An Algebraically Closed Field, Its Multiplicative Group, and Its Isomorphic SubgroupDate: 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 suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ 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. |
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