List all abelian groups of order 60.
List all \(\mathbb{Q}[x]\)-modules that are annihilated by \(x^8 - 1\) and have dimension 4 when thought of as vector spaces over \(\mathbb{Q}\).
Solution
All classifications below are up to isomorphism.
1. The abelian groups of order \(60\) are\(C_{60}\) and \(C_{30}\times C_2\), where \(C_n=\mathbb Z/n\mathbb Z\).
In an abelian group \(A\) of order \(60=4\cdot3\cdot5\), each Sylow subgroup is normal and hence unique by Sylow conjugacy [@DF04]. Multiplication from the product of these subgroups to \(A\) is a homomorphism because the factors commute. Its kernel is trivial: if \(uvw=1\) with \(u,v,w\) in the order-\(4\), order-\(3\), and order-\(5\) factors, then \(u=(vw)^{-1}\) has order dividing both \(4\) and \(15\), so \(u=1\). Now \(v=w^{-1}\) has order dividing both \(3\) and \(5\), so \(v=w=1\). The domain and codomain both have order \(60\), proving that multiplication is an isomorphism. The order-\(3\) and order-\(5\) factors are cyclic. An abelian group of order \(4\) either contains an element of order \(4\), in which case it is \(C_4\), or has every nonidentity element of order \(2\), in which case it is a two-dimensional vector space over \(\mathbb F_2\) and is \(C_2\times C_2\).
Thus the possibilities are \(C_4\times C_3\times C_5\cong C_{60}\) and \(C_2\times C_2\times C_3\times C_5 \cong C_2\times C_{30}\). The displayed identifications combine cyclic factors of coprime orders; a tuple of generators has order equal to the product of those orders. The two groups are not isomorphic, since only the first contains an element of order \(4\).
2. Every module in part (b) has a unique multiplicitydescription\[ M\cong E_1^{\oplus a}\oplus E_2^{\oplus b} \oplus E_4^{\oplus c}\oplus E_8^{\oplus d}, \qquad a+b+2c+4d=4, \]where the multiplicities are nonnegative integers and\[ E_1=\mathbb Q[x]/(x-1),\quad E_2=\mathbb Q[x]/(x+1),\quad E_4=\mathbb Q[x]/(x^2+1),\quad E_8=\mathbb Q[x]/(x^4+1). \]Each quotient has its natural \(\mathbb Q[x]\)-module structure.
Factor \[ f=x^8-1=(x-1)(x+1)(x^2+1)(x^4+1). \] The quadratic has no rational root and is irreducible. The substitution \(x=y+1\) changes \(x^4+1\) to \(y^4+4y^3+6y^2+4y+2\), which is Eisenstein at \(2\) [@DF04]. Since substitution by \(y+1\) is an invertible change of polynomial variable, \(x^4+1\) is also irreducible. Together with the two linear factors, these are distinct monic irreducibles and are therefore pairwise coprime.
The Chinese remainder theorem gives \[ R=\mathbb Q[x]/(f)\cong E_1\times E_2\times E_4\times E_8 \] [@DF04]. In particular, each \(E_j\) is a field. Their dimensions over \(\mathbb Q\) are respectively \(1,1,2,4\): for a monic polynomial of degree \(r\), polynomial division gives the basis \(1,x,\ldots,x^{r-1}\) in its quotient.
Because \(fM=0\), the action on \(M\) factors through \(R\). Let \(e_j\in R\) be the element whose coordinate in \(E_j\) is \(1\) and whose other coordinates are \(0\). Then \(\sum_j e_j=1\), \(e_j^2=e_j\), and \(e_je_k=0\) for \(j\ne k\). Thus every \(m\in M\) equals \(\sum_j e_jm\). If a sum of elements of the \(e_jM\) is zero, multiplying by \(e_k\) shows that its \(k\)th term is zero. Consequently \(M=\bigoplus_j e_jM\).
Each \(e_jM\) is a vector space over \(E_j\). It is finite-dimensional, since any linearly independent family over \(E_j\) is also linearly independent over \(\mathbb Q\) and \(\dim_{\mathbb Q}M=4\). Choose bases of sizes \(a,b,c,d\). This proves the claimed decomposition and dimension equation. Conversely, every direct sum satisfying that equation has dimension \(4\) and is annihilated by \(f\).
A \(\mathbb Q[x]\)-module isomorphism commutes with the action of each \(e_j\) and hence restricts to an \(E_j\)-linear isomorphism on the corresponding summand. It therefore preserves all four multiplicities. Equal multiplicities give isomorphic direct sums, proving uniqueness.
3. There are exactly ten modules in part (b), namely
| \((a,b,c,d)\) | Module |
|---|---|
| \((4,0,0,0)\) | \(E_1^{\oplus4}\) |
| \((3,1,0,0)\) | \(E_1^{\oplus3}\oplus E_2\) |
| \((2,2,0,0)\) | \(E_1^{\oplus2}\oplus E_2^{\oplus2}\) |
| \((1,3,0,0)\) | \(E_1\oplus E_2^{\oplus3}\) |
| \((0,4,0,0)\) | \(E_2^{\oplus4}\) |
| \((2,0,1,0)\) | \(E_1^{\oplus2}\oplus E_4\) |
| \((1,1,1,0)\) | \(E_1\oplus E_2\oplus E_4\) |
| \((0,2,1,0)\) | \(E_2^{\oplus2}\oplus E_4\) |
| \((0,0,2,0)\) | \(E_4^{\oplus2}\) |
| \((0,0,0,1)\) | \(E_8\) |
The equation in step 2 implies \(d\leq1\). For \(d=1\), the other multiplicities vanish, giving the last row. For \(d=0\), one has \(c\leq2\). If \(c=2\), then \(a=b=0\). If \(c=1\), then \(a+b=2\), giving the three pairs \((2,0),(1,1),(0,2)\). If \(c=0\), then \(a+b=4\), giving the five pairs displayed in the first five rows. These cases are exhaustive and disjoint, and uniqueness in step 2 shows that no two rows are isomorphic.