## Modules of fractions

Let $R$ be an integral domain with field of fractions $Q$. Let $N$ be a left unital $R$-module, and let $U = R^\times$ denote the set of all nonzero elements of $R$. Define a relation $\sigma$ on $U \times N$ as follows: $(u_1, n_1)\ \sigma\ (u_2,n_2)$ if and only if $u_1 \cdot n_2 = u_2 \cdot n_1$. This $\sigma$ is clearly an equivalence relation; denote by $U^{-1}N$ the set $(U \times N)/\sigma$ of $\sigma$-equivalence classes.

1. Prove that the operation $+$ on $U^{-1}N$ given by $[(u,n)] + [(v,m)] = [(uv, v \cdot n + u \cdot m)]$ is well defined, and that $(U^{-1}N, +)$ is an abelian group. Prove further that the operation $r \cdot [(u,n)] = [(u, r \cdot n)]$ makes $U^{-1}N$ into a left $R$-module.
2. Prove that $U^{-1}N \cong_R Q \otimes_R N$.
3. Prove that $(1/d) \otimes n = 0$ in $Q \otimes_R N$ if and only if $r \cdot n = 0$ for some $r \in R$.
4. Let $A$ be an abelian group. Prove that $\mathbb{Q} \otimes_\mathbb{Z} A = 0$ if and only if $A$ is torsion (that is, every element of $A$ has finite order).

1. Suppose $[(a_1,n_1)]\ \sigma\ [(a_2,n_2)]$ and $[(b_1,m_1)]\ \sigma\ [(b_2,m_2)]$. Then $a_1 \cdot n_2 = a_2 \cdot n_1$ and $b_1 \cdot m_2 = b_2 \cdot m_1$, so that $b_1b_2a_1 \cdot n_2 = b_1b_2a_2 \cdot n_1$ and $a_1a_2b_1 \cdot m_2 = a_1a_2b_2 \cdot m_1$. Adding these equations together, we see that $a_1b_1(b_2 \cdot n_2 + a_2 \cdot m_2) = a_2b_2(b_1 \cdot n_1 + a_1 \cdot m_1)$. Thus $[(a_1b_1, b_1 \cdot n_1 + a_1 \cdot m_1)] = [(a_2b_2, b_2 \cdot n_2 + a_2 \cdot m_2)]$. In particular, the operator $+$ on $U^{-1}N$ is well defined.

Since $[(a,n)] + ([(b,m)] + [(c,t)]) = [(a,n)] + [(bc,b \cdot t + c \cdot m)]$ $= [(abc, ab \cdot t + ac \cdot m + bc \cdot n)]$ $= [(ab, a \cdot m + b \cdot n)] + [(c,t)]$ $= ([(a,n)] + [(b,m)]) + [(c,t)]$, $+$ is associative.

Since $[(a,n)] + [(1,0)] = [(a1, 1 \cdot n + a \cdot 0)]$ $= [(a,n)]$, $[(1,0)]$ is a left identity (similarly, a right identity). Moreover, note that $[(a,0)] = [(1,0)]$ for all $a \in U$.

Since $[(a,n)] + [(a,-n)] = [(aa, a \cdot n + a \cdot (-n))]$ $= [(a^2,0)]$ $= [(1,0)]$, every element of $U^{-1}N$ has an additive inverse. So $U^{-1}N$ is a group with respect to $+$.

Since $[(a,n)] + [(b,m)]$ $= [(ab, a \cdot m + b \cdot n)]$ $= [(ba, b \cdot n + a \cdot m)]$ $= [(b,m)] + [(a,n)]$, $U^{-1}N$ is an abelian group.

Now suppose $(a,n)\ \sigma\ (b,m)$. Then $a \cdot m = b \cdot n$, so that $a \cdot (r \cdot n) = b \cdot (r \cdot m)$ (since $R$ is commutative). Thus $(a, r \cdot n)\ \sigma\ (b, r \cdot m)$, and so the action $r \cdot [(a,n)] = [(a, r \cdot n)]$ is well defined.

Now $r \cdot ([(a,n)] + [(b,m)])$ $= r \cdot [(ab, a \cdot m + b \cdot n)]$ $= [(ab, a \cdot (r \cdot m) + b \cdot (r \cdot n))]$ $= [(a,r \cdot n)] + [(b, r \cdot m)]$, $rs \cdot [(a,n)]$ $= [(a, rs \cdot n)]$ $= [(a,r \cdot (s \cdot n))]$ $= r \cdot (s \cdot [(a,n)])$, and $(r+s) \cdot [(a,n)] = [(a, (r+s) \cdot n)]$ $= [(a^2, a \cdot (r \cdot n) + a \cdot (s \cdot n))]$ $= [(a,r \cdot n)] + [(a, s \cdot n)]$ $= r \cdot [(a,n)] + s \cdot [(a,n)]$. So $U^{-1}N$ is a left $R$-module with respect to this action.

