Abstract Algebra Theory and Applications - Computer Science ...

19.1 EXTENSION FIELDS 323Example 1. For example, letF = Q( √ 2 ) = {a + b √ 2 : a, b ∈ Q}and let E = Q( √ 2 + √ √ √3 ) be the smallest field containing both Q and2 + 3. Both E and F are extension fields of the rational numbers. Weclaim√that E is an√extension√field of F .√To see√this, we√need√only show that2 is in E. Since 2 + 3 is in E, 1/( 2 + 3 ) = 3 − 2 must also bein E. Taking linear combinations of √ 2 + √ 3 and √ 3 − √ √ √2, we find that2 and 3 must both be in E.Example 2. Let p(x) = x 2 + x + 1 ∈ Z 2 [x]. Since neither 0 nor 1 isa root of this polynomial, we know that p(x) is irreducible over Z 2 . Wewill construct a field extension of Z 2 containing an element α such thatp(α) = 0. By Theorem 15.13, the ideal 〈p(x)〉 generated by p(x) is maximal;hence, Z 2 [x]/〈p(x)〉 is a field. Let f(x) + 〈p(x)〉 be an arbitrary element ofZ 2 [x]/〈p(x)〉. By the division algorithm,f(x) = (x 2 + x + 1)q(x) + r(x),where the degree of r(x) is less than the degree of x 2 + x + 1. Therefore,f(x) + 〈x 2 + x + 1〉 = r(x) + 〈x 2 + x + 1〉.The only possibilities for r(x) are then 0, 1, x, and 1 + x. Consequently,E = Z 2 [x]/〈x 2 + x + 1〉 is a field with four elements and must be a fieldextension of Z 2 , containing a zero α of p(x). The field Z 2 (α) consists ofelements0 + 0α = 01 + 0α = 10 + 1α = α1 + 1α = 1 + α.Notice that α 2 + α + 1 = 0; hence, if we compute (1 + α) 2 ,(1 + α)(1 + α) = 1 + α + α + (α) 2 = α.Other calculations are accomplished in a similar manner. We summarizethese computations in the following tables, which tell us how to add andmultiply elements in E.+ 0 1 α 1 + α0 0 1 α 1 + α1 1 0 1 + α αα α 1 + α 0 11 + α 1 + α α 1 0