \otimes K^n$ for arbitrary $m,n$.

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  • $\begingroup$ Thanks, this solution was also instrucive to me since I’m new to abstract algebra $\endgroup$ – user555729 Jun 22 '18 at 11:45