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An Algebraically Closed Field, Its Multiplicative Group, and Its Isomorphic Subgroup

Date: 02/05/2012 at 13:45:28
From: Hector.
Subject: infinite subgroups of the multiplicative group of a field.

Let K be an algebraically closed field.

Suppose that its multiplicative group K* has a subgroup H which is
isomorphic to K*.

Is H = K*?

Date: 02/10/2012 at 15:14:26
From: Doctor Vogler
Subject: Re: infinite subgroups of the multiplicative group of a field.

Hi Hector,

Thanks for writing to Dr. Math.

The short answer to your question is: No, not necessarily. And
unfortunately, I don't think that there is an easy way to prove this.

Now, if the isomorphism between H and K* is required to be continuous,
then the answer is probably yes. The trouble is that algebraically closed
fields are very big, so handling them as abstract groups is very messy. So
we probably can't construct a counterexample without the Axiom of Choice,
for example.

Allow me to illustrate what I mean. The simplest algebraically closed
field is of course the complex numbers K = C. It is useful that we can
describe the multiplicative group K* = C* rather well. In fact, 

   C* = R x R/Z
That is, the multiplicative group of the (nonzero) complex number is
isomorphic to the direct product of the additive group of real numbers
with the additive group of real numbers mod integers. (That is, the
quotient of the group of real numbers with the subgroup of integers.) The
isomorphism takes a complex number z to the real number |z| (the absolute
value, or norm, or modulus, of z) and the real number arg(z)/2*pi (the
imaginary part of the natural logarithm of z, divided by 2*pi).

As an illustration, let's consider automorphisms of C* (isomorphisms of C*
and C*). Recalling that R is the additive group of real numbers, we know
that multiplication by any nonzero real number r is an automorphism of R.
So if we multiply the first component of R x R/Z by a nonzero real number
r, and leave unchanged the second component (the R/Z part), then what does
that do to a complex number? Well, the argument stays the same, but the
magnitude is raised to the power r. So that means that the isomorphism is

   z --> z*|z|^(r - 1)

And its inverse is

   z --> z*|z|^(1/r - 1)

Now, I'm going to describe an isomorphism of R with a proper subgroup of
R, which you can extend to an isomorphism of R x R/Z with a proper
subgroup of the same thing by leaving the R/Z component unchanged. I claim
that this provides an isomorphism of C* with a proper subgroup H of C*.

Here's where the Axiom of Choice comes in, and where things start to get a
little messy. The group R is a vector space (of infinite dimension) over
the field of rational numbers Q. That means that there is a set (an
infinite, uncountable set) of basis elements for R over Q. Let's call this
set B. And then every element of R is a (rational) linear combination of
finitely many of those basis elements, in a unique way.

Well, since the set B is infinite, there is a map s: B -> B that is
injective but not surjective. For example, you start by making a map from
the natural numbers to B (i.e., choose a "first" element of B, and then a
second element, a third, and so on to infinity). Since B is uncountable,
this cannot cover all elements of B, but that's okay. Now define s to take
any element that was numbered to the *next* element that was numbered, and
to take any element that was not numbered to itself. Then s maps nothing
to the first element, but everything else gets mapped somewhere, so no two
things get mapped to the same one (and nothing else besides the first
element is missed).

Now since every real number is a unique rational linear combination of
elements of B, we can extend s from a map on B to a group homomorphism of
R by taking any element of R, writing it as a rational linear combination
of elements of B, and then applying s to each of those elements of B. And
this group homomorphism s: R -> R is the thing we needed. So we're done!

If that last part was confusing, I can provide another example that might
be more illustrative. The additive group of polynomials in x with rational
coefficients is an infinite-dimensional vector space over the rational
numbers Q. Unlike the real numbers, the dimension (or size of the basis
set B) is countable, making it much easier to understand. The natural
basis is

   B = {1, x, x^2, x^3, x^4, x^5, ...}

This also gives a natural ordering of the basis, so we might as well take
the "first" and "second" elements, and so on, as I listed them. You should
convince yourself that every polynomial with rational coefficients is a
rational linear combination of the elements of the basis B; and that B has
to contain infinitely many elements.

Now, this set s takes an element of B to its "next" element. Given our
ordering of B above, that means that

   s(x^i) = x^(i + 1)

Then we extend s to polynomials like so:

   s(3 + 5/2*x + 7*x^2) = 3*s(1) + 5/2*s(x) + 7*s(x^2)
                        = 3*x    + 5/2*x^2  + 7*x^3.

In fact, you can check that the map s, in this case, is just
multiplication by the polynomial x. And it shouldn't be too hard to
realize that s is a group homomorphism that is injective but not
surjective, which means that s is an isomorphism between the additive
group of polynomials in x with rational coefficients and the subgroup of
polynomials with a constant term of zero.

We've done something similar with the real numbers R, only the basis is
uncountable and can't be written out explicitly.

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further

- Doctor Vogler, The Math Forum 

Date: 02/14/2012 at 08:16:11
From: Hector.
Subject: Thank you (infinite subgroups of the multiplicative group of a field.)

Thanks a lot, Dr. Math. You have been of great help.
Associated Topics:
College Modern Algebra

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