Let $R$ be a commutative ring; let $M$, $N$ be $R$-modules and let $f\colon M\to N$ be an injective homomorphism of $R$-modules. Is $f\otimes {\rm id}\colon M\otimes_RN\to N\otimes_RN$ injective?
$\begingroup$
$\endgroup$
5
-
$\begingroup$ I'm sorry, the trivial "counterexample" I had in mind was not one. I removed my downvote. $\endgroup$– YCorCommented Nov 21, 2018 at 17:20
-
6$\begingroup$ Still not far from my initial expectation: choose $R$ a PID with a nonzero irreducible element $b$, and $M=R$, and $N=R\oplus (R/bR)$. Define $f:M\to N$ by $r\mapsto br\oplus 0$, clearly injective. Then the nonzero element $1\otimes (0\oplus 1)$ belongs to the kernel of $f\otimes\mathrm{id}$, as it's mapped to $(b\oplus 0)\otimes (0\oplus 1)=b(1\oplus 0)\otimes (0\oplus 1)=(1\oplus 0)\otimes b(0\oplus 1)=(1\oplus 0)\otimes (0\oplus b)=0$. $\endgroup$– YCorCommented Nov 21, 2018 at 17:47
-
$\begingroup$ After counterexamples given by others, the question is: Which $R$-modules $N$ are $N$-flat (in the terminology of Bourbaki's AC.I)? Clearly, free modules have this property. Any other modules? $\endgroup$– Fred RohrerCommented Nov 21, 2018 at 18:58
-
$\begingroup$ Sorry, read: flat (not free). $\endgroup$– Fred RohrerCommented Nov 21, 2018 at 21:10
-
$\begingroup$ Injective modules and simple modules are further examples of self-flat modules (i.e., modules that are flat w.r.t. themselves). The Uriya's example shows that self-flatness is not preserved by direct sums. $\endgroup$– Fred RohrerCommented Nov 22, 2018 at 12:38
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Take $N=\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ and $M=2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ (with $R=\mathbb{Z}$).
It is easy to check that $(2,0+2\mathbb{Z})\otimes (0,1+2\mathbb{Z})$ is a nonzero element in the kernel of $M\otimes N\to N\otimes N$.
-
2$\begingroup$ This is essentially contained in my example. $\endgroup$– YCorCommented Nov 21, 2018 at 21:34
-
$\begingroup$ @YCor Correct. Sorry. I did not read all the comments before I posted the answer. $\endgroup$ Commented Nov 22, 2018 at 10:28