There is 2 c automorphisms of complex numbers

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1 University of Silesia, Institute of Mathematics Wisła, February 7, 2011

2 Some facts from field theory Every field has its algebraic closure. It is unique up to isomorphism; hence one can denote the algebraic closure of K by K. If K is an infinite field, then K has the same cardinality as K. Let L be a field extension of a field K and S L. If a 1,..., a n S, P is a polynomial with coefficients in K and the polynomial equation P(a 1,..., a n ) = 0 is satisfied only for the zero polynomial P, then the set S is called algebraically independent (over K). Let L be a field extension of a field K. Using the lemma of Kuratowski- Zorn one can show the existence of a maximal algebraically independent subset of L; such a subset is called a transcendence basis (of the extension L over K).

3 Examples The transcendence basis of Q( 2) over Q is the empty set. A transcendence basis of Q(π) over Q is the set {π}. A transcendence basis of Q(π, e) over Q seems to be {π, e}, but it is not known. Even, it is not known if both π e and π + e are irrational [3]. As we shall see later, transcendence bases of C over Q are uncountable.

4 Lemma Let S be a transcendence basis of C over Q. Then C is algebraic over the field Q(S). Proof. For an indirect proof suppose α C \ Q(S) to be transcendental over Q(S). Consider a 1,..., a n S and a polynomial P Q[x 1,..., x n+1 ] with P(a 1,..., a n, α) = 0. Then arranging the polynomial P according to powers of α we obtain m P i (a 1,..., a n )α i = 0, i=0 where P i Q[x 1,..., x n ], i = 0, 1,..., m. By the transcendence of α and algebraic independence of S we have P i = 0, i = 0, 1,..., m, and consequently, P = 0. Hence the set S {α} is algebraically independent over Q, which contradicts the maximality of S.

5 Theorem Every transcendence basis of C (over Q) has the cardinality of the continuum. Proof. Let S be an arbitrary transcendence basis of C and card S = m. By the foregoing lemma C Q(S) C = C, and Q(S) = C. Consequently, card Q(S) = c. Let us note that { } f (a1,..., a n ) Q(S)= g(a 1,..., a n ) : f, g Q[x 1,..., x n ], g 0, a 1,..., a n S, n N. The inclusion follows from the fact that the right-hand side set is a field containing the set Q S, and the opposite inclusion follows from the fact that any field containing S has to include the fractions of such polynomials.

6 Proof ctd. Note that m is an infinite cardinal number. Indeed, suppose m to be finite. Then the set Q(S) would be countable and also Q(S), which is not the case. Let us denote the collection of all subsets K of S with card K = n by K n. Then the set Q(K) is countable for every K K n. Moreover card K n m n = m for any n N and card Q(K) mℵ 0 = m. K K n Furthermore Q(S) = n=1 K K n Q(K). Hence c = card Q(S) ℵ 0 m = m. On the other hand S C, thus m c. Finally, card S = m = c.

7 Lemma If A is a set of the cardinality of the continuum, then the set S(A) of all permutations of A has the cardinality of 2 c. Proof. Without loss of generality assume A = R. Let us denote card S(R) = m. Now for a function f : (0, 1) {0, 1} define a function g : R R in the following way { x for x R \ B, g(x) = x for x B, where B = f 1 ({1}) ( f 1 ({1})). One can see that g S(R). In this fashion we obtained a mapping Φ: {0, 1} (0,1) S(R) which is injective. Hence 2 c = card {0, 1} (0,1) m. On the other hand Consequently, m = 2 c. m card R R = c c = 2 c.

8 Lemma Let K be a subfield of an algebraically closed field L. If ϕ: K L is a homomorphism and α is algebraic over K, then ϕ can be extended to a homomorphism ϕ : K(α) L. Sketch of the proof. Denote by f ϕ the polynomial which is obtained from a polynomial f K[x] by sending each coefficient a of the polynomial f to ϕ(a). Let p be the minimal polynomial of α and β be a root of p ϕ. Consider x K(α). It is of the form f (α), where f K[x], and the mapping ϕ : x f ϕ (β), x K(α) is desired extension of ϕ. For details see [2].

9 Theorem There exists 2 c automorphisms of the field C. Proof. Let S be an arbitrary transcendence basis of C over Q. Consider a permutation f : S S. It defines an isomorphism ϕ: Q(S) Q(S) sending α to f (α), α S. We shall show that ϕ can be extended to an automorphism of C. Let X = { (F, ψ): Q(S) F C, ψ : F C field homomorphism, ψ Q(S) = ϕ } and order X as usual (F 1, ψ 1 ) (F 2, ψ 2 ) F 1 F 2, ψ 2 F1 = ψ 1. X is non-empty, because of (Q(S), ϕ) X. If Γ X is a chain, then it has an upper bound in X. Indeed, define F = (F,ψ) Γ F and ϕ: F C in the following way: If (F, ψ) Γ, then ϕ F = ψ. One can easily see that ( F, ϕ) is an upper bound for Γ.

10 Proof ctd. By the lemma of Kuratowski-Zorn there is a maximal element (F, ϕ ) in X. We have F = C: If it were α C \ F, then, by the previous lemma, ϕ could be extended to F (α), which would contradict the maximality of (F, ϕ ). Furthermore, ϕ as a non-zero homomorphism is injective. It remains to show that the range of ϕ is C. Indeed, ϕ : C ϕ (C) is an isomorphism. Thus fields C and ϕ (C) are isomorphic, in particular ϕ (C) is algebraically closed. Hence C = Q(S) ϕ (C) = ϕ (C) C and ϕ (C) = C. It shows that one can construct 2 c automorphisms of the field C. On the other hand, the number of these automorphisms cannot be greater than card C C = c c = 2 c.

11 Concluding remarks Although there is so many automorphisms of C, it is quite impossible to give an example aside from the identity or the complex conjugation, beacause they are necessarily non-measurable [1] and the AC has to be applied to prove their existence. There are automorphisms of C swapping 3 with 3 and leaving 17 unchanged. There are non-trivial endomorphisms of C whose range is properly contained in C. It can be proven in the following way: Consider a mapping f from a transendence basis of C over Q into itself which is injective, but not surjective. Say α is not in its range. Then f extends to an endomorphism of C whose range does not include α. Hence there are many proper subfields of C isomorphic to C.

12 Marek Kuczma, An Introduction to the Theory of Functional Equations and Inequalities, PWN, Uniwersytet Śląski, Warszawa-Kraków-Katowice, Serge Lang, Algebra, PWN, Warszawa, James S. Milne, Fields and Galois Theory, 2008, available at: Paul B. Yale, Automorphisms of Complex Numbers, available at: mathdl.maa.org/images/upload library/22/ford/ PaulBYale.pdf.

Fields and Galois Theory. Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory.

Fields and Galois Theory. Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory. Fields and Galois Theory Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory. This should be a reasonably logical ordering, so that a result here should