If we have a module $M$ over a ring $Z$, and we consider the the canonical module homomorphism $M→Q⊗M$ over $Z$, is it true that the kernel of this map is the torsion submodule of $M$? Why?
Torsion submodule equal to kernel of canonical map
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$\begingroup$
modules
tensor-products
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2Akhil, you should have made that an answer, as the only answer that is posted here is trivial. – 2010-11-30
1 Answers
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Hint: If $m\in M$ is a torsion element, it is killed by some integer, say $r$. Then, $1\otimes m=\frac{r}{r}\otimes m=\frac{1}{r}\otimes rm=\frac{1}{r}\otimes 0=0$ where the second equality follows from the bilinearity of the tensor product.
I was working under the assumption that $Z$ is the ring of integers and $Q$ the ring of rationals. As Yuval points out, everything works the same way if you think of $Z$ as an arbitrary domain and $Q$ as its fraction field.
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0Yes. Tha$n$kyou. Someo$n$e e$n$ded up showin$g$ me essentially this idea in Dummit and Foote, where it is a homewor$k$ problem. I was originally trying to follow an idea from Atiyah and Macdonald, which constructs the quotient field as a direct limit, and it's rather frustrating. This is much nicer. – 2010-11-28