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From: Arturo Magidin on 14 Jun 2010 15:51
The origin of this question comes from a colleague, who asked "Can you
have two nonisomorphic rings R and S, such that the additive group of
R is isomorphic to the additive group of S, and the multiplicative
semigroup of R is isomorphic to the multiplicative semigroup of S"?

My answer, after some thought, was "yes", with the following example.
Let F be a field, and let R=F[x], S=F[x,y]. The isomorphism of (R,+)
and (S,+) can be done explicitly, as they are vector spaces of F of
dimension aleph_0, and one can give an explicit bijection by ordering
the monomials in R by degree, and the monomials in S by total degree
and lexicographically among those of the same degree.

To prove the isomorphism between (R,*) and (S,*), I argued as follows:
the cardinality of the set of monic irreducible polynomials in F[x] is
|F|*aleph_0: for infinite F, there are at least |F| monic irreducibles
(the linear polynomials), and since |F[x]|=|F|, there are at most |F|
of them. For finite F, there are at least aleph_0 monic irreducibles
(at least one for each positive integer n, given by the finite
extensions and the primitive element theorem), and |F[x]|=aleph_0, so
you get aleph_0 again. Since |F[x,y]|=|F|*aleph_0 as well, this gives
that the cardinality of the set of irreducibles in S is the same. If M
is the free commutative monoid on a set with |F|*aleph_0 elements,
then (R,*) and (S,*) are both isomorphic to F x M, since they are both
UFDs with isomorphic (in fact, identical) group of units.

That R and S are not isomorphic follows from any number of arguments
(R is a PID, S is not; the Krull dimension of S is one more than that
of R, etc).

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.

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?

(I think that replacing F with any UFD will also give counterexamples
to the original question, but then we may need choice to select
representatives from the equivalence classes of irreducible elements,
so let's stick to fields where there is an obvious choice of
representative).

--
Arturo Magidin
From: Rupert on 15 Jun 2010 03:11
On Jun 15, 5:51 am, Arturo Magidin <magi...(a)member.ams.org> wrote:
> The origin of this question comes from a colleague, who asked "Can you
> have two nonisomorphic rings R and S, such that the additive group of
> R is isomorphic to the additive group of S, and the multiplicative
> semigroup of R is isomorphic to the multiplicative semigroup of S"?
>
> My answer, after some thought, was "yes", with the following example.
> Let F be a field, and let R=F[x], S=F[x,y]. The isomorphism of (R,+)
> and (S,+) can be done explicitly, as they are vector spaces of F of
> dimension aleph_0, and one can give an explicit bijection by ordering
> the monomials in R by degree, and the monomials in S by total degree
> and lexicographically among those of the same degree.
>
> To prove the isomorphism between (R,*) and (S,*), I argued as follows:
> the cardinality of the set of monic irreducible polynomials in F[x] is
> |F|*aleph_0: for infinite F, there are at least |F| monic irreducibles
> (the linear polynomials), and since |F[x]|=|F|, there are at most |F|
> of them.
>
> For finite F, there are at least aleph_0 monic irreducibles
> (at least one for each positive integer n, given by the finite
> extensions and the primitive element theorem), and |F[x]|=aleph_0, so
> you get aleph_0 again. Since |F[x,y]|=|F|*aleph_0 as well, this gives
> that the cardinality of the set of irreducibles in S is the same. If M
> is the free commutative monoid on a set with |F|*aleph_0 elements,
> then (R,*) and (S,*) are both isomorphic to F x M, since they are both
> UFDs with isomorphic (in fact, identical) group of units.
>
> That R and S are not isomorphic follows from any number of arguments
> (R is a PID, S is not; the Krull dimension of S is one more than that
> of R, etc).
>
> 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.

> 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.

> (I think that replacing F with any UFD will also give counterexamples
> to the original question, but then we may need choice to select
> representatives from the equivalence classes of irreducible elements,
> so let's stick to fields where there is an obvious choice of
> representative).
>
> --
> Arturo Magidin

From: Arturo Magidin on 15 Jun 2010 13:20
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...

--
Arturo Magidin
From: Rupert on 15 Jun 2010 17:06
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...
>
> --
> Arturo Magidin

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.
From: Arturo Magidin on 15 Jun 2010 23:50
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
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