2.9 C ASOS DE PRUEBAS
2.9.6 Caso de prueba Gestionar Usuario
We begin with a theorem of Dedekind, who was the first to make a systematic study of field monomorphisms.
To motivate the theorem and its proof, we consider a special case. Suppose that Kand L are subfields of C, and let λ and µ be monomorphisms K → L. We claim that λ cannot be a constant multiple of µ unless λ = µ. By ‘constant’ here we mean an element of L. Suppose that there exists a ∈ L such that
µ (x) = aλ (x) (10.1)
for all x ∈ K. Replace x by yx, where y ∈ K, to get µ (yx) = aλ (yx) Since λ and µ are monomorphisms,
µ (y)µ (x) = aλ (y)λ (x)
137
Multiplying (10.1) by λ (y), we also have
λ (y)µ (x) = aλ (y)λ (x) Comparing the two, λ (y) = µ(y) for all y, so λ = µ.
In other words, if λ and µ are distinct monomorphisms K → L, they must be linearly independentover L.
Next, suppose that λ1, λ2, λ3are three distinct monomorphisms K → L, and as-sume that they are linearly dependent over L. That is,
a1λ1+ a2λ2+ a3λ3= 0 for aj∈ L. In more detail,
a1λ1(x) + a2λ2(x) + a3λ3(x) = 0 (10.2) for all x ∈ K. If some aj= 0 then we reduce to the previous case, so we may assume all aj6= 0.
Substitute yx for x in (10.2) to get
a1λ1(yx) + a2λ2(yx) + a3λ3(yx) = 0 (10.3) That is,
[a1λ1(y)]λ1(x) + [a2λ2(y)]λ2(x) + [a3λ3(y)]λ3(x) = 0 (10.4) Relations (10.2) and (10.4) are independent—that is, they are not scalar multiples of each other—unless λ1(y) = λ2(y) = λ3(y), and we can choose y to prevent this.
therefore we may eliminate one of the λjto deduce a linear relation between at most two of them, contrary to the previous case. Specifically, there exists y ∈ K such that λ1(y) 6= λ3(y). Multiply (10.2) by λ3(y) and subtract from (10.4) to get
[a1λ1(y) − a1λ3(y)]λ1(x) + [a2λ2(y) − a2λ3(y)]λ2(x) = 0 Then the coefficient of λ1(x) is a1(λ1(y) − λ3(y)) 6= 0, a contradiction.
Dedekind realised that this approach can be used inductively to prove:
Lemma 10.1 (Dedekind). If K and L are subfields of C, then every set of distinct monomorphisms K→ L is linearly independent over L.
Proof. Let λ1, . . . , λnbe distinct monomorphisms K → L. To say these are linearly independent over L is to say that there do not exist elements a1, . . . , an∈ L such that
a1λ1(x) + · · · + anλn(x) = 0 (10.5) for all x ∈ K, unless all the ajare 0.
Assume the contrary, so that (10.5) holds. At least one of the ai is non-zero.
Among all the valid equations of the form (10.5) with all ai6= 0, there must be at least one for which the number n of non-zero terms is least. Since all λj are non-zero, n 6= 1. We choose notation so that equation (10.5) is such as expression. Hence
Linear Independence of Monomorphisms 139 we may assume that there does not exist an equation like (10.5) with fewer than n terms. From this we deduce a contradiction.
Since λ16= λn, there exists y ∈ K such that λ1(y) 6= λn(y). Therefore y 6= 0. Now (10.5) holds with yx in place of x, so
a1λ1(yx) + · · · + anλn(yx) = 0 for all x ∈ K, whence
a1λ1(y)λ1(x) + · · · + anλn(y)λn(x) = 0 (10.6) for all x ∈ K. Multiply (10.5) by λ1(y) and subtract (10.6), so that the first terms cancel: we obtain
a2[λ2(x)λ1(y) − λ2(x)λ2(y)] + · · · + an[λn(x)λ1(y) − λn(x)λn(y)] = 0 The coefficient of λn(x) is an[λ1(y) − λn(y)] 6= 0, so we have an equation of the form (10.5) with fewer terms. Deleting any zero terms does not alter this statement. This contradicts the italicised assumption above.
Consequently no equation of the form (10.5) exists, so and the monomorphisms are linearly independent.
