Give an example of two finitely generated \(\mathbb Z\)-modules \(M\) and \(N\) that are not isomorphic as \(\mathbb Z\)-modules but satisfy \[ \mathbb Q\otimes_{\mathbb Z}M\cong \mathbb Q\otimes_{\mathbb Z}N \] as \(\mathbb Q\)-modules.
Let \(M\) be a finitely generated \(\mathbb R[x]\)-module, described using the classification of finitely generated modules over a PID. Give a corresponding description of \[ \mathbb C[x]\otimes_{\mathbb R[x]}M \] as a \(\mathbb C[x]\)-module.
Solution
1. For part (a), take\[ M=\mathbb Z, \qquad N=\mathbb Z\oplus\mathbb Z/2\mathbb Z. \]Then \(M\not\cong N\) as \(\mathbb Z\)-modules, but\[ \mathbb Q\otimes_{\mathbb Z}M \cong \mathbb Q \cong \mathbb Q\otimes_{\mathbb Z}N. \]
The module \(N\) has nonzero torsion while \(M\) is torsionfree, so they are not isomorphic. Also \[ \mathbb Q\otimes_{\mathbb Z}(\mathbb Z/2\mathbb Z)=0, \] because \(2\) is invertible in \(\mathbb Q\). Hence \[ \mathbb Q\otimes_{\mathbb Z}N \cong \mathbb Q\oplus0 \cong\mathbb Q. \]
2. Write the elementary-divisor decomposition of the finitely generated \(\mathbb R[x]\)-module \(M\) as\[ M\cong \mathbb R[x]^r \oplus \bigoplus_{i=1}^s \mathbb R[x]/(p_i(x)^{e_i}), \]where each \(p_i\) is monic irreducible in \(\mathbb R[x]\) and \(e_i\ge1\).
The ring \(\mathbb R[x]\) is a PID, so this is the elementary-divisor form of the structure theorem for finitely generated modules over a PID.
3. Extension of scalars gives\[ \mathbb C[x]\otimes_{\mathbb R[x]}M \cong \mathbb C[x]^r \oplus \bigoplus_{i=1}^s \mathbb C[x]/(p_i(x)^{e_i}). \]
Tensor product commutes with finite direct sums, and \[ \mathbb C[x]\otimes_{\mathbb R[x]}\mathbb R[x]\cong\mathbb C[x]. \] For every ideal \((a)\subseteq\mathbb R[x]\), right exactness applied to \[ \mathbb R[x]\xrightarrow{\cdot a}\mathbb R[x]\to\mathbb R[x]/(a)\to0 \] gives \[ \mathbb C[x]\otimes_{\mathbb R[x]}\mathbb R[x]/(a) \cong \mathbb C[x]/(a)\mathbb C[x]. \] Apply this with \(a=p_i^{e_i}\).
4. If \(p_i(x)=x-a\) with \(a\in\mathbb R\), then the corresponding summand remains\[ \mathbb C[x]/((x-a)^{e_i}). \]
A real linear irreducible remains linear over \(\mathbb C\).
5. If \(p_i\) is an irreducible quadratic over \(\mathbb R\), write\[ p_i(x)=(x-z_i)(x-\overline z_i), \qquad z_i\in\mathbb C\setminus\mathbb R. \]Then\[ \mathbb C[x]/(p_i^{e_i}) \cong \mathbb C[x]/((x-z_i)^{e_i}) \oplus \mathbb C[x]/((x-\overline z_i)^{e_i}). \]
The two linear factors are distinct, so their powers are coprime. The Chinese remainder theorem gives \[ \mathbb C[x]/\big((x-z_i)^{e_i}(x-\overline z_i)^{e_i}\big) \cong \mathbb C[x]/((x-z_i)^{e_i}) \oplus \mathbb C[x]/((x-\overline z_i)^{e_i}). \]
6. Thus scalar extension preserves the free rank, preserves each real-linear primary summand, and splits every real-quadratic primary summand into the two conjugate complex primary summands described in
5.
5. Every monic irreducible polynomial over \(\mathbb R\) has degree \(1\) or \(2\).