I am learning about tensor products of modules, but there is a question which makes me very confused about it!

If $E$ is a right $R$-module and $F$ is a left $R$-module, then suppose we have a balanced map (or bilinear map) $E\times F\to E\otimes F$. If some element $x\otimes y \in E\otimes F$ is $0$, then can we say $x$ or $y$ must be equal to $0$? I know if $x = 0$ or $y = 0$, then $x\otimes y$ is $0$. Are there other cases where $x\otimes y$ is $0$? Can someone give me a specific example?

Really thank you!

  • $\begingroup$ I don't understand the sentence beginning "If some elements..." I think you're asking whether $x\otimes y$ can be zero if $x$ and $y$ are not zero. If this is the case, I have given an answer below. If you intended to ask a different question, please edit your post to make it clear. $\endgroup$ – Michael Albanese Mar 7 '16 at 4:50
  • $\begingroup$ yes, that's what I mean! $\endgroup$ – python3 Mar 7 '16 at 4:50
  • $\begingroup$ Given that is what you intended to ask, I have edited your question in the hopes of making your question clear. Feel free to change it back if I have misunderstood what you meant. $\endgroup$ – Michael Albanese Mar 7 '16 at 4:56
  • $\begingroup$ You might also be interested in the question math.stackexchange.com/questions/1044952/… $\endgroup$ – dfsfljn Mar 7 '16 at 5:17

By the universal property of tensor product, an elementary tensor $x\otimes y$ equals zero if and only if for every $R$-bilinear map $B:E\times F\to M$, $B(x,y)=0$. While this may seem like a difficult thing to check, in practice it is usually not so bad.

As an example, we will show that $\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}=0$. Let $x\otimes y\in\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}$ be an elementary tensor. Then by the bilinearity of the canonical map $\mathbb{Z}/5\mathbb{Z}\times\mathbb{Q}\to\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}$, we have $$ x\otimes y=x\otimes 5y/5=5(x\otimes y/5)=5x\otimes y=0\otimes y=0. $$ This shows that all elementary tensors are zero, and thus since the tensor product is generated by elementary tensors, $\mathbb{Z}/5\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}=0$.

We can show that an elementary tensor $x\otimes y$ is nonzero by giving an $R$-bilinear map $B:E\times F\to M$ such that $B(x,y)\neq 0$. As an example, consider $E=F=\mathbb{Z}$ as $\mathbb{Z}$-modules and the $\mathbb{Z}$-bilinear map $B:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ given by multiplication: $B(x,y)=xy$. Then if $x,y\neq 0$, $B(x,y)\neq 0$, so that $x\otimes y\neq 0$ in $\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}$ when $x,y\neq 0$.

  • $\begingroup$ This is helpful when showing $m\otimes n=0$ but it is hard to use this to show $m\otimes n$ is not zero. $\endgroup$ – Arun Kumar Mar 7 '16 at 5:10
  • $\begingroup$ OK, I'll edit my answer to give an example of showing that $m\otimes n\neq 0$. $\endgroup$ – dfsfljn Mar 7 '16 at 5:11
  • $\begingroup$ @dfsfljn How does the universal property guarantee that if and only if statement with which you started? $\endgroup$ – Tanner Strunk Jun 23 '18 at 5:57
  • $\begingroup$ @dfsfljn, how did you do $ \ 5(x \otimes y/5)=5x \otimes y$ ? Is it like $a B(x,y)=B(ax,ay)$ ? $\endgroup$ – M. A. SARKAR Aug 17 at 16:09
  • $\begingroup$ @M.A.SARKAR that is a mistake, it should be $5x\otimes y/5$ $\endgroup$ – Heath Winning Oct 21 at 19:37

It is not possible to find a nice characterization of when simple tensors are zero.

To give an example where $a, b$ are nonzero but $a\otimes b=0$, consider $\Bbb Z/6\Bbb Z\otimes_\Bbb Z \Bbb Z$ where $2\otimes 3=2\cdot3\otimes 1=0\otimes 1=0$.

  • 2
    $\begingroup$ In $M \otimes_R N$, a simple tensor $m \otimes n$ is zero exactly when all $R$-bilinear maps $M \times N \rightarrow P$ from $M \times N$ to $R$-modules vanish on $(m,n)$. For example, a $\mathbf Z$-bilinear map $B \colon \mathbf Z/(6) \times \mathbf Z \rightarrow A$ to a $\mathbf Z$-module $A$ (that is, an abelian group $A$) has $B(2,3) = 3B(2,1) = B(6,1) = B(0,1) = 0$. This is related to your example that $2 \otimes 3 = 0$ in $\mathbf Z/(6) \otimes_{\mathbf Z} \mathbf Z$. $\endgroup$ – KCd Jul 16 at 6:57
  • $\begingroup$ @KCd Yes, that’s true! But I think when I wrote this, the bar I had in mind for what would count as “a nice way” was more along the lines of “just by looking at it”. $\endgroup$ – rschwieb Jul 16 at 10:27

I found this question while seeking to answer a related one. I think it's worth saying that $0\otimes n = (0\cdot 0)\otimes n = 0\cdot(0\otimes n) = 0\otimes(0\cdot 0) = 0\otimes 0 = 0\in M\otimes_A N$ where $M$ and $N$ are left $A$-modules (or right--I can never remember which is which).

(You may also note that $0\cdot n = (1 + (-1))\cdot n = n - n = 0$.)


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