The Degree of an Extension
6.2 The Tower Law
The next theorem lets us calculate the degree of a complicated extension if we know the degrees of certain simpler ones.
Theorem 6.4 (Short Tower Law). If K, L, M are subfields of C and K ⊆ L ⊆ M, then
[M : K] = [M : L][L : K]
Note:For those who are happy with infinite cardinals this formula needs no extra explanation; the product on the right is just multiplication of cardinals. For those who are not, the formula needs interpretation if any of the degrees involved is infinite. This interpretation is the obvious one: if either [M : L] or [L : K] = ∞ then [M : K] = ∞;
and if [M : K] = ∞ then either [M : L] = ∞ or [L : K] = ∞.
Proof. Let (xi)i∈I be a basis for L as vector space over K and let (yj)j∈J be a basis for M over L. For all i ∈ I and j ∈ J we have xi∈ L, yj ∈ M. We shall show that (xiyj)i∈I, j∈J is a basis for M over K (where xiyj is the product in the subfield M).
Since dimensions are cardinalities of bases, the theorem follows.
First, we prove linear independence. Suppose that some finite linear combination of the putative basis elements is zero; that is,
∑
i, jki jxiyj= 0 (ki j∈ K)
The Tower Law 81
Since the coefficients ∑iki jxilie in L and the yjare linearly independent over L,
∑
iki jxi= 0
Repeating the argument inside L we find that ki j= 0 for all i ∈ I, j ∈ J. So the ele-ments xiyjare linearly independent over K.
Finally we show that the xiyjspan M over K. Any element x ∈ M can be written
for λi j∈ K. Putting the pieces together, x=
∑
i, j
λi jxiyj as required.
Example 6.5. Suppose we wish to find [Q(√ 2,√
2 are linearly independent over Q. Suppose that p + q√
2 = 0, where p, q ∈ Q. If q6= 0 then√
2 = p/q, which is impossible since√
2 is irrational. Therefore q = 0.
But this implies p = 0. we argue much as above: if
(p + q√
which is possible only if either a = 0 or b = 0. But then√ The theorem even furnishes a basis for Q(√
2,√
3) over Q : form all possible pairs of products from the two bases {1,√
2} and {1,√
3}, to get the ‘combined’ basis {1,√
2,√ 3,√
6}.
By induction on n we easily parlay the Short Tower Law into a useful generali-sation:
Corollary 6.6 (Tower Law). If K0⊆ K1⊆ · · · ⊆ Knare subfields of C, then [Kn: K0] = [Kn: Kn−1][Kn−1: Kn−2] · · · [K1: K0]
In order to use the Tower Law we have to get started. The degree of a simple extension is fairly easy to find:
Proposition 6.7. Let K(α) : K be a simple extension. If it is transcendental then [K(α) : K] = ∞. If it is algebraic then [K(α) : K] = ∂ m, where m is the minimal polynomial of α over K.
Proof. For the transcendental case it suffices to note that the elements 1, α, α2, . . . are linearly independent over K. For the algebraic case, we appeal to Lemma 5.14.
For example, we know that C = R(i) where i has minimal polynomial t2+ 1, of degree 2, Hence [C : R] = 2, which agrees with our previous remarks.
Example 6.8. We now illustrate a technique that we shall use, without explicit refer-ence, whenever we discuss extensions of the form Q(√
α1, . . . ,√
αn) : Q with rational αj. The technique can be used to prove a general theorem about such extensions, see Exercise 6.15. The question we tackle is: find [Q(√
2,√
It is ‘obvious’ that each factor equals 2, but it takes some effort to prove it. As a cautionary remark: the degree [Q(√
6,√
The Tower Law 83 We argue as in Example 6.5. Squaring,
3 = (p2+ 2q2) + 2pq which is impossible for the same reason. Therefore√
3 6∈ Q(√
(c) Finally, we claim that√
5 6∈ Q(√ 2,√
3). Here we need a new idea. Suppose
√ Adding the first two equations, we get p + q√
2 = 0 or p + q√ 2 =√
5. The first implies that p = q = 0. The second implies that p2+ 2q2+ 2pq√
2 = 5, which is easily seen to be impossible. Adding the first and third, r√
3 = 0 or r√ 3 =√
5, so r= 0. Finally, s = 0 since s√
6 =√
5 is impossible by Exercise 1.3.
