Field Extensions

We are familiar with the observation that the equation x² = 2 cannot be solved within the rational field, but has the solutions ± √3 in the field ℝ of real numbers. In fact its solutions lie within a much smaller field than ℝ, namely, the extension

of ℚ. It is not perhaps quite obvious that this is a field, but it is easy to verify the subfield conditions. If a + b √3, c + d√3 ∈ ℚ(√3), then

and (if c + d√3 ≠ 0)

where

Similarly there are actually many fields lying between ℚ and ℂ, not only ℝ. One simple example is the field ℚ(i) = {p + qi | p, q ∈ ℚ} which has already been mentioned a couple of times. A similar example is ℚ(√2) = {p + q√2 | p,q ∈ ℚ}.

Definition 8.1.2 Let F be a field, A subset K of F that is itself a field under the operations of F is called a subfield of F. In this context, F is called an extension field of K. If K ≠ F, we say that K is a proper subfield of F. A field containing no proper subfields is called a prime field.  □

To be more precise, an extension field of a field F is a pair (K, ι) made by a field K and a monomorphic (of field) ι from F to K. Generally we identify an element a of F by its image ι(a) in K, considering F as subfield of K.

Characteristic of a field

We already introduced the notion of characteristic of a field and that of a prime subfield. We have proved in Theorem 4.15.6 that the prime subfield of every field with characteristic zero is the rational field, whilst the prime subfield of every field with characteristic p is the field ℤ_p. Thus every field with characteristic zero is the extension the the rationals (it embeds the rational field), while every field of characteristic p is the extension of the field ℤ_p.

Every field ia an integral domain. Thus in order to find the characteristic of field F, we should find the order of the unit element 1 of F when regarded as a member of the additive group of F, (F,+), which is generated by the multiplicative identity 1.

If F is an extension field of K, then multiplication in F defines a scalar multiplication, if we consider the elements of K as scalars and the elements of F as vectors. It is easy to check that the necessary axioms hold for this scalar multiplication, and since F is an abelian group under addition, we see that F is a vector space over K. We crystalyze this fact as a proposition.

Proposition 8.1.3. If F is an extension field of K, then F is a vector space over K.  □

Example 8.1.4. Every element of ℂ (considered as a vector space over ℝ) is a linear combination of 1 and i because every element can be written in the formal a1 + bi, with a, b ∈ ℝ. Thus the set {1, i} spans ℂ over ℝ, so ℂ is an extension of degree 2 on ℝ. The set {1 + i, 5i, 2 + 3i} also spans ℂ because any a + bi ∈ ℂ is a linear combination of these three elements with coefficients in ℝ:

{1, i} represents a basis of the vector field since it is linear inddependent; the set {1 + i, 5i, 2 + 3i} is not linearly independent and so it's not a basis.  ■

The field ℝ(x) of rational functions with coefficients in ℝ has infinite dimension on ℝ, since it contains the rational functions 1, x, x², ..., xⁱ, whilst is an extension of degree 1 on itself.

Knowing that an extension field is a vector space over the base field allows us to make use of the concept of the dimension of a vector space.

Proposition 8.1.5. Let F be an extension field of K. If the dimension of F as a vector space over K is finite, then F is said to be a finite extension of K. The dimension of F as a vector space over K is called the degree of F over K, and is denoted by [F : K]. □

ℂ ⊇ ℝ is finite, and [ℂ : ℝ] = 2 because {1, i} is an ℝ-basis of ℂ.

A tower of extensions is a sequence of successive field extensions. In a tower of extensions, the degree is multiplicative:

8.1.6 Mutliplication Theorem Let L be a finite extension of K and let K in turn be a finite extension of F. Then

1. L is a finite extension of F

2. [L : F] = [L : K][K : F].

Schematically the situation is the following

Proof. Let [L : K] = m, [K : F] = n and let X = {x₁, x₂, ..., x_n} a basis of L over the field K and let Y = {y₁, y₂, ..., y_n} a basis of K over the field F. We shall prove that the set of mn products {x_iy_j}_{i = 1,...,m, j = 1, ..., n} is a basis of L over F, proving both assertions of the theorem.

1. The x_iy_j are generators. Let l an arbitrary element of L. Then

   $$l=\sum _{i=1}^m{k}_i{x}_i,k_i\in K$$

   since the x₁, x₂, ..., x_n are a basis of L over K. Since y₁, y₂, ..., y_n is a basis of K over the field F, every k_i can be written a linear combination of the y_i with coefficients in F:

   $$k_i=\sum _{j=1}^n{a}_{ij}{y}_j$$

   It follows that

   $$l=\sum _{i=1}^m\left(\sum _{j=1}^n{a}_{ij}{y}_j\right)x_i,=\sum _{i,j}{a}_{ij}{y}_j{x}_i$$

2. x_iy_j are a linearly independent set over F. Suppose that

   $$l=\sum _{i,j}{a}_{ij}(y_j{x}_i)=0,a_{ij}\in F$$

   We have then

   $$0=\sum _{i=1}^m\underset{\in K}{\underbrace{\left(\sum _{j=1}^n{a}_{ij}{y}_j\right)}}{x}_i$$

   From the independence of the x_i over K,

   $$\sum _{j=1}^n{a}_{ij}{y}_j=0$$

   Since the y_j are independent over F, it must follow a_ij = 0 for every i,j. □

Corollary 8.1.7 Let L a finite extension of F and K a subfield of L embedding F. Then

Proof. If L is a finite extension of F, then every othter subfield of L is a finite extension of F (think to them as vector space). If L has finite degree on F, that is if L has finite dimension as vector space over F, then it has for sure finite dimension over K (by extending the scalar field, there's a reduction of independent elements, i.e. the dimension decreases see Example 1.4.7), that is [L : K] < ∞. From the theorem

The last result can be seen as an analogous for field extension of the Lagrange's theorem for finite groups.

Corollary 8.1.8 If K is a field extension of F which has a degree a prime number p, then there aren't intermediate field betweem K and F.