Let $p$ be a prime number and define $\mathbb{Z}_{(p)}=\{\frac{m}{n}:m,n \in \mathbb{Z}, n \not\in p\mathbb{Z}\}$.

Let $M$ be a $\mathbb{Z}$-module and define an equivalence relation $\sim$ in $M \times (\mathbb{Z} \backslash p\mathbb{Z})$ by $(m,r) \sim (m',r') \iff sr'm=srm'$ for some $s \in \mathbb{Z} \backslash p\mathbb{Z}$.

Denote by $\frac{m}{r}$ the equivalence class that contains $(m,r)$.

Define $M_p=\{\frac{m}{r}: m \in M, r \in \mathbb{Z} \backslash p\mathbb{Z}\}$ the set of these equivalence classes.

We know that $M_p$ is a $\mathbb{Z}_{(p)}$-module with respect to the usual operations $+$ and $\cdot$ (like how we sum and multiply two fractions).

Define $\alpha_p:M \rightarrow M_p$ by $m \mapsto \frac{m}{1}$, which is a homomorphism of $\mathbb{Z}$-modules.

Let $N$ be a $\mathbb{Z}_{(p)}$-module and $\alpha:M \rightarrow N$ a homomorphism of $\mathbb{Z}$-modules.

Prove that there exists a unique homomorphism of $\mathbb{Z}_{(p)}$-modules $\beta:M_p \rightarrow N$ such that $\beta(\frac{m}{r})=\frac{1}{r}\alpha(m)$.

My attempt:

Existence: Let $\beta:M_p \rightarrow N$ be defined as $\beta(\frac{m}{r})=\frac{1}{r}\alpha(m)$.

Then $\beta$ is well defined and is a homomorphism of $\mathbb{Z}_{(p)}$-modules (easy to check).

So this proves existence.

For uniqueness I have no idea, can someone help me trying to figure out how to do it, please?


  • $\begingroup$ I think you want a unique $\beta$ such that $\beta\circ\alpha_p=\alpha$, that is, $\beta(m/1)=\alpha(m)$, otherwise the question is a non-sense like you are asking "show that there is a unique map $f:A\to A$ such that $f(a)=a$". (Btw, it seems this is the second time when posted this question.) $\endgroup$ – user26857 Oct 28 '15 at 22:25
  • $\begingroup$ Yes you are right, this is why I posted the question, because the uniqueness seemed trivial. (I deleted the previous question btw) $\endgroup$ – Leafar Oct 28 '15 at 22:36

The question here is in fact the following:

Show that there is a unique homomorphism of $\mathbb{Z}_{(p)}$-modules $\beta:M_p \rightarrow N$ such that $\beta\circ\alpha_p=\alpha$, that is, $\beta(\frac{m}{1})=\alpha(m)$.

This is a kind of universal property of the module of fractions. (Of course, one can define $\beta$ as you did and thus showing the existence.)

  • $\begingroup$ See also here, Proposition 3.8. $\endgroup$ – user26857 Oct 28 '15 at 22:48

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