Having proved the claim, we immediately deduce that [Q(√
Linear algebra is at its most powerful when dealing with finite-dimensional vec-tor spaces. Accordingly we shall concentrate on field extensions that give rise to such vector spaces.
Definition 6.9. A finite extension is one whose degree is finite.
Proposition 6.7 implies that any simple algebraic extension is finite. The converse is not true, but certain partial results are: see Exercise 6.16. In order to state what is true we need:
Definition 6.10. An extension L : K is algebraic if every element of L is algebraic over K.
Algebraic extensions need not be finite, see Exercise 6.11, but every finite exten-sion is algebraic. More generally:
Lemma 6.11. An extension L : K is finite if and only if L = K(α, . . . , αr) where r is finite and each αiis algebraic over K.
Proof. Induction using Theorem 6.4 and Proposition 6.7 shows that any extension of the form K(α1, . . . , αs) : K for algebraic αjis finite.
Conversely, let L : K be a finite extension. Then there is a basis {α1, . . . , αs} for Lover K, whence L = K(α1, . . . , αs). Each αjis clearly algebraic.
EXERCISES
6.1. Find the degrees of the following extensions:
(a) C : Q (b) R(√
5) : R
(c) Q(α) : Q where α is the real cube root of 2 (d) Q(3,√
5,√ 11) : Q (e) Q(√
6) : Q
(f) Q(α) : Q where α7= 3 6.2. Show that every element of Q(√
5,√
7) can be expressed uniquely in the form p+ q√
5 + r√ 7 + s√
35
where p, q, r, s ∈ Q. Calculate explicitly the inverse of such an element.
6.3. If [L : K] is a prime number show that the only fields M such that K ⊆ M ⊆ L are K and L themselves.
6.4. If [L : K] = 1 show that K = L.
6.5. Write out in detail the inductive proof of Corollary 6.6.
6.6. Let L : K be an extension. Show that multiplication by a fixed element of L is a linear transformation of L considered as a vector space over K. When is this linear transformation nonsingular?
Exercises 85 6.7. Let L : K be a finite extension, and let p be an irreducible polynomial over K.
Show that if ∂ p does not divide [L : K], then p has no zeros in L.
6.8. If L : K is algebraic and M : L is algebraic, is M : K algebraic? Note that you may not assume the extensions are finite.
6.9. Prove that Q(√ 3,√
5) = Q(√ 3 +√
5). Try to generalise your result.
6.10* Prove that the square roots of all prime numbers are linearly independent over Q. Deduce that algebraic extensions need not be finite.
6.11 Find a basis for Q(
q (1 +√
3)) over Q and hence find the degree of Q(
If K = Q, how can we verify the hypotheses on the ajby looking at their prime factorisations?
6.15* Let L : K be an algebraic extension and suppose that K is an infinite field. Prove that L : K is simple if and only if there are only finitely many fields M such that K ⊆ M ⊆ L, as follows.
(a) Assume only finitely many M exist. Use Lemma 6.11 to show that L : K is finite.
(b) Assume L = K(α1, α2). For each β ∈ K let Jβ = K(α1+ β α2). Only finitely many distinct Jβ can occur: hence show that L = Jβ for some β . (c) Use induction to prove the general case.
(d) For the converse, let L = K(α) be simple algebraic, with K ⊆ M ⊆ L.
Let m be the minimal polynomial of α over K, and let mM be the min-imal polynomial of α over M. Show that mM|m in L[t]. Prove that mM determines M uniquely, and that only finitely many mMcan occur.
6.16 Mark the following true or false.
(a) Extensions of the same degree are isomorphic.
(b) Isomorphic extensions have the same degree.
(c) Every algebraic extension is finite.
(d) Every transcendental extension is not finite.
(e) Every element of C is algebraic over R.
(f) Every extension of R that is a subfield of C is finite.
(g) Every algebraic extension of Q is finite.