Finitely generated modules over Z and scalar extension from R[x] to C[x]

The two modules are not isomorphic because \(N\) contains a nonzero element of order \(2\), whereas \(M\) is torsion-free. On the other hand, \[ \mathbb Q\otimes_{\mathbb Z}\mathbb Z\cong\mathbb Q, \] while \[ \mathbb Q\otimes_{\mathbb Z}\mathbb Z/2\mathbb Z=0. \] Indeed, if \(\bar1\) denotes the generator of \(\mathbb Z/2\mathbb Z\), then \[ 1\otimes\bar1 =\frac12\otimes 2\bar1 =0, \] and hence every pure tensor vanishes. Therefore \[ \mathbb Q\otimes_{\mathbb Z}N \cong \mathbb Q\oplus0 \cong\mathbb Q \cong \mathbb Q\otimes_{\mathbb Z}M. \]

The ring \(\mathbb R[x]\) is a PID. Thus the structure theorem gives an elementary-divisor decomposition \[ M\cong \mathbb R[x]^r \oplus \bigoplus_{a\in\mathbb R}\bigoplus_{e\ge1} \left(\mathbb R[x]/((x-a)^e)\right)^{m_{a,e}} \oplus \bigoplus_{z\in S}\bigoplus_{e\ge1} \left(\mathbb R[x]/(q_z(x)^e)\right)^{n_{z,e}}, \] where only finitely many multiplicities are nonzero, \(S\) contains one representative from each conjugate pair of nonreal complex numbers, and \[ q_z(x):=(x-z)(x-\bar z) =x^2-2\operatorname{Re}(z)x+|z|^2 \in\mathbb R[x]. \] These are exactly the monic irreducible polynomials over \(\mathbb R\): the linear factors and the quadratics \(q_z\).

For every \(f\in\mathbb R[x]\), \[ \mathbb C[x]\otimes_{\mathbb R[x]}\mathbb R[x]/(f) \cong \mathbb C[x]/(f). \] Also \[ \mathbb C[x]\otimes_{\mathbb R[x]}\mathbb R[x]^r \cong\mathbb C[x]^r. \] Hence the real linear blocks remain \[ \mathbb C[x]/((x-a)^e). \] For a nonreal \(z\), the ideals \[ ((x-z)^e) \quad\text{and}\quad ((x-\bar z)^e) \] are comaximal in \(\mathbb C[x]\), and \[ q_z(x)^e=(x-z)^e(x-\bar z)^e. \] The Chinese remainder theorem therefore gives \[ \mathbb C[x]/(q_z^e) \cong \mathbb C[x]/((x-z)^e) \oplus \mathbb C[x]/((x-\bar z)^e). \] Consequently \[ \mathbb C[x]\otimes_{\mathbb R[x]}M \] has free rank \(r\), the same elementary divisors \((x-a)^e\) with multiplicities \(m_{a,e}\) for real \(a\), and for each nonreal conjugate pair \(\{z,\bar z\}\) it has both elementary divisors \((x-z)^e\) and \((x-\bar z)^e\), each with multiplicity \(n_{z,e}\).

Suppose \[ \mathbb C[x]\otimes_{\mathbb R[x]}M \cong \mathbb C[x]\otimes_{\mathbb R[x]}N. \] Since \(\mathbb C[x]\) is a PID, the structure theorem over \(\mathbb C[x]\) implies that the two complexified modules have the same free rank and exactly the same elementary divisors with the same multiplicities.

By 2, the real elementary-divisor data can be recovered uniquely from this complex data. For each real \(a\), every block \((x-a)^e\) comes from the identical real block, so its multiplicity is read off directly. For each nonreal \(z\), the complexification of a real \(q_z^e\)-block contributes exactly one \((x-z)^e\)-block and one \((x-\bar z)^e\)-block. Thus the common multiplicity of the conjugate pair recovers the multiplicity of the real irreducible block \[ q_z(x)^e=((x-z)(x-\bar z))^e. \] The free rank is also unchanged by scalar extension.

Therefore \(M\) and \(N\) have the same free rank and the same elementary divisors over \(\mathbb R[x]\), with the same multiplicities. The structure theorem for finitely generated modules over the PID \(\mathbb R[x]\) then yields \[ M\cong N. \]