Tensor product and the pure elements From my understanding, one important thing about tensor product is that not all elements are pure elements, that is in $M\otimes N$ not all elements are in the form of $m\otimes n$ ,they are linear combinations of pure elements. 
My professor mentioned that while doing a problem thinking the tensor product only contains pure elements is a very common mistake. But I don't quite see the importance of this, if we can check some property $P$ for all pure elements, then shouldn't it hold for $M\otimes N$?
For example, if $m\otimes n = 0$ for each $m\in M, n\in N$, then $M\otimes N = 0$.
Can you give me an example where checking all pure elements is not good enough for $M\otimes N$. 
 A: Some properties are preserved by sums, and those can be checked on pure elements.  However, occasionally, you want to look at nonlinear things.
For example, consider the alternating algebra of a module $\bigwedge^{\bullet}M$, which is the quotient of $\bigoplus_n \bigotimes_{1}^n M$ (where the multipication is tensor product) by the ideal generated by $m\otimes m$, $m\in M$.  We call the image of pure tensor products pure elements of the quotient.  The relation forces (outside of characteristic 2) that $m\wedge n=-n\wedge m$, and so it is easy to check that the square of any pure element is zero.  One might incorrectly assume that the square of ANY element is zero.  However, this is not the case.  If you take $x=m_1\wedge m_2 +m_3\wedge m_4$, then $x\wedge x = 2m_1 \wedge m_2 \wedge m_3 \wedge m_4$, which is generally nonzero.  
A: In a kind of trivial way, as hinted-at by @Hurkyl in the comments, to make things more formal, the pure elements do not form a subspace, i.e., it is not an identity that given $a,b,c,d$ and $a \otimes b, c \otimes d$ , there are $e,f$ with $$a\otimes b + c \otimes d = e \otimes f $$.
So the pure elements by themselves do not form a subspace. But, at least in the case of free objects ( like when M,N are vector spaces) , you may have a subspace generated by basis elements $v_1,..,v_n \in V ; w_1,..,w_m \in W$.
