Let us take a closer look at the field Q[x]/(x² - 2). Since this is a field of remainders after division by a quadratic polynomial, the elements of the field are all the linear polynomials Q[x]/(x² - 2) = {a + bx : a,b Î Q}. We already know how the a’s and b’s operate, they are just rational numbers. What is new and different about this field is the x. If we multiply x ´ x = x² and then ask for the remainder after division by x² - 2 we get x² = 1(x² - 2) + 2, so the remainder is 2 and we have x² = 2 in Q[x]/(x² - 2). Our extension field Q[x]/(x² - 2) differs from the base field Q only in the addition of a new element x, which satisfies x² = 2, or in other words x = Ö2. So we have proved Q[x]/(x² - 2) @ Q(Ö2), where Q(Ö2) is the field generated by the rational numbers Q with the real number Ö2 adjoined.

This is actually the standard pattern. We can look at field extensions generated either by adjoining a new element to the field or by moving from field to the polynomial domain and then "modding out" by an irreducible polynomial. These will turn out to be the same thing if the polynomial used to "mod out" is the minimal polynomial of the element you adjoin. In a graduate algebra course we would spend some time on this point, but for our purposes we will focus mainly on some applications of the study of extension fields. In particular, we will prove the impossibility of trisecting an angle with straightedge and compass, or of duplicating a cube with the same tools. These were classical questions of Greek geometry, but it turns out their answers depend on the theory of extension fields.

The usual approach in secondary and college algebra is just to toss in all the real numbers, or even all the complex numbers, and not to worry about Ö and ³ Ö and the like when looking for answers to polynomial equations. This is not the historical approach. The Greeks were very concerned about the philosophical implications of the existence of irrational numbers and kept careful track of whether they were manipulating "numbers" (which had to be rational) or "lengths" (which could be irrational). It was not until the 15^th and 16^th centuries that mathematicians started to manipulate real numbers without worry. It is no accident that the 15^th and 16^th centuries also saw tremendous developments in algebraic reasoning. When you are first working with algebra and polynomials, you usually don’t want to worry about what field you are working in. But in the 19^th century, mathematicians discovered that when they wanted to understand the theory of polynomial equations, it did help to be very careful about such things. And now that our historical interlude is over, back to the properties of extension fields.

Definition: A basis for an extension field E over a base field F is a set of elements {e₁, e₂, ... ,e_n} that satisfies the following two properties

1. For every element e Î E, there exist f₁, f₂, ... , f_n Î F such that e = f₁e₁ + f₂e₂ + ... + f_ne_n.
2. If f₁e₁ + f₂e₂ + ... + f_ne_n = 0, then f₁ = f₂ = ... = f_n = 0.

The first rule says that for any e Î E, there is at least one way it can be constructed as a linear combination of the e_i’s. A consequence of the second rule is that there is at most one way to construct any e. So together the rules say a basis is a set of elements such that any element of the extension field can be constructed in exactly one way as a linear combination of the basis elements where the coefficients are drawn from the base field.

Example: The pair {1, x} is a basis for Q[x]/(x² - 2). Every element can be written as a + bx with a,b Î Q, the base field, in exactly one way.

A basis for an extension field is not unique. For example, if we consider the extension field Q(Ö2,Ö3) over Q, one possible basis is {1, Ö2, Ö3, Ö6}. Another basis is {1, Ö2+Ö3, (Ö2+Ö3)², (Ö2+Ö3)³}. We won’t bother to prove this, but it isn’t that hard to check. One thing to note is that while the bases look quite different, they both have 4 elements in them. While there are many possible choices for the elements to make up a basis, it turns out that how many elements you need is forced on you by the extension.

Theorem: Suppose E is an extension field over F. If e₁, ... , e_n and g₁, ... , g_m are both bases for E over F, then n = m.

Proof: Suppose m ³ n. We will show it is possible to replace the e_i by g_i while keeping a basis. This process is straightforward, but involves some messy algebra. Since g₁ Î E, we can write g₁ = f₁₁e₁ + … + f_1ne_n for some choice of f_ij Î F. Not all the f_ij can be 0, else g₁ would be 0. Say f₁₁ ¹ 0. Then we can solve our for e₁ in our expression for g₁ to get e₁ = (1/f₁₁)g₁ - (f₁₂/f₁₁)e₂ - … - (f_1n/f₁₁)e_n. Now I claim g₁, e₂, …, e_n is another basis for E. To check this we must check our two conditions. To verify the first condition, suppose e Î E. Then since e₁, …, e_n is a basis, we can find x₁, …, x_n Î F so that e = x₁e₁ + … + x_ne_n = (x₁/f₁₁)g₁ + (x₂ - x₁f₁₂/f₁₁)e₂ + … + (x_n - x₁f_1n/f₁₁)e_n and condition 1 is satisfied. To check the second condition, we assume x₁g₁ + x₂e₂ + … + x_ng_n = 0. Then x₁f₁₁e₁ + (x₂ + x₁f₁₂)e₂ + … + (x_n + x₁f_1n)e_n = 0 and since the e_j form a basis, we must have x₁f₁₁ = 0, x₂ + x₁f₁₂ = 0, …, x_n + x₁f_1n = 0. Since f₁₁ ¹ 0, the first equality implies x₁ = 0. Once we have x₁ = 0, the other equations then become x₂ = 0, …, x_n = 0, and the second condition for a basis is also satisfied.

Now we repeat our process and to replace another of the e_j with g₂. Since g₂ Î E, we can write g₂ = f₂₁g₁ + f₂₂e₂ + … + f_2ne_n. Now one of f₂₂, …, f_2n must not be 0, since if all these terms were 0 then we would have g₂ = f₂₁g₁, from which we could get f₂₁g₁ + (-1)g₂ = 0, in contradiction of condition 2 of the g_j forming a basis. So let's say f₂₂ ¹ 0. Then we can write e₂ = -(f₂₁/f₂₂)g₁ + (1/f₂₂)g₂ - (f₂₃/f₂₂)e₃ - … - (f_2n/f₂₂)e_n and run through the same process as before to show g₁, g₂, e₃, …, e_n is also a basis. We proceed in this manner until we finally replace all the e_j and conclude g₁, …, g_n is a basis.

Now if m > n, we have g_n+1 Î E and we can write g_n+1 = x₁g₁ + … + x_ng_n for some x₁, …, x_n Î F, since g₁, …, g_n is a basis. But we can rewrite this as x₁g₁ + … + x_ng_n + (-1)g_n+1 = 0. But this contradicts condition 2 of the g_j being a basis since not all the coefficients are 0 (in particular, the coefficient of g_n+1 is -1 ¹ 0). So it can't happen that m > n, and the two bases must have the same number of elements. This is the end of the proof.

Now that we know that all bases must have the same number of elements, we are justified in making the following definition.

Definition: If E is an extension field of F, the degree of the extension, deg[E : F], is the number of elements in a basis for E over F.

It should be noted that not all extensions have bases. For example, Q(p) doesn't have a basis over Q. The problem here is that there is no finite set of elements that satisfies condition 1 for a basis. We will stick to nice finite degree extensions as much as possible. On the other hand, the extension field Q(p) does come up in the question of squaring a circle, which is why I wanted to point out the limits of our definition of degree of extension.

In the next lecture we will extend our study of degree of extension to towers of extension fields. Our big theorem will be that if K is an extension field of E which is an extension field of F, then deg[K : F] = deg[K : E]΄deg[E : F]. With this result we will be ready to tackle trisecting the angle and the other construction problems of antiquity.