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Closure of the rational polynomials

Posted: Sun May 18, 2008 9:26 am UTC
by merc
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

Posted: Sun May 18, 2008 10:17 am UTC
by antonfire
Closure in what sense?

Re: Closure of the rational polynomials

Posted: Sun May 18, 2008 3:10 pm UTC
by Robin S
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.

Re: Closure of the rational polynomials

Posted: Sun May 18, 2008 5:12 pm UTC
by NathanielJ
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

Posted: Sun May 18, 2008 5:41 pm UTC
by Robin S
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 non-polynomial functions in the closure.

Re: Closure of the rational polynomials

Posted: Mon May 19, 2008 5:40 am UTC
by merc
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?

Re: Closure of the rational polynomials

Posted: Mon May 19, 2008 9:48 am UTC
by Ended
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 non-real 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.

Re: Closure of the rational polynomials

Posted: Mon May 19, 2008 11:07 am UTC
by Nimz
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 x2 + (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.)