# Showing function with domain tensor product is injective? Way to do so

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?

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"Canonical middle linear map"? What in the world is that? –  DonAntonio Mar 13 at 5:45
@DonAntonio It seems to be equivalent to the universal property of the tensor product, whatever it is. –  PVAL Mar 13 at 5:47

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 $\mathbb{Q}\cong\mathbb{Q}\otimes\mathbb{Q}$.