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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?

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What is Q? Also is Z an arbitrary ring or the ring of integers? – Nuno Nov 16 '10 at 23:26
Hint: $\mathbb{Q} \otimes M$ is the localization of $M$ at the multiplicative subset $\mathbb{Z}-0 \subset \mathbb{Z}$. Now you can use the definition of localization. – Akhil Mathew Nov 17 '10 at 0:40
Q is the quotient field of Z. – Yuval Filmus Nov 17 '10 at 1:16
Akhil, you should have made that an answer, as the only answer that is posted here is trivial. – Jon Beardsley Nov 30 '10 at 20:58
up vote 2 down vote accepted

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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You can replace "integer" by $r \in Z$. – Yuval Filmus Nov 17 '10 at 1:16
It only proves the trivial inclusion. The point here is to prove that if $1 \otimes m = 0$, then $m$ is a torsion element. – Nuno Nov 17 '10 at 2:37
I think Akhil's comment addresses this issue. – Timothy Wagner Nov 17 '10 at 5:56
Yes Akhil's hint does provide a way to prove the other part. Sorry if my comment sound rude, I just thought your post was a hint to prove everything, which is not. – Nuno Nov 17 '10 at 20:05
Yes. Thankyou. Someone ended up showing me essentially this idea in Dummit and Foote, where it is a homework 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. – Jon Beardsley Nov 28 '10 at 21:42

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