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From: Rupert on 16 Jun 2010 00:07
On Jun 16, 1:50 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Jun 15, 4:06 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
>
>
>
>
> > On Jun 16, 3:20 am, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
> > > On Jun 15, 2:11 am, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
> > > [.Construction snipped. at some point it requires arguing that |F[x]|=|
> > > F[x,y]|=|F|aleph_0 for field F .]
>
> > > > > Now, the isomorphism between additive structures can be given
> > > > > constructively/explicitly, without having to invoke the Axiom of
> > > > > Choice. Can the same be done with the isomorphism of the
> > > > > multiplicative structures? I'm not sure if I'm invoking AC above; I
> > > > > may be in arguing some of the cardinalities or the inequalities
> > > > > between cardinalities.
>
> > > > I think you might be using the fact that |X|^2=|X|, which is
> > > > equivalent to the Axiom of Choice.
>
> > > Of course! I've run into that one before, and it always seems to
> > > escape me that I need Axiom of Choice in general.
>
> > > For arbitrary infinite field F, the argument that |F[x]|=|F| that is
> > > the "obvious" one is to note that F[x] can be put in bijection with
> > > the disjoint union of F, F^2, F^3, F^4,..., F^n, etc. This is a
> > > countable union of sets of *the same cardinality as F* (here is where
> > > AC is used, I guess), so has cardinality |F|.
>
> > > > > If not in general, can it be done for some specific fields, say F=Q,
> > > > > where we have explicit well-orderings of Q; or C, where the set of
> > > > > monic irreducibles of F[x] is easy?
>
> > > > Yes. The equalities you need can be proven without AC in these cases,
> > > > I believe.
>
> > > For Q, the problem above does not arise. For C, it may be difficult to
> > > argue that the set of monic irreducibles in C[x,y] is at most
> > > denumerable...
>
> > I thought that you wanted to prove that the set of monic irreducibles
> > in C[x,y] had the same cardinality as C. I would think this should be
> > all right. We can prove that |C|^2=|C| and |C|*aleph_0=|C| without
> > using AC.
>
> If we can prove |C^2| = |C| without AC, then I'm indeed fine.
> Iterating this we can show |C[x]|=|C|, and from there that |C[x,y]|=|
> C[x][y]|=|C|; then showing |C| <= # monic irreducibles in C[x.y] <= |
> C[x,y]| is trivial, so that would do it.
>
> --
> Arturo Magidin- Hide quoted text -
>
> - Show quoted text -

Well |C|=2^(aleph_0) so |C|
^2=(2^(aleph_0))^2=(2^(aleph_0*2))=2^aleph_0=|C|.
From: Arturo Magidin on 16 Jun 2010 00:13
On Jun 15, 11:07 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
> On Jun 16, 1:50 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
>
>
> > On Jun 15, 4:06 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
> > > On Jun 16, 3:20 am, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
> > > > On Jun 15, 2:11 am, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
> > > > [.Construction snipped. at some point it requires arguing that |F[x]|=|
> > > > F[x,y]|=|F|aleph_0 for field F .]
>
> > > > > > Now, the isomorphism between additive structures can be given
> > > > > > constructively/explicitly, without having to invoke the Axiom of
> > > > > > Choice. Can the same be done with the isomorphism of the
> > > > > > multiplicative structures? I'm not sure if I'm invoking AC above; I
> > > > > > may be in arguing some of the cardinalities or the inequalities
> > > > > > between cardinalities.
>
> > > > > I think you might be using the fact that |X|^2=|X|, which is
> > > > > equivalent to the Axiom of Choice.
>
> > > > Of course! I've run into that one before, and it always seems to
> > > > escape me that I need Axiom of Choice in general.
>
> > > > For arbitrary infinite field F, the argument that |F[x]|=|F| that is
> > > > the "obvious" one is to note that F[x] can be put in bijection with
> > > > the disjoint union of F, F^2, F^3, F^4,..., F^n, etc. This is a
> > > > countable union of sets of *the same cardinality as F* (here is where
> > > > AC is used, I guess), so has cardinality |F|.
>
> > > > > > If not in general, can it be done for some specific fields, say F=Q,
> > > > > > where we have explicit well-orderings of Q; or C, where the set of
> > > > > > monic irreducibles of F[x] is easy?
>
> > > > > Yes. The equalities you need can be proven without AC in these cases,
> > > > > I believe.
>
> > > > For Q, the problem above does not arise. For C, it may be difficult to
> > > > argue that the set of monic irreducibles in C[x,y] is at most
> > > > denumerable...
>
> > > I thought that you wanted to prove that the set of monic irreducibles
> > > in C[x,y] had the same cardinality as C. I would think this should be
> > > all right. We can prove that |C|^2=|C| and |C|*aleph_0=|C| without
> > > using AC.
>
> > If we can prove |C^2| = |C| without AC, then I'm indeed fine.
> > Iterating this we can show |C[x]|=|C|, and from there that |C[x,y]|=|
> > C[x][y]|=|C|; then showing |C| <= # monic irreducibles in C[x.y] <= |
> > C[x,y]| is trivial, so that would do it.

