0
$\begingroup$

Show $\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q}\cong \mathbb{Q}$ as groups.

I used the universal property of the canonical middle linear map to get a homomorphism $$f:\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q}\rightarrow \mathbb{Q} \text{ by } f(m/n\otimes p/q)=mp/nq.$$

I showed that $f$ is injective and surjective. For injectivity I showed by assuming $f(a\otimes b)=f(c\otimes d)$ and then showed that $a\otimes b=c\otimes d$. Surjectivity is quick to show. So this completes the proof.

I also constructed an inverse map to $f$, where $f^{-1}$ does the mapping $m/n\mapsto m/n\otimes 1/1$, so this means $f$ is an isomorphism of groups.

Basically my methods have been explicitly check the function is surjective injective, and explicitly construct the inverse function. I was wondering if there is a way to solve this using exact sequences? We know if we have an exact sequence of left $R$ modules $$A\overset{f}{\rightarrow}B\overset{g}{\rightarrow}C\rightarrow 0$$ then for any right $R$ module $D$, $$D\otimes_R A\overset{1_D\otimes f}{\rightarrow}D\otimes_R B\overset{1_D\otimes g}{\rightarrow}D\otimes_R C\rightarrow 0$$ is an exact sequence of abelian groups where $1_D\otimes f$ maps $d\otimes a\mapsto d\otimes f(a)$.

I am trying to find maybe efficient ways to solve this type of question. Specifically, showing a function is injective if the domain is the tensor product, and if I can somehow use exact sequences in a way to do this?

$\endgroup$
  • $\begingroup$ "Canonical middle linear map"? What in the world is that? $\endgroup$ – DonAntonio Mar 13 '14 at 5:45
  • $\begingroup$ @DonAntonio It seems to be equivalent to the universal property of the tensor product, whatever it is. $\endgroup$ – PVAL-inactive Mar 13 '14 at 5:47
0
$\begingroup$

Take the exact sequence with inclusion and projection: $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Q}\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0.$$ $S^{-1}R$ is a flat $R$-module so $$0\rightarrow \mathbb{Q}\otimes\mathbb{Z}\rightarrow \mathbb{Q}\otimes \mathbb{Q}\rightarrow \mathbb{Q}\otimes\mathbb{Q}/\mathbb{Z} \rightarrow 0\qquad (1)$$ is an exact sequence. $\mathbb{Q}\otimes \mathbb{Z}\cong \mathbb{Q}$. Then $$\frac{p}{q}\otimes\frac{m}{n}+\mathbb{Z}=\frac{npm}{nq}\otimes \frac{1}{n}+\mathbb{Z}=\frac{pm}{nq}\otimes\frac{n}{n}+\mathbb{Z} =0$$ so $\mathbb{Q}\otimes \mathbb{Q}/\mathbb{Z}=0$ implying $(1)$ gives an isomorhism from $\mathbb{Q}\otimes\mathbb{Z}\cong \mathbb{Q}\otimes\mathbb{Q}$ so we use transitivity of $\cong$ to get $\mathbb{Q}\cong\mathbb{Q}\otimes\mathbb{Q}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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