Abstract Algebra Theory and Applications - Computer Science ...

324 CHAPTER 19 FIELDS· 0 1 α 1 + α0 0 0 0 01 0 1 α 1 + αα 0 α 1 + α 11 + α 0 1 + α 1 αThe following theorem, due to Kronecker, is so important and so basicto our understanding of fields that it is often known as the FundamentalTheorem of Field Theory.Theorem 19.1 Let F be a field and let p(x) be a nonconstant polynomialin F [x]. Then there exists an extension field E of F and an element α ∈ Esuch that p(α) = 0.Proof. To prove this theorem, we will employ the method that we usedto construct Example 2. Clearly, we can assume that p(x) is an irreduciblepolynomial. We wish to find an extension field E of F containing an elementα such that p(α) = 0. The ideal 〈p(x)〉 generated by p(x) is a maximal idealin F [x] by Theorem 15.13; hence, F [x]/〈p(x)〉 is a field. We claim thatE = F [x]/〈p(x)〉 is the desired field.We first show that E is a field extension of F . We can define a homomorphismof commutative rings by the map ψ : F → F [x]/〈 p(x) 〉, whereψ(a) = a + 〈p(x)〉 for a ∈ F . It is easy to check that ψ is indeed a ringhomomorphism. Observe thatandψ(a) + ψ(b) = (a + 〈p(x)〉) + (b + 〈p(x)〉) = (a + b) + 〈p(x)〉 = ψ(a + b)ψ(a)ψ(b) = (a + 〈p(x)〉)(b + 〈p(x)〉) = ab + 〈p(x)〉 = ψ(ab).To prove that ψ is one-to-one, assume thata + 〈p(x)〉 = ψ(a) = ψ(b) = b + 〈p(x)〉.Then a − b is a multiple of p(x), since it lives in the ideal 〈p(x)〉. Sincep(x) is a nonconstant polynomial, the only possibility is that a − b = 0.Consequently, a = b and ψ is injective. Since ψ is one-to-one, we canidentify F with the subfield {a + 〈p(x)〉 : a ∈ F } of E and view E as anextension field of F .