Moving Up the Ladder of Subfields of a Splitting Field

25 Apr

Given a number field K with a chain of subfields K_1, K_2, …, K, each of which is contained in the previous one, one can try to pass from the rational numbers Q to K by successively breaking symmetries of present in a given subfield to move to the next one up. For this, the following theorem is important:

Theorem: Suppose that L is the splitting field of a polynomial with rational coefficients, and suppose that M is a subfield of L. Let G be the Galois group of K, and let H be a subgroup of G such that M is symmetric with respect to H. Let ‘a’ be an element of L. Let

$\large \sigma_1, \sigma_2, \ldots, \sigma_n$

be the elements of H. Define a polynomial P(x) by

$\large P(x) = (x - \sigma_{1}(a))(x - \sigma_{2}(a))\cdots(x - \sigma_{n}(x))$

Then P(x) is a polynomial with coefficients in M and with ‘a’ as a root.

Proof: H contains the trivial permutation, so ‘a’ is a root of the polynomial. The coefficients of P(x) are invariant with respect to H, because if

$\large \sigma \in H$

Then

$\sigma: H \to H$

defined by

$\sigma(\sigma_{i}(y))$

permutes the elements of H, so the coefficients of P(x) are symmetric with respect to H, so by the Fundamental Theorem of Galois Theory the coefficients are in M.

Thus, we can pass from the rational numbers Q to K by (successively, for each i) expressing an element of K_i in terms of roots of polynomials with coefficients in K_(i – 1).