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.
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