Andrej Bauer wrote:
Reinhard Boerger wrote:
For each natural number n let p(n) be the n-th prime and consider the poynomials f_n:=x(s)+x(t) and g_n:=x(s)x(t)+p(n) in Q(S)
What precisely is the status of the expressions "x(s)+x(t)" and "x(s)x(t)" here? It is not clear to me that, given a two-element set which possibly cannot be ordered, we can form a sum or a product indexed by the set. I am afraid you're covertly ordering the two element set when you name its elements s and t, then form the sum x(s)+x(t). After all, a polynomial in Q[S] is going to be an (equivalence class of) sequence of pairs (coefficient, monomial). So if we could form x(s)+x(t) we could (perhaps) also order s and t by looking at the sequence which represents x(s)+x(t). Am I raising any doubts in your mind? I think some details need to be worked out here.
You are right; without choice we should be careful, and several classically equal things become distinct. I am used to the following way of defining the polynomial ring over a ring A in variables x(s) for all s from a given set S: Monomials are maps from S to the set of nonnegative integers, which map all but finitely many elements to 0. The images under this map are considered as exponents, and monomials are multiplied by adding the exponents, and x(s) is the monomial that maps s to 1 and all other elements to 0. Then x(s)x(t) is literally the same as x(t)x(s). Analogously, we can define polynomials as maps from the set of all monomials to A, which are 0 almost every where; the images are interpreted as coefficients. Then we also have x(s)+x(t)=x(t)+x(s). However, also another definition should turm the polynomial ring into a commutative rings; so the elements could also be defined as equivalence classes of (first-order) terms. The point is that x(s)x(t) should bet taken as an element of the (commutative) polynomial ring, not as a term; the terms x(s)x(t)and x(t)x(s) are indeed different. The algebraic closure of Q (or of a finite field) can be constructed by successively adding zeros of polynomials. Since the set of all polynomials can be numbered, an isomorphism between two different algebraic closures can be found, if on each step we find a bijection between the finte sets of zeros of a given polynomials in both closures. This choice usually depends on the privious choices; so countable dependent choice in finite sets (or König's Lemma) should render the desired isomorphism. On the other hand, the situation is very specific, I do not see how König's Lemma should follow from the existence of isomorphism between algebraic closures of Q. So I considered an easier case: The choices of square roots of different primes are independent. So the isomorphism between K and K' in my example can be obtained uniquely if for every prime you have bijectin between the set of the two square roots of a prime p in K and the set of the square roots of p in K'. ButK is ordered by construction; so you obtain the bijection if fir each p you choose which one of the two square roots of p shall be mapped to the positive square root in K': If you have the isomorphism, you get an ordering of K' and therefore a choice of positive square roots. So had to build the field K' for a given sequence of two-element sets without using a choice for these sets. Of course the result is not satifactory; it was just my first idea when I read John Baez' mail. Since there are also irreducible polynomials of degrees >2, I think that the sock axiom (i.e. countable choise for sequences of two-element sets, i.e. choosing a left one for countably many pairs of symmetric socks) should not imply the uniqueness of the algebraic closure of Q up to isomorphism. In order to getthe converse, we should need a way to extend an isomporphism between K and K' to an isomorphism between algebraic closures without using choice again. Greetings Reinhard