Why are the tensor products over Q and Z_n equal?

[quasar987]

Given R a ring and M,N two R-modules, we may form their tensor product over Z or R. They can be defined as the group presentations
< A x B | (a + a',b)=(a,b) + (a',b), (a,b+b')=(a,b) + (a,b') >,
< A x B | (a + a',b)=(a,b) + (a',b), (a,b+b')=(a,b) + (a,b'), (ra,b)=(a,rb) >
respectively and the image of (a,b) in the quotient is written $a\otimes b$.

Tensoring over R will generally yield a smaller group since there is the additional relation $(ra)\otimes b = a\otimes (rb)$. But sometimes tensoring over R or Z give the same group.

I have read in a textbook that this happens in the case where R = Z_n or Q. For Z_n, it is quite clear, since in this case, the relation $(ra)\otimes b = a\otimes (rb)$ is implied by the relations $(a + a') \otimes b = a\otimes b + a'\otimes b$ and $a \otimes (b+b') = a\otimes b + a\otimes b'$.

But for R = Q, I do not quite see why the two tensor products are equal. Thx!

 

[mathwonk]

I am a little puzzled. It seems to me that any Z/nZ module is entirely torsion. I.s. as a Z module, every element is annihilated by n. But when tensoring by Q, all torsion elements are annihilated, since (1)(atensb) = (n/n)(atensb) = (a/n)tens(nb) = (a/n)tens(0) = 0.

Thus tensoring any Z/nZ modules over Q should give zero. But if say n is prime, then Z/nZ is a field, so tensoring non zero vector spaces over Z/nZ should never annihilate any non zero vector space.

 

[quasar987]

Um, you talk abour tensoring Z/Zn-modules over Q. This doesn't really make sense I think because Q is not a subring of Z/Zn so the action of Q is ill-defined.

 

[mathwonk]

i guess i misread your question. you seem to be asking about tensoring over Z and over Q giving the same result.

(A module can be a module over two rings even though neither is a subring of the other, although the only module that is a module over both Q and Z/n is zero, but never mind. I presume you want a non zero example.)

let me think about it. ... OK, just tensor Z/n with Z/n. That apparently gives the same result over both Z and Z/n.and Q tensor Q seems to be Q, whether tensored over Z or Q. you can probably check this in a few minutes using the basic properties of tensors, i.e. (ab)tens(x) = atens(bx) if b is in the ring you are tensoring over.

 

[Landau]

I don't quite understand the question. Are you asking for an example of Q-modules whose tensor product over Q is the same as their tensor product over Z? In that case, mathwonk's example is valid:

$$\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Q}=\mathbb{Q}=\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}$$

The first equality is trivial, for the second the obvious map $\frac{a}{b}\otimes_\mathbb{Z} \frac{p}{q}\mapsto \frac{ap}{bq}$ is easily checked to be an isomorphism.

 

[quasar987]

I don't quite understand the question. Are you asking for an example of Q-modules whose tensor product over Q is the same as their tensor product over Z?

Almost! I am asking for a proof that the tensor product of any Q-module over Q is the same as their tensor product over Z.

The statement is taken from Hatcher's. See pages 34-35 of http://www.math.cornell.edu/~hatcher/AT/ATch3.pdf (pages 218-219 of the book). The question is at the bottom of page 34 and top of page 35.

 

[Landau]

Ah, right, the proof of that fact (which I hadn't realized myself) is the same as showing that the map in my last post is an isomorphism. Basically, you need to show that if R resp. S is a right resp. left Q-module, then for r in R, s in S and q in Q:

$$rq\otimes_{\mathbb{Z}}s=r\otimes_{\mathbb{Z}}qs$$

This is done with the following 'trick'. Write q=a/b, and $\otimes=\otimes_{\mathbb{Z}}$; then:

$$r\frac{a}{b}\otimes s=r\frac{1}{b}\otimes as=r\frac{1}{b}\otimes \frac{bas}{b}=r\frac{b}{b}\otimes \frac{as}{b}=r\otimes \frac{a}{b}s$$

 

[quasar987]

Geez, I should have seen that. :(

Thanks friend!

 

[Landau]

Yeah, as Hatcher says: "It is an easy algebra exercise to see that..." :P

You're welcome!

 

[mathwonk]

just playing. but this seems to be true because every Z linear map of Q vector spaces is also Q linear. hence I believe it follows that Z bilinear and Q bilinear maps are also the same. Then since the tensor product over Q represents the functor of Q bilinear maps, while that over Z represents the Z bilinear maps, and those functors are the same, then their representing modules are the same. blah blah blah...

 

[Landau]

just playing. but this seems to be true because every Z linear map of Q vector spaces is also Q linear.

Yeah, so I guess the underlying reason is that Q is (by definition) the quotient field of Z. For every ring homomorphism f:Z\to Q there is a unique ring homomorphism Q\to Z extending it.