Over a commutative ring, the tensor product of a module by a free module of finite rank is a direct power

Let R be a commutative ring and let N be a free R-module of finite rank n; say B = \{e_i\}_{i=1}^n is a basis for N.

  1. Let M be any nonzero R-module. (Since R is commutative, M is naturally an (R,R)-bimodule.) Show that M \otimes_R N \cong_R M^n. In particular, every element of M \otimes_R N can be written uniquely in the form \sum_{i=1}^n m_i \otimes e_i.
  2. Show that if \sum m_i \otimes n_i = 0, where the n_i are merely assumed to be R-linearly independent in N, it need not be the case that the m_i are all zero. (That is, it is crucial that the n_i generate N as an R-module.)

Define a mapping \varphi_B : M \times N \rightarrow M^n by \varphi_B(m, \sum r_ie_i) = (r_i \cdot m). (This is well defined since B is a basis for N.) Moreover, \varphi_B(m_1 + m_2, \sum r_ie_i) = (r_i(m_1 + m_2)) = (r_i m_1) + (r_i m_2) = \varphi_b(m_1, \sum r_ie_i) + \varphi_b(m_2, \sum r_ie_i), \varphi(m, (\sum r_i e_i) + (\sum s_i e_i)) = \varphi_B(m, \sum (r_i+s_i)e_i) = ((r_i+s_i)m) = (r_im) + (s_im) = \varphi_B(m, \sum r_ie_i) + \varphi_B(m, \sum s_i e_i), and \varphi_B(m \cdot a, \sum r_i e_i) = (r_i (ma)) = (ar_i m) = \varphi_B(m, \sum ar_ie_i) = \varphi_B(m, a \sum r_ie_i), so that \varphi_B is bilinear. By the universal property of tensor products, \varphi_B induces a unique group homomorphism (indeed, R-module homomorphism) \Phi_B : M \otimes_R N \rightarrow M^n such that \Phi_B(m \otimes \sum r_i e_i) = (r_im).

Now define \Psi_B : M^n \rightarrow M \otimes_R N by (m_i) \mapsto \sum m_i \otimes e_i. Clearly \Psi_B is an R-module homomorphism.

Moreover, note that (\Psi_B \circ \Phi_B)(m \otimes \sum r_i e_i) = \Psi_B(r_im)) = \sum r_im \otimes e_i = \sum m \otimes r_ie_i = m \otimes \sum r_ie_i. Similarly, (\Phi_B \circ \Psi_b)(m_i) = \Phi_B(\sum m_i \otimes e_i) = \sum \Phi_B(m_i \otimes e_i) = (m_i). Thus \Phi_B and \Psi_B are mutual inverses, and we have M \otimes_R N \cong_R M^n.

In particular, \sum m_i \otimes e_i = 0 if and only if m_i = 0 for all i.

Now for the counterexample, let R = \mathbb{Z}, let N = \mathbb{Z}^1 be the free \mathbb{Z}-module of rank 1, and let M = \mathbb{Z}/(2) be a \mathbb{Z}-module in the usual way. Consider the simple tensor \overline{1} \otimes 2 in M \otimes_R N; note that 1 \neq 0, but that \overline{1} \otimes 2 = \overline{1} \otimes 2 \cdot 1 = \overline{1} \cdot 2 \otimes 1 = \overline{2} \otimes 1 = 0 \otimes 1 = 0.

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