According to Theorem 30.23, the element 1 + of Z 2 (a) of Example 29.19 is

According to Theorem 30.23, the element 1 + α of Z₂(a) of Example 29.19 is algebraic over Z₂. Find the irreducible polynomial for 1 + α in Z₂[x].

Data from in Example 29.19

The polynomial p(x) = x² + x + 1 in Z₂[x] is irreducible over Z₂ by Theorem 23.10, since neither element 0 nor element 1 of Z₂ is a zero of p(x). By Theorem 29.3, we know that there is an extension field ∈ of Z₂ containing a zero α of x² + x + 1. By Theorem 29.18, Z₂(α) has as elements 0 + 0α, 1 + 0α, 0 + 1α, and 1 + 1α, that is, 0, 1, α, and 1 + α. This gives us a new finite field, of four elements! The addition and multiplication tables for this field are shown in Tables 29.20 and 29.21. For example, to compute (1 + α)(l + α) in Z₂(α), we observe that since p(α) = α² + α + 1 = 0, then α² = - α - 1 = α + 1. 

Therefore, (1 + α)(1 +α)= 1 + α + α+ α² = 1 + α² = 1 + α + 1 = α.