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Definition 6.5.1. Let K be a field and let f(x) = a0 + a1 x + · · · + anxn be a polynomial in K[x] of degree n>0. An extension field F of K is called a splitting field for f(x) over K if there exist elements r1, r2, . . . , rn in F such that
In the above situation we usually say that f(x) splits over the field F. The elements r1, r2, . . . , rn are roots of f(x), and so F is obtained by adjoining to K a complete set of roots of f(x).
Theorem 6.4.2. Let f(x) be a polynomial in K[x] of degree n>0. Then there exists a splitting field F for f(x) over K, with [F:K] n!.
Lemma 6.4.3. Let : K -> L be an isomorphism of fields. Let F be an extension field of K such that F = K(u) for an algebraic element u in F. Let p(x) be the minimal polynomial of u over K. If v is any root of the image q(x) of p(x) under , and E=L(v), then there is a unique way to extend to an isomorphism : F -> E such that (u) = v and (a) = (a) for all a in K.
Theorem 6.4.5. Let f(x) be a polynomial over the field K. The splitting field of f(x) over K is unique up to isomorphism.
1. Find the splitting field over Q for the polynomial x4 + 4. Solution
2. Find the degree of the splitting field over Z2 for the polynomial (x3 + x + 1)(x2 + x + 1). Solution
3. Find the degree [F:Q], where F is the splitting field of the polynomial x3 - 11 over the field Q of rational numbers. Solution
4. Determine the splitting field over Q for x4 + 2. Solution
5. Determine the splitting field over Q for x4 + x2 + 1. Solution
6. Factor x6 - 1 over Z7; factor x5 - 1 over Z11. Solution
Solutions to the problems | Forward to §6.5 | Back to §6.3 | Up | Table of Contents