2. Define $\varphi : Q \times N \rightarrow U^{-1}N$ by $\varphi(\frac{a}{b}, n) = [(b,a \cdot n)]$. Note that if $a/b = c/d$, then $ad = bc$, so that $ad \cdot n = bc \cdot n$, and thus $(b, a \cdot n) = (d, c \cdot n)$. That is, $\varphi$ is well-defined. Moreover, we have $\varphi(\frac{a}{b} + \frac{c}{d}, n) = \varphi(\frac{ad+bc}{bd}, n)$ $= [(bd, ad \cdot n + bc \cdot n)]$ $= [(b, a \cdot n)] + [(d, c \cdot n)]$ $= \varphi(\frac{a}{b}, n) + \varphi(\frac{c}{d}, n)$, $\varphi(\frac{a}{b}, n+m) = [(b, a \cdot n + a \cdot m)]$ $= [(b^2, ba \cdot m + ba \cdot n)]$ $= [(b, a \cdot n)] + [(b, a \cdot m)]$ $= \varphi(\frac{a}{b},n) + \varphi(\frac{c}{d}, m)$, and $\varphi(\frac{a}{b} \cdot r, n)$ $= \varphi(\frac{ar}{b}, n)$ $= [(b, ar \cdot n)]$ $= \varphi(\frac{a}{b}, r \cdot n)$. Thus $\varphi$ is $R$-balanced, and so induces an $R$-module homomorphism $\Phi : Q \otimes_R N \rightarrow U^{-1}N$ such that $\Phi(\frac{a}{b}, n) = [(b, a \cdot n)]$.

Now define $\Psi : U^{-1}N \rightarrow Q \otimes_R N$ by $\Psi([(a,n)]) = \frac{1}{a} \otimes n$. If $(a,n)\ \sigma\ (b,m)$, then $a \cdot m = b \cdot n$. So $\frac{1}{ab} \otimes a \cdot m = \frac{1}{ab} \otimes b \cdot m$, so that $\frac{1}{a} \otimes n = \frac{1}{b} \otimes m$. That is, $Psi$ is well-defined. Moreover, note that $\Psi([(a,n)] + [(b,m)]) = \Psi([(ab, a \cdot m + b \cdot n)])$ $= \frac{1}{ab} \otimes (a \cdot m + b \cdot n)$ $= \frac{1}{a} \otimes n + \frac{1}{b} \otimes m$ $= \Psi([(a,n)]) + \Psi([(b,m)])$. Thus $\Psi$ is a group homomorphism.

Now $(\Psi \circ \Phi)(\frac{a}{b} \otimes n) = \Psi([(b, a \cdot n)])$ $= \frac{1}{b} \otimes a \otimes n$ $= \frac{a}{b} \otimes n$ and $(\Phi \circ \Psi)([(a,n)]) = \Phi(\frac{1}{a} \otimes n) = [(a,n)]$. Thus $\Phi$ and $\Psi$ are mutual inverses, and so $\Phi$ is a group isomorphism, hence an $R$-module isomorphism. So $Q \otimes_R N \cong_R U^{-1}N$.

3. Suppose $\frac{1}{d} \otimes n = 0$ in $Q \otimes_R N$. Then $\Phi(\frac{1}{d} \otimes n) = [(d,n)] = [(d,0)]$. In particular, we have $d \cdot n = d \cdot 0 = 0$. Conversely, suppose $r \cdot n = 0$ with $r \neq 0$ and let $d \in R$ be nonzero. Then $\frac{1}{d} \otimes n = \frac{r}{rd} \otimes n$ $= \frac{1}{rd} \otimes r \cdot n$ $= \frac{1}{rd} \otimes 0 = 0$.
4. Suppose $\mathbb{Q} \otimes_\mathbb{Z} A = 0$. In particular, every simple tensor $\frac{1}{d} \otimes a$ is zero, so that by part (3) above, there exists a nonzero $r \in \mathbb{Z}$ such that $r \cdot a = 0$. Thus $A$ is torsion. conversely, if $A$ is torsion, then for every element $a \in A$, there exists $r \in \mathbb{Z}$ nonzero so that $r \cdot a = 0$. But then the simple tensor $\frac{1}{d} \otimes a$ is zero, so that $\mathbb{Q} \otimes_\mathbb{Z} A = 0$.