Prove or disprove each of the following:
If \(R[x]\) is a PID, then \(R\) is a field.
All cyclic subgroups of order 15 in \(S_8\) are conjugate.
All torsion-free \(\mathbb{Z}\)-modules are free.
All degree two field extensions are Galois.
Solution
The assertions are respectively true, true, false, and false.
1. If \(R[x]\) is a principal ideal domain, then \(R\) is a field.
The constant polynomials identify \(R\) with a subring of the commutative integral domain \(R[x]\), so \(R\) is a commutative integral domain with \(1\ne0\). Fix \(0\ne a\in R\). The ideal \((a,x)\) is generated by some \(f\in R[x]\). As \(a=fg\) for a nonzero \(g\in R[x]\), additivity of degrees over a domain gives \(\deg f=\deg g=0\). Thus \(f\) is a nonzero constant. The identity \(x=fh\) for some \(h\in R[x]\) shows, by comparing coefficients of \(x\), that \(1=fh_1\) for an element \(h_1\in R\). Hence \(f\) is a unit, and \((a,x)=R[x]\).
Write \(1=au(x)+xv(x)\) with \(u,v\in R[x]\). Evaluation at \(x=0\) gives \(1=au(0)\). Every nonzero \(a\in R\) is therefore invertible.
2. All cyclic subgroups of order \(15\) in \(S_8\) are conjugate.
The order of a permutation is the least common multiple of its disjoint-cycle lengths [@DF04]. A permutation of order \(15\) in \(S_8\) must have a cycle with length divisible by \(5\). Every cycle length divides \(15\) and is at most \(8\), so this cycle has length \(5\). Only three letters remain. The order must also be divisible by \(3\), so those letters form a \(3\)-cycle. Thus every generator of a cyclic subgroup of order \(15\) has cycle type \((5)(3)\).
For two such generators, match the five letters in their cyclic orders and then match the three remaining letters in their cyclic orders. The resulting permutation conjugates one generator to the other. It consequently conjugates their generated subgroups. The subgroup generated by \((1\,2\,3\,4\,5)(6\,7\,8)\) shows that such subgroups exist.
3. The \(\mathbb Z\)-module \(\mathbb Q\) is torsion-free but not free.
For a nonzero integer \(n\) and \(q\in\mathbb Q\), the equation \(nq=0\) implies \(q=0\), so the module is torsion-free. Moreover, \(2\mathbb Q=\mathbb Q\), since \(q=2(q/2)\).
If \(\mathbb Q\) were free over \(\mathbb Z\), its nonzero underlying module would have a nonempty basis. A basis vector \(e\) could not belong to twice the module: in the unique finite basis expansion of \(2m\), every coefficient is even, whereas the coefficient of \(e\) in itself is \(1\). This contradicts \(2\mathbb Q=\mathbb Q\).
4. In characteristic \(2\) a quadratic extension need not be Galois.
Let \(F=\mathbb F_2(t)\) with \(t\) an indeterminate. The polynomial \(X^2-t\) has no root in \(F\). Indeed, a root \(A(t)/B(t)\) with nonzero polynomials \(A,B\) would imply \(A(t)^2=tB(t)^2\). The degrees of the two sides would be respectively even and odd, an impossibility. A quadratic over a field is reducible exactly when it has a root, so \(X^2-t\) is irreducible.
Set \(K=F(u)\) where \(u^2=t\). The minimal polynomial of \(u\) is \(X^2-t\), giving \([K:F]=2\). In characteristic \(2\), \[ X^2-t=(X-u)^2. \] Thus the minimal polynomial has a repeated root, and \(K/F\) is not separable. In particular it is not Galois [@DF04].