Determine the number of conjugacy classes of \(16 \times 16\) matrices with entries in \(\mathbb{Q}\) and minimal polynomial \((x^2+1)^2(x^3+2)^2\).
Solution
There are exactly \(\boxed{4}\) conjugacy classes. Put \(p=x^2+1\) and \(q=x^3+2\).
1. The classes correspond bijectively to integer tuples\[ (a,b,c,d),\qquad a,c\geq1,\quad b,d\geq0, \qquad 4a+2b+6c+3d=16. \]
The polynomial \(p\) has no rational root and is irreducible. The polynomial \(q\) is Eisenstein at two and is also irreducible [@DF04]. They are distinct monic irreducibles.
For a matrix \(A\), let \(x\) act on \(V=\mathbb Q^{16}\) by \(A\). The structure theorem over the PID \(\mathbb Q[x]\) gives a unique multiset of elementary divisors [@DF04]. Since the annihilator is generated by the minimal polynomial \(p^2q^2\), the only possible summands are \[ \mathbb Q[x]/(p^2),\quad \mathbb Q[x]/(p),\quad \mathbb Q[x]/(q^2),\quad \mathbb Q[x]/(q). \] Indeed, the annihilator of a direct sum is the intersection of the summands’ annihilators, so its monic generator is the least common multiple of their elementary divisors. It equals \(p^2q^2\) exactly when at least one \(p^2\) summand and at least one \(q^2\) summand occur.
Let their multiplicities in the displayed order be \(a,b,c,d\). Their rational dimensions are \(4,2,6,3\), respectively, giving the equation. Conversely, any tuple satisfying these conditions gives a module of dimension sixteen with exactly the specified minimal polynomial. Matrix conjugacy is equivalent to isomorphism of these modules, since a module isomorphism is exactly an invertible rational linear map intertwining the two operators. Thus uniqueness of the elementary divisors gives the asserted bijection.
2. The four possible tuples and representatives are
| \((a,b,c,d)\) | Representative |
|---|---|
| \((1,3,1,0)\) | \(C_{p^2}\oplus C_p^{\oplus3}\oplus C_{q^2}\) |
| \((1,0,1,2)\) | \(C_{p^2}\oplus C_{q^2}\oplus C_q^{\oplus2}\) |
| \((2,1,1,0)\) | \(C_{p^2}^{\oplus2}\oplus C_p\oplus C_{q^2}\) |
| \((1,0,2,0)\) | \(C_{p^2}\oplus C_{q^2}^{\oplus2}\) |
Here \(C_f\) is the companion matrix of the monic polynomial \(f\).
Since \(a\geq1\), one has \(6c\leq12\), so \(c=1\) or \(2\). If \(c=2\), the equation becomes \(4a+2b+3d=4\); it forces \(a=1,b=d=0\). If \(c=1\), then \(4a+2b+3d=10\), so \(a=1\) or \(2\). For \(a=2\), one gets \(2b+3d=2\), hence \(b=1,d=0\). For \(a=1\), one gets \(2b+3d=6\). The integer \(d\) is even and lies between zero and two, giving \((b,d)=(3,0)\) or \((0,2)\). These disjoint cases give exactly the four rows.
1.Different rows have different elementary divisors,so they are not conjugate. No additional tuple,and hence no additional conjugacy class, is possible.