Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
What is Q? Also is Z an arbitrary ring or the ring of integers? –  Nuno Nov 16 '10 at 23:26
5  
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

1 Answer 1

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.

share|improve this answer
    
You can replace "integer" by $r \in Z$. –  Yuval Filmus Nov 17 '10 at 1:16
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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