Example 10.2. Let K = Q(α) where α =√3
2 ∈ R. There are three monomorphisms K→ C, namely
λ1(p + qα + rα2) = p + qα + rα2 λ2(p + qα + rα2) = p + qωα + rω2α2 λ3(p + qα + rα2) = p + qω2α + rω α2
where p, q, r ∈ Q and ω is a primitive cube root of unity. We prove by ‘bare hands’
methods that the λj are linearly independent. Suppose that a1λ1(x) + a2λ2(x) + a3λ3(x) = 0 for all x ∈ K. Set x = 1, α, α2respectively to get
a1+ a2+ a3 = 0 a1+ ωa2+ ω2a3 = 0 a1+ ω2a2+ ωa3 = 0
The only solution of this system of linear equations is a1= a2= a3= 0.
For our next result we need two lemmas. The first is a standard theorem of linear algebra, which we quote without proof.
Lemma 10.3. If n > m then a system of m homogeneous linear equations ai1x1+ · · · + ainxn= 0 1 ≤ i ≤ m
in n unknowns x1, . . . , xn, with coefficients ai jin a field K, has a solution in which the xiare all in K and are not all zero.
This theorem is proved in most first-year undergraduate linear algebra courses, and can be found in any text of linear algebra, for example Anton (1987).
The second lemma states a useful general principle.
Lemma 10.4. If G is a group whose distinct elements are g1, . . . , gn, and if g∈ G, then as j varies from1 to n the elements ggjrun through the whole of G, each element of G occurring precisely once.
Proof. If h ∈ G then g−1h= gj for some j and h = ggj. If ggi= ggj then gi= g−1ggi= g−1ggj= gj. Thus the map gi7→ ggiis a bijection G → G, and the result follows.
We also recall some standard notation. We denote the cardinality of a set S by
|S|. Thus if G is a group, then |G| is the order of G. For example, |Sn| = n! and
|An| = n!/2.
We now come to the main theorem of this chapter, whose proof is similar to that of Lemma 10.1, and which can be motivated in a similar manner.
Theorem 10.5. Let G be a finite subgroup of the group of automorphisms of a field K, and let K0be the fixed field of G. Then[K : K0] = |G|.
Proof. Let n = |G|, and suppose that the elements of G are g1, . . . , gn, where g1= 1.
We prove separately that [K : K0] < n and [K : K0] > n are impossible.
(1) Suppose that [K : K0] = m < n. Let {x1, . . . , xm} be a basis for K over K0. By Lemma 10.3 there exist y1, . . . , yn∈ K, not all zero, such that
y1g1(xi) + · · · + yngn(xi) = 0 (10.7) for i = 1, . . . , m. Let x be any element of K. Then
x= α1x1+ · · · + αmxm where α1, . . . , αm∈ K0. Hence
y1g1(x) + · · · + yngn(x) = y1g1
∑
l
αlxl
!
+ · · · + yngn
∑
l
αlxl
!
=
∑
l
αl[y1g1(xl) + · · · + yngn(xl)]
= 0
using (10.7). Hence the distinct monomorphisms g1, . . . , gnare linearly dependent, contrary to Lemma 10.1. Therefore m ≥ n.
(2) Next, suppose for a contradiction that [K : K0] > n. Then there exists a set of n + 1 elements of K that are linearly independent over K0; let such a set be {x1, . . . , xn+1}.
By Lemma 10.3 there exist y1, . . . , yn+1∈ K, not all zero, such that for j = 1, . . . , n y1gj(x1) + · · · + yn+1gj(xn+1) = 0 (10.8)
Linear Independence of Monomorphisms 141 We subject this equation to a combinatorial attack, similar to that used in proving Lemma 10.1. Choose y1, . . . , yn+1so that as few as possible are non-zero, and renum-ber so that
y1, . . . , yr6= 0, yr+1, . . . , yn+1= 0 Equation (10.8) now becomes
y1gj(x1) + · · · + yrgj(xr) = 0 (10.9) Let g ∈ G, and operate on (10.9) with g. This gives a system of equations
g(y1)ggj(x1) + · · · + g(yr)ggj(xr) = 0
By Lemma 10.4, as j varies, this system of equations is equivalent to the system g(y1)gj(x1) + · · · + g(yr)gj(xr) = 0 (10.10) Multiply (10.9) by g(y1) and (10.10) by y1and subtract, to get
[y2g(y1) − g(y2)y1]gj(x2) + · · · + [yrg(y1) − g(yr)y1]gj(xr) = 0
This is a system of equations like (10.9) but with fewer terms, which gives a contra-diction unless all the coefficients
yig(y1) − y1g(yi) are zero. If this happens then
yiy−11 = g(yiy−11 )
for all g ∈ G, so that yiy−11 ∈ K0. Thus there exist z1, . . . , zr ∈ K0 and an element k∈ K such that yi= kzifor all i. Then (10.9), with j = 1, becomes
x1kz1+ · · · + xrkzr= 0
and since k 6= 0 we may divide by k, which shows that the xiare linearly dependent over K0. This is a contradiction.
