0
$\begingroup$

can tensor product of two free non zero module over commutative ring with unity be zero?

And can tensor product of two non zero vector spaces be zero space?

$\endgroup$
1
$\begingroup$

Let $U$ and $V$ be free modules over a nontrivial commutative ring $R$ and let $\alpha:U\to R$ and $\beta:V\to R$ be two $R$-linear maps which have $1$ in their image; such things are easily seen to exist using freeness. Then using the properties of tensor products you can show that there is a morphism of abeelian groups $f:U\otimes_RV\to R$ such that for each $u\in U$ and each $v\in V$ we have $f(u\otimes v)=\alpha(u)\beta(v)$.

Now the hypothesis on $\alpha$ and $\beta$ tells us there exist $u_0\in U$ and $v_0\in V$ with $\alpha(u_0)=1$ and $\beta(v_0)=1$, and then $f(u_0\otimes v_0)=1$. This implies, of course, that $u_0\otimes v_0$ is a nonzero element of $U\otimes_RV$.

$\endgroup$
  • $\begingroup$ If $U$ is a right $R$-module and $V$ is a left $R$-module, and then we allow $R$ to be a possibly non-commutative,, exactly the same argument works. $\endgroup$ – Mariano Suárez-Álvarez May 23 '16 at 4:37
  • $\begingroup$ So will tensor product equal to LC of $e_i \otimes e_j$ where $e_i$ basis of $U$ and $e_j$ basis of $V$ $\endgroup$ – Sushil May 23 '16 at 4:41
  • $\begingroup$ While that is true, that most certainly does not follow from what I wrote... $\endgroup$ – Mariano Suárez-Álvarez May 23 '16 at 4:42
  • $\begingroup$ Okay but can you please tell why that will be true? $\endgroup$ – Sushil May 23 '16 at 4:47
  • $\begingroup$ I suggest you ask another question. Comments to an answer is a very bad place to ask new questions! $\endgroup$ – Mariano Suárez-Álvarez May 23 '16 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.