Let \(R\) be a principal ideal domain and \(A\) and \(B\) be finitely generated \(R\)-modules. Show that if \(A \oplus A \cong B \oplus B\) then \(A \cong B\).
Solution
1. A finitely generated \(R\)-module \(M\) has a decomposition\[ M\cong R^{r_M}\oplus \bigoplus_{p}\bigoplus_{e\geq1} \bigl(R/(p^e)\bigr)^{m_M(p,e)}, \]where \(p\) runs through one representative of each associate class ofprime elements of \(R\), the integers \(r_M\) and \(m_M(p,e)\) arenonnegative, and only finitely many \(m_M(p,e)\) are nonzero.These integers are uniquely determined by \(M\).
This is the elementary-divisor form of the structure theorem for finitely generated modules over a PID [@DF04]. Choosing one representative of each associate class fixes the indexing of the invariants for both \(A\) and \(B\). When \(R\) is a field, there are no prime elements and the displayed torsion sum is empty.
2. The assumed isomorphism implies equality of all invariants of\(A\) and \(B\), and hence \(A\cong B\).
Taking a direct sum of two copies doubles the free rank and every elementary-divisor multiplicity. Thus the uniqueness in step 1, applied to \(A\oplus A\cong B\oplus B\), gives \[ 2r_A=2r_B,\qquad 2m_A(p,e)=2m_B(p,e)\quad\text{for every }p,e. \] These are equalities of finite nonnegative integers, so division by \(2\) gives \(r_A=r_B\) and \(m_A(p,e)=m_B(p,e)\) for every pair \(p,e\). The decompositions of \(A\) and \(B\) in step 1 therefore have precisely the same summands with the same multiplicities. Matching those summands gives an \(R\)-module isomorphism \(A\cong B\).