Use Dedekinds Theorem to factorise the following principal i

Use Dedekind’s Theorem to factorise the following principal ideals in the ring of integers of the following fields. (Note that these rings are UFDs.) ? (a) Q(?3): ?2?,?3?,?5?,?10?,?30?.

Solution

) The field Ef¯ = Af /P = Fp[¯x1, x¯2, . . . , x¯n] is a splitting field for ¯f, where ¯x is the residue class of x modulo P. The group Gf¯ = Gal(Ef¯/Fp) is cyclic generated by the automorphism ¯x 7 x¯ p . Let DP = { Gf | (P) = P}. This is a subgroup of Gf called the decomposition group at P. Every automorphism DP induces an automorphism ¯ Gf¯ = Gal(Ef¯), where ¯(¯x) = (x). The homomorphism : DP Gf¯ sending to ¯ is injective. We now show that it is surjective by showing that the fixed field of (DP ) has Fp as its fixed field. Let a Af . Then, by the Chinese Remainder Theorem, there an element x Af such that x a mod P and x 0 mod 1 (P) for all Gf , / DP . Then g = Q Gf (X (x) Z[X] and ¯g = Xm Q DP (X ¯(¯a)) Fp[X]. It follows that the conjugates of ¯a are all of the form ¯(¯a) which implies that the fixed field of (DP ) is Fp. Let P DP be the unique element such that ¯P (¯x) = ¯x p . Then P is the unique element of Gf such that P (x) x p for every x Af . Since the homomorphism x 7 x¯ maps the roots of f bijectively onto the roots of ¯f we see that the groups DP and Gf¯, when viewed as permutation groups of the roots of f, ¯f respectively, are also isomorphic as permutation groups. Since the cycle decompostion of ¯ is determined by the orbits of the action of Gf¯ on the roots of ¯f and since the group Gf¯ acts transitively on the roots of each polynomial gi , we obtain the stated cycle decomposition of P . If Rf is the ring of integers of Ef , i.e., the elements of Ef which are integral over Z and Q is a prime ideal of Rf such that Q Z = pZ then, as above, one can prove the existence of a unique automorphism sQ Gf such that sQ(x) x p (mod Q) for all x Rf . This automorphism is called the Frobenius automorphism at Q. Since the elements of Gf are uniquely determined by their restriction to Af , we see that sQ = P , where P = Q Af . If Q0 is any ideal of Rf such that Q0 Z = Q Z and x Q0 then Q Gf (x) Q0 Z Q which shows that (x) Q for some Gf . Hence Q0 S Gf (Q). By the following Lemma, we have Q0 (Q) and hence Q0 = (Q) for some Gf Since D(Q) = DQ 1 , it follows that sQ0 = Q1 . Thus two Frobenius automorphisms at primes over the same prime p of Z are conjugate. The conjugacy class of sQ is called the Frobenius class at p. If G is abelian, this class reduces to a single element called the Frobenius automorphism at p.