Re: Choice and algebraic closure

Re: Choice and algebraic closure

[Reinhard Boerger]

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

Choice and algebraic closure

[Reinhard Boerger]

Hallo,

in the PS to his mail on duality, John Baez wrote:

> Briefly, while the existence of an algebraic closure of Q
> can be shown without choice, it uniqueness-up-to-isomorphism
> seems to require choice.  Also, while arithmetic operations
> in Qbar are computable, they seem to present interesting challenges.

it would look interesting to me to show that uniqueness up to isomorphism
requires some kind of choice, e.g. by proving that it implies some choice
principle. I started thinking about this and came to the following
observation:

Let K be the field obtained from Q by adjoining all square roots of
(positive) primes or - equivalently - of all positive rational numbers.
Then an algebraic closureof K is obviously the same as an algebraic
closure of Q - even without choice. Now consider an arbitrary sequence of
two element sets S(n), w.l.o.g. pairwise disjoint, and let S be their
union. Now consider the polynomial ring Q(S) over Q in variables x(s) for
all s in S. 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)
where S(n) consists of the two elements s and t. Now let K' be the factor
ring obtained from Q(S) by dividing out all f_n and g_n. If we have a
choice function which assigns to each n an element s(n) in S(N), then
there is a unique isomorphism from K' to K that maps each s(n) to the
(positive)  square root of p(n). Conversely, if we have an arbitrary
isomorphism j from K' to K, we get a choice function which chooses for the
each n unique s(n) in S(n) with j(n)>0.  Thus existence of an isomorphism
is equivalent to existence of a choice function.

If the existence of an algebraic closure of K' could be shown without
choice, then an isomorphism fom this algebraic closure to the set of
algebraic complex numbers would restrict to an isomorphism between K and
K' and thus render a chioce function for the S(n). Thus the
uniqueness-up-to-isomorphism would imply choice for countable families of
two-element sets. Maybe refinement oft his argument could even be used to
get stronger choice principles.



                                                    Greetings
                                                    Reinhard Boerger