> Well |C|=2^(aleph_0) so |C|
> ^2=(2^(aleph_0))^2=(2^(aleph_0*2))=2^aleph_0=|C|.

I'm wary of just invoking cardinal arithmetic because things like |
X^2| = |X| do not hold in general without AC. Does the fact that the
power set of a cardinal is bijectable with a cardinal require AC, for
example?

But we could just go from |R^2| = |R| to |R^n| = |R| and |C|=|R^2| to |
C^2| = |(R^2)x(R^2)| = |R^4| = |R| = |R^2|=|C|, and from there to |C^n|
=|C|. This does not seem to me to require AC, or even comparability to
cardinals and cardinal arithmetic.

--
Arturo Magidin
From: Rupert on 17 Jun 2010 05:22
On Jun 16, 2:13 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Jun 15, 11:07 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
>
>
> > On Jun 16, 1:50 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
> > > On Jun 15, 4:06 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
> > > > On Jun 16, 3:20 am, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
> > > > > On Jun 15, 2:11 am, Rupert <rupertmccal...(a)yahoo.com> wrote:
>
> > > > > [.Construction snipped. at some point it requires arguing that |F[x]|=|
> > > > > F[x,y]|=|F|aleph_0 for field F .]
>
> > > > > > > Now, the isomorphism between additive structures can be given
> > > > > > > constructively/explicitly, without having to invoke the Axiom of
> > > > > > > Choice. Can the same be done with the isomorphism of the
> > > > > > > multiplicative structures? I'm not sure if I'm invoking AC above; I
> > > > > > > may be in arguing some of the cardinalities or the inequalities
> > > > > > > between cardinalities.
>
> > > > > > I think you might be using the fact that |X|^2=|X|, which is
> > > > > > equivalent to the Axiom of Choice.
>
> > > > > Of course! I've run into that one before, and it always seems to
> > > > > escape me that I need Axiom of Choice in general.
>
> > > > > For arbitrary infinite field F, the argument that |F[x]|=|F| that is
> > > > > the "obvious" one is to note that F[x] can be put in bijection with
> > > > > the disjoint union of F, F^2, F^3, F^4,..., F^n, etc. This is a
> > > > > countable union of sets of *the same cardinality as F* (here is where
> > > > > AC is used, I guess), so has cardinality |F|.
>
> > > > > > > If not in general, can it be done for some specific fields, say F=Q,
> > > > > > > where we have explicit well-orderings of Q; or C, where the set of
> > > > > > > monic irreducibles of F[x] is easy?
>
> > > > > > Yes. The equalities you need can be proven without AC in these cases,
> > > > > > I believe.
>
> > > > > For Q, the problem above does not arise. For C, it may be difficult to
> > > > > argue that the set of monic irreducibles in C[x,y] is at most
> > > > > denumerable...
>
> > > > I thought that you wanted to prove that the set of monic irreducibles
> > > > in C[x,y] had the same cardinality as C. I would think this should be
> > > > all right. We can prove that |C|^2=|C| and |C|*aleph_0=|C| without
> > > > using AC.
>
> > > If we can prove |C^2| = |C| without AC, then I'm indeed fine.
> > > Iterating this we can show |C[x]|=|C|, and from there that |C[x,y]|=|
> > > C[x][y]|=|C|; then showing |C| <= # monic irreducibles in C[x.y] <= |
> > > C[x,y]| is trivial, so that would do it.
> > Well |C|=2^(aleph_0) so |C|
> > ^2=(2^(aleph_0))^2=(2^(aleph_0*2))=2^aleph_0=|C|.
>
> I'm wary of just invoking cardinal arithmetic because things like |
> X^2| = |X| do not hold in general without AC. Does the fact that the
> power set of a cardinal is bijectable with a cardinal require AC, for
> example?
>

It depends how you define "cardinal". If we are doing without AC then
we can define the cardinality of a set S to be the set of all sets
equipollent with S of least possible rank, and then every set has a
cardinal even if that set may not be well-orderable.

> But we could just go from |R^2| = |R| to |R^n| = |R| and |C|=|R^2| to |
> C^2| = |(R^2)x(R^2)| = |R^4| = |R| = |R^2|=|C|, and from there to |C^n|
> =|C|. This does not seem to me to require AC, or even comparability to
> cardinals and cardinal arithmetic.
>

Yes, that's right.

