Closure of the rational polynomials
Moderators: gmalivuk, Moderators General, Prelates
Closure of the rational polynomials
In my algebra class, the professor mentioned that the closure of Q[x] isn't very well understood. My question then is, what is known about it? Does anyone know of any good references, preferably in the online world? Can anyone give a more intuitive description of which real numbers are contained in it?
Re: Closure of the rational polynomials
Closure in what sense?
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
Re: Closure of the rational polynomials
More to the point, how can real numbers be contained in the closure of a set of functions, whatever the sense?
Unless you mean the closure of the set of roots of rational polynomials, in which case I'm fairly sure it's all the reals.
Unless you mean the closure of the set of roots of rational polynomials, in which case I'm fairly sure it's all the reals.
This is a placeholder until I think of something more creative to put here.
 NathanielJ
 Posts: 882
 Joined: Sun Jan 13, 2008 9:04 pm UTC
Re: Closure of the rational polynomials
Robin S wrote:More to the point, how can real numbers be contained in the closure of a set of functions, whatever the sense?
Unless you mean the closure of the set of roots of rational polynomials, in which case I'm fairly sure it's all the reals.
If that's what he meant, then it's definitely all the reals since that set is exactly the algebraic numbers, which contain the rational numbers, which are dense in the real numbers.
I assume that when he said "which real numbers..." he meant "which functions...". Unfortunately, I don't know the answer.
Re: Closure of the rational polynomials
In the topology of pointwise convergence, all continuous functions and quite a few discontinuous ones (possibly those of Baire class 1) are in the closure of the polynomials. Under uniform convergence, I'm not sure if there are even any nonpolynomial functions in the closure.
This is a placeholder until I think of something more creative to put here.
Re: Closure of the rational polynomials
I was talking about the smallest field in which every finite polynomial over the rationals splits completely. "Algebraic closure," I believe. Sorry about the confusion.
I'm pretty sure that it's not all the reals though. A counting argument shows that there are fewer elements in this field (that's the term that we used in class) than there are in the reals. As an example, pi isn't in the field.
So with more clarity, does anyone have any good resources/have any good ways of thinking about the algebraic closure of Q?
I'm pretty sure that it's not all the reals though. A counting argument shows that there are fewer elements in this field (that's the term that we used in class) than there are in the reals. As an example, pi isn't in the field.
So with more clarity, does anyone have any good resources/have any good ways of thinking about the algebraic closure of Q?

 Posts: 1459
 Joined: Fri Apr 20, 2007 3:27 pm UTC
 Location: The Tower of Flints. (Also known as: England.)
Re: Closure of the rational polynomials
The algebraic closure of Q is the field of algebraic numbers by definition.
It's not all the reals; in fact, as you mention, it's countable so in a sense it contains hardly any of the reals. It's also worth noting that it contains nonreal numbers as well (i, for example). I suppose it's not well understood in that its complement in R, the transcendental numbers, are hard to work with. In general it's very hard to show that a given number is transcendental, i.e., not algebraic.
It's not all the reals; in fact, as you mention, it's countable so in a sense it contains hardly any of the reals. It's also worth noting that it contains nonreal numbers as well (i, for example). I suppose it's not well understood in that its complement in R, the transcendental numbers, are hard to work with. In general it's very hard to show that a given number is transcendental, i.e., not algebraic.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
dubsola
dubsola
Re: Closure of the rational polynomials
Ended wrote:In general it's very hard to show that a given number is transcendental, i.e., not algebraic.
For instance, pi and e are both known to be transcendental, and it is also known that at least one of (pi + e) and (pi * e) is transcendental, but, last I checked, it isn't known if either of those is algebraic. (It's known that at least one of them is transcendental because x^{2} + (pi + e) x + (pi * e) = (x + pi)(x + e). Since the roots of that polynomial are transcendental, at least one of the coefficients must be transcendental as well.)
LOWA
Who is online
Users browsing this forum: No registered users and 12 guests