# an element of a tensor product of modules not equal to any generator. [duplicate]

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?

## marked as duplicate by Jendrik Stelzner, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 22 '18 at 15:13

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 $m,n$.