Page 293 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Figure 31.3

From left to right, constructions of a 17-gon, 257-gon and 65537-gon.

Other Constructions

It should be stressed that the concept of constructible as discussed in this article applies specifically
to compass and straightedge construction. More constructions become possible if other tools are
allowed. The so-called neusis constructions, for example, make use of a marked rulers.The
constructions are a mathematical idealization and are assumed to be done exactly.

Example: Determine the group of all automorphisms of a field with 4 elements.
Solution: The automorphism group consists of two elements: the identity mapping and the
Frobenius automorphism.

As you know this field with 4 elements can be constructed as F = Z [x] / < x +x+1 >. Letting a be
2
2
the coset of x, we have F = {0, 1, a, 1+a}. Any automorphism of F must leave 0 and 1 fixed, so the
only possibility for an automorphism other than the identity is to interchange a and 1+a. Is this
an automorphism? Since x +x+1 0, we have x -x-1 x+1, so a = 1+a and (1+a) = 1+2a+a = a. Thus
2
2
2
2
2
the function that fixes 0 and 1 while interchanging a and 1+a is in fact the Frobenius automorphism
of F.
Example: Let F be the splitting field in C of x +1.
4
(i) Show that [F:Q] = 4.
Solution: The polynomial x -1 factors over Q as x -1 = (x -1)(x +1) = (x-1)(x+1)(x +1)(x +1). The
8
4
4
4
2
8
factor x +1 is irreducible over Q by Eisenstein’s criterion. The roots of x +1 are thus the primitive
4
4
8th roots of unity, ± 2 / 2 ± 2 / 2i, and adjoining one of these roots also gives the others,
together with i. Thus, the splitting field is obtained in one step, by adjoining one root of x +1, so
4
its degree over Q is 4.
It is clear that the splitting field can also be obtained by adjoining first 2 and then i, so it can
also be expressed as Q( 2 , i).

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