Abstract Algebra Theory and Applications - Computer Science ...

19.2 SPLITTING FIELDS 333Proof. Let F be an algebraically closed field. If p(x) ∈ F [x] is a nonconstantpolynomial, then p(x) has a zero in F , say α. Therefore, x−α must bea factor of p(x) and so p(x) = (x − α)q 1 (x), where deg q 1 (x) = deg p(x) − 1.Continue this process with q 1 (x) to find a factorizationp(x) = (x − α)(x − β)q 2 (x),where deg q 2 (x) = deg p(x) − 2. The process must eventually stop since thedegree of p(x) is finite.Conversely, suppose that every nonconstant polynomial p(x) in F [x] factorsinto linear factors. Let ax − b be such a factor. Then p(b/a) = 0.Consequently, F is algebraically closed.□Corollary 19.14 An algebraically closed field F has no proper algebraicextension E.Proof. Let E be an algebraic extension of F ; then F ⊂ E. For α ∈ E, theminimal polynomial of α is x − α. Therefore, α ∈ F and F = E. □Theorem 19.15 Every field F has a unique algebraic closure.It is a nontrivial fact that every field has a unique algebraic closure. Theproof is not extremely difficult, but requires some rather sophisticated settheory. We refer the reader to [3], [4], or [7] for a proof of this result.We now state the Fundamental Theorem of Algebra, first proven byGauss at the age of 22 in his doctoral thesis. This theorem states thatevery polynomial with coefficients in the complex numbers has a root in thecomplex numbers. The proof of this theorem will be given in Chapter 21.Theorem 19.16 (Fundamental Theorem of Algebra) The field of complexnumbers is algebraically closed.19.2 Splitting FieldsLet F be a field and p(x) be a nonconstant polynomial in F [x]. We alreadyknow that we can find a field extension of F that contains a root of p(x).However, we would like to know whether an extension E of F containing allof the roots of p(x) exists. In other words, can we find a field extension ofF such that p(x) factors into a product of linear polynomials? What is the“smallest” extension containing all the roots of p(x)?