Re: Ordinal numbers and rings

Arturo Magidin schrieb:


I suspect the author wants to use the fact that although for successor
ordinals A_{alpha} and A_{alpha'} need not be related (other than
being extensions of A), if you can find a limit ordinal gamma such
that alpha,alpha' <= gamma < beta, then A_{alpha} and A_{alpha'} are
both contained in A_{gamma}, and you can go from there.



That said, however, I think the author is wrong (or at least, that his
assertion is incomplete). For example, let beta = omega = w, the natural
number, which is a limit ordinal. Let A = Q, B=algebraic closure of Q,
and define A_n for natural number n as follows:

A_0 = Q

A_n = Q(sqrt(p_n)), where p_n is the n-th prime.

Then A_n is a ring extension of A for every n, and it is a tight
extension, since the only nonzero ideal of A_n is A_n itself, which
intersect A in a nonzero ideal (namely A).

This means that A_{n+1} is a tight extension of A for every ordinal
n<w, and by vacuity A_gamma is the union of all A_alpha with
alpha<gamma, when gamma is a limit ordinal strictly less than omega.

However, the union of the A_n is not a ring, since for example
sqrt(2)+sqrt(3) is not defined and is not even in any of the A_n.

Yes, nice observation.. I think maybe there is another typographical
error or maybe I didn't copied the text correctly, unfortunately I have
the paper in office now and its almost weekend. I believe the second
condition was that A_{alpha+1} is a tight extension of A_alpha for
every ordinal alpha<beta (otherwise I don't see why the ordinals are at
all being used here!).. he is forming some kind of recursive argument.

Your example wouldn't work given that condition, because A_{n+1} for
instance isnt an extension of A_n ..

Now let me see that supposing we assume this if
\/ A_alpha
for alpha < beta is a ring ....

if a is in A_alpha and a' is in A_alpha' for alpha,alpha'<=gamma<beta
and without loss of generality alpha<alpha' (well-ordering of ordinals?
not equal because otherwise its trivial) and we can assume that
alpha'=gamma and that gamma is the infimum (exists?) of limit ordinals
greater than alpha , because Im trying to show that a is in A_gamma,
and A_gamma is in fact a ring.. a in A_gamma is clear...

a' is in A_gamma implies that there is an ordinal beta strictly less
than gamma such that alpha<=beta<gamma and a' is in A_gamma,

but since gamma is the infimum of limit ordinals greater than alpha,
then beta is not a limit ordinal (is this true?) and beta=alpha+n (is
this true?) for some natural number n. Which means that
A_beta=A_(alpha+n) is a ring extension of A_alpha, and since a and a'
are both contained in A_beta, we can use aa' and a+a' in A_beta ring
structure.

Im not an expert in transfinite induction (in fact I never used it in
my mathematical career so far..), I'm assuming that this can also be
done using tranfinite induction?...

assume for all ordinals beta between alpha and gamma, A_beta is a ring.
Then because gamma is the infimum of the limit ordinals greater than
alpha, beta+1 =/= beta, and so A_(beta+1)

I'm still trying to understand why the author is trying ot use ordinal
numbers.. is this because he wants to account ALL the rings that lie
between A and B? But I have to admit, I feel very weird when using
ordinal arithmetic in algebra or ring theory.. Im not at all
comfortable or totally convienced on using them.

Sincerely,
Jose Capco

.

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