- List all abelian groups of size 90.
- List all \(\mathbf{R}[x]\) modules \(M\) such that \(\dim_{\mathbf{R}} M=4\) and \(M\) is annihilated by \(x^4-1\).
Solution
1. Up to isomorphism, the abelian groups of order \(90\) are \(\ZZ/90\) and \(\ZZ/3\times\ZZ/30\).
Since \(90=2\cdot3^2\cdot5\), the structure theorem for finite abelian groups gives a unique decomposition into cyclic groups of prime-power order whose exponents for each prime \(p\) form a partition of the exponent of \(p\) in \(90\) [@DF04]. The \(2\)-part and \(5\)-part are \(\ZZ/2\) and \(\ZZ/5\); the \(3\)-part is \(\ZZ/9\) or \(\ZZ/3\times\ZZ/3\). By the Chinese remainder theorem, \(\ZZ/2\times\ZZ/9\times\ZZ/5\cong\ZZ/90\) and \(\ZZ/2\times\ZZ/3\times\ZZ/3\times\ZZ/5\cong\ZZ/3\times\ZZ/30\).
2. Up to isomorphism, the \(\RR[x]\)-modules \(M\) with \(\dim_\RR M=4\) and \((x^4-1)M=0\) are the nine modules\[ M_{a,b,c}=\bigl(\RR[x]/(x-1)\bigr)^{a}\oplus\bigl(\RR[x]/(x+1)\bigr)^{b}\oplus\bigl(\RR[x]/(x^2+1)\bigr)^{c}, \qquad a,b,c\ge0,\ a+b+2c=4. \]
Over \(\RR\), \(x^4-1=(x-1)(x+1)(x^2+1)\) with pairwise distinct irreducible factors. The module \(M\) is finitely generated and torsion over the principal ideal domain \(\RR[x]\), so by the structure theorem it is a direct sum of cyclic modules \(\RR[x]/(p^k)\) with \(p\) monic irreducible, and the multiset of these summands is determined by \(M\) [@DF04]. Each \(p^k\) divides \(x^4-1\), which is squarefree, so \(k=1\) and \(p\in\{x-1,x+1,x^2+1\}\). Hence \(M\cong M_{a,b,c}\) for unique \(a,b,c\ge0\), and \(\dim_\RR M=a+b+2c\). The solutions of \(a+b+2c=4\) are the five triples with \(c=0\), the three with \(c=1\) and \(a+b=2\), and \((0,0,2)\). Each \(M_{a,b,c}\) is annihilated by \(x^4-1\) and has dimension \(4\).
3. Q.E.D.
Step 1 answers part (a) and step 2 answers part (b).