> --
> Arturo Magidin

From: Robert E. Beaudoin on 20 Jun 2010 21:17
On 06/16/10 00:13, Arturo Magidin wrote:
> On Jun 15, 11:07 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>> On Jun 16, 1:50 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
>>
>>
>>
>>> On Jun 15, 4:06 pm, Rupert <rupertmccal...(a)yahoo.com> wrote:
>>
>>>> On Jun 16, 3:20 am, Arturo Magidin <magi...(a)member.ams.org> wrote:
>>
>>>>> On Jun 15, 2:11 am, Rupert <rupertmccal...(a)yahoo.com> wrote:
>>
>>>>> [.Construction snipped. at some point it requires arguing that |F[x]|=|
>>>>> F[x,y]|=|F|aleph_0 for field F .]
>>
>>>>>>> Now, the isomorphism between additive structures can be given
>>>>>>> constructively/explicitly, without having to invoke the Axiom of
>>>>>>> Choice. Can the same be done with the isomorphism of the
>>>>>>> multiplicative structures? I'm not sure if I'm invoking AC above; I
>>>>>>> may be in arguing some of the cardinalities or the inequalities
>>>>>>> between cardinalities.
>>
>>>>>> I think you might be using the fact that |X|^2=|X|, which is
>>>>>> equivalent to the Axiom of Choice.
>>
>>>>> Of course! I've run into that one before, and it always seems to
>>>>> escape me that I need Axiom of Choice in general.
>>
>>>>> For arbitrary infinite field F, the argument that |F[x]|=|F| that is
>>>>> the "obvious" one is to note that F[x] can be put in bijection with
>>>>> the disjoint union of F, F^2, F^3, F^4,..., F^n, etc. This is a
>>>>> countable union of sets of *the same cardinality as F* (here is where
>>>>> AC is used, I guess), so has cardinality |F|.
>>
>>>>>>> If not in general, can it be done for some specific fields, say F=Q,
>>>>>>> where we have explicit well-orderings of Q; or C, where the set of
>>>>>>> monic irreducibles of F[x] is easy?
>>
>>>>>> Yes. The equalities you need can be proven without AC in these cases,
>>>>>> I believe.
>>
>>>>> For Q, the problem above does not arise. For C, it may be difficult to
>>>>> argue that the set of monic irreducibles in C[x,y] is at most
>>>>> denumerable...
>>
>>>> I thought that you wanted to prove that the set of monic irreducibles
>>>> in C[x,y] had the same cardinality as C. I would think this should be
>>>> all right. We can prove that |C|^2=|C| and |C|*aleph_0=|C| without
>>>> using AC.
>>
>>> If we can prove |C^2| = |C| without AC, then I'm indeed fine.
>>> Iterating this we can show |C[x]|=|C|, and from there that |C[x,y]|=|
>>> C[x][y]|=|C|; then showing |C| <= # monic irreducibles in C[x.y] <= |
>>> C[x,y]| is trivial, so that would do it.
>
>> Well |C|=2^(aleph_0) so |C|
>> ^2=(2^(aleph_0))^2=(2^(aleph_0*2))=2^aleph_0=|C|.
>
> I'm wary of just invoking cardinal arithmetic because things like |
> X^2| = |X| do not hold in general without AC. Does the fact that the
> power set of a cardinal is bijectable with a cardinal require AC, for
> example?
>
> But we could just go from |R^2| = |R| to |R^n| = |R| and |C|=|R^2| to |
> C^2| = |(R^2)x(R^2)| = |R^4| = |R| = |R^2|=|C|, and from there to |C^n|
> =|C|. This does not seem to me to require AC, or even comparability to
> cardinals and cardinal arithmetic.
>
> --
> Arturo Magidin

You do want to be a bit careful here: for example it is consistent with
ZF that the power set of \omega is not well-orderable, and so not in
one-one correspondence with any ordinal. (Of course as Rupert has
pointed out if you define "cardinal" in other ways than "initial
ordinal" then you can change the answer to your question.) More
importantly, without AC a countable union of sets of infinite
cardinality \kappa need not have cardinality \kappa, so I don't think
the argument you sketch in your last paragraph is quite enough to get
the result you want. (Showing each C^n is in one-one correspondence
with C doesn't by itself show that the union of all C^n is.) But it's
close: without assuming AC it is still easy to construct a bijection
between \omega and \omega x \omega, and so between P(\omega) and the
product of \omega copies of P(\omega). Then (with the aid of the Cantor
Schroeder Bernstein Theorem, which does not require AC) it is clear that
for every \alpha between one and \omega inclusive, there is a bijection
between P(\omega) and the product of \alpha copies of P(\omega). It's
also easy to get a bijection between R and P(\omega), and so between C
and the product of \omega copies of C (without AC). As the set S of
monic irreducible elements of C[x,y] is injectable into that product,
the CSB Theorem again implies that S is in one-one correspondence with
C. (Whew!) So your main result can be proved for C as well as Q
without invoking AC.

Hope that helps.

Robert E. Beaudoin

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