Therefore [K : K0] is not less than n and not greater than n, so [K : K0] = n = |G|
as required.
Corollary 10.6. If G is the Galois group of the finite extension L : K, and H is a finite subgroup of G, then
[H†: K] = [L : K]/|H|
Proof. By the Tower Law, [L : K] = [L : H†][H†: K], so [H†: K] = [L : K]/[L : H†].
But this equals [L : K]/|H| by Theorem 10.5.
Examples 10.7. We illustrate Theorem 10.5 by two examples, one simple, the other more intricate.
(1) Let G be the group of automorphisms of C consisting of the identity and complex conjugation. The fixed field of G is R, for if x − iy = x + iy (x, y ∈ R) then y = 0, and conversely. Hence [C : R] = |G| = 2, a conclusion which is manifestly correct.
(2) Let K = Q(ζ ) where ζ = exp(2πi/5) ∈ C. Now ζ5= 1 and Q(ζ ) consists of all elements
p+ qζ + rζ2+ sζ3+ tζ4 (10.11) where p, q, r, s,t ∈ Q. The Galois group of Q(ζ ) : Q is easy to find, for if α is a Q-automorphism of Q(ζ ) then
(α(ζ ))5= α(ζ5) = α(1) = 1,
so that α(ζ ) = ζ , ζ2, ζ3, or ζ4. This gives four candidates for Q-automorphisms:
α1 : p + qζ + rζ2+ sζ3+ tζ4 7→ p + qζ + rζ2+ sζ3+ tζ4
α2 : 7→ p + sζ + qζ2+ tζ3+ rζ4
α3 : 7→ p + rζ + tζ2+ qζ3+ sζ4
α4 : 7→ p + tζ + sζ2+ rζ3+ qζ4
It is easy to check that all of these are Q-automorphisms. The only point to bear in mind is that 1, ζ , ζ2, ζ3, ζ4are not linearly independent over Q. However, their linear relations are generated by just one: ζ + ζ2+ ζ3+ ζ4= −1, and this relation is preserved by all of the candidate Q-automorphisms.
Alternatively, observe that ζ , ζ2, ζ3, ζ4 all have the same minimal polynomial t4+ t3+ t2+ t + 1 and use Corollary 5.13.
We deduce that the Galois group of Q(ζ ) : Q has order 4. It is easy to find the fixed field of this group: it turns out to be Q. Therefore, by Theorem 10.5, [Q(ζ ) : Q] = 4. At first sight this might seem wrong, for equation (10.11) expresses each element in terms of five basic elements; the degree should be 5. In support of this contention, ζ is a zero of t5− 1. The astute reader will already have seen the source of this dilemma: t5− 1 is not the minimal polynomial of ζ over Q, since it is reducible. The minimal polynomial is, as we have seen, t4+ t3+ t2+ t + 1, which has degree 4. Equation (10.11) holds, but the elements of the supposed ‘basis’ are linearly dependent. Every element of Q(ζ ) can be expressed uniquely in the form
p+ qζ + rζ2+ sζ3
where p, q, r, s ∈ Q. We did not use this expression because it lacks symmetry, making the computations formless and therefore harder.
EXERCISES
10.1 Check Theorem 10.5 for the extension C(t1, . . . ,tn) : C(s1, . . . , sn) of Chapter 8 Section 8.7.
Exercises 143 10.2 Find the fixed field of the subgroup {α1, α4} for Example 10.7(2). Check that
Theorem 10.5 holds.
10.3 Parallel the argument of Example 10.7(2) when ζ = e2πi/7. 10.4 Find all monomorphisms Q → C.
10.5 Mark the following true or false.
(a) If S ⊆ T is a finite set and |S| = |T |, then S = T . (b) The same is true of infinite sets.
(c) There is only one monomorphism Q → Q.
(d) If K and L are subfields of C, then there exists at least one monomorphism K→ L.
(e) Distinct automorphisms of a field K are linearly independent over K.
(f) Linearly independent monomorphisms are distinct.