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{1+i, 1−i} is a normal basis of C/R while {1,i} is not

Example

The extension C/R (The complex numbers as R[x]/(x2+1), with the real embedding and imaginary unit i) is finite Galois of degree two with Gal⁡(C/R)={id, z↦z‾} (Real and imaginary parts, complex conjugation, and modulus). For it:

  1. the conjugate list (1+i, 1−i) is a normal basis (Normal bases of a finite Galois extension);
  2. (1,i) is an R-basis of C that is not a conjugate list of any element;
  3. (i,−i) is the conjugate list of i but is not a basis.

The last two show that the two conditions in the definition of a normal basis are independent of each other.

Facts & Assumptions

Given: The complex field with i2=−1 and conjugation a+bi‾=a−bi for a,b∈R (Real and imaginary parts, complex conjugation, and modulus); in R one has 2≠0.

[L1]

C=R(i) is a simple algebraic extension with power basis 1,i and [C:R]=2 (C/R has power basis 1,i and degree 2, The degree [K:F]=dim⁡FK of a finite field extension).

[L2]

Every field automorphism of C fixing R pointwise is either the identity or complex conjugation, and these two are distinct (The only real-field automorphisms of C are the identity and complex conjugation, Relative field automorphisms and Aut⁡(K/F)).

[L4]

A finite extension K/F with G=Aut⁡(K/F) is Galois exactly when ∣G∣=[K:F] (Equivalent characterizations of a finite Galois extension, Finite Galois extensions and Gal⁡(K/F)).

[L6]

Every finite Galois extension of an infinite field has a normal basis (Every finite Galois extension of an infinite field has a normal basis).

Verification

technique · direct
1.1L1L2L3L4

By [L2] and [L3] the group Aut⁡(C/R) has exactly the two elements id and conjugation, so its order is 2=[C:R] by [L1]; hence C/R is finite Galois with that Galois group by [L4].

2.1step 1.1L1given

The conjugate list of 1+i is (1+i, 1−i), whose members have coordinate lists (1,1) and (1,−1) in the ordered basis (1,i) of [L1]. For a,b∈R, a(1+i)+b(1−i)=(a+b)+(a−b)i vanishes exactly when a+b=0 and a−b=0, hence when 2a=0, that is a=0 and then b=0.

2.2step 1.1L1given

(1,i) is a basis by [L1], but no z∈C has conjugate list (z,z‾) with underlying set {1,i}: such a z would lie in {1,i}, and the set for z=1 is {1} while for z=i it is {i,−i}, neither of which is {1,i}.

3.1step 1.1step 2.1L1L5L6

So the linear map R2→C sending (a,b) to a(1+i)+b(1−i) has trivial kernel; both spaces have dimension two by [L1], so [L5] makes it bijective and (1+i,1−i) an ordered R-basis of C. Being the conjugate list of 1+i, it is a normal basis, in agreement with [L6].

4.1step 1.1step 3.1step 2.2L5given∎

(i,−i) is the conjugate list of i, since i‾=−i, and its two members are distinct; but 1⋅i+1⋅(−i)=0 is a vanishing combination with nonzero coefficients, so the list is not independent and by [L5] is not a basis. With steps 3.1 and 2.2 this establishes all three claims.

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