From: rgep@can.pmms.cam.ac.uk (Richard Pinch)
Newsgroups: sci.math
Subject: Re: irreducible polynomials dense in rationals
Date: 27 Oct 1995 12:53:52 GMT
Summary: Proof by direct construction
Keywords: Eisenstein's criterion

In article <xlHHYR4.laurahelen@delphi.com> 
Laura Helen <laurahelen@delphi.com> writes:
>Suppose you want to show that irreducible polynomials over the rationals
>are dense in all polynomials over the rationals.
>[...]
>So, if you can show that a polynomial in Q_p [x] which is irreducible
>over Q_p is also irreducible over Q, you would have shown that
>irreducible polynomials are dense in rational polynomials.

Surely this is obvious?  If there are factors over Q, they are also
factors over Q_p?  Or did I miss something?

Here's another proof of the main result.  Given a rational polynomial
$f = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$, choose $p$ a large 
prime and write 
$$
g = (1+1/p^{n+1})x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0
$$
Then 
$$
p g(px) = (p^{n+1}+1) x^n + p^n a_{n-1} x^{n-1} + \cdots + p^2 a_1 x + p a_0
$$
If $p$ does not divide $a_0$ then this is irreducible by Eisenstein.  So
for $p$ sufficiently large, $g$ is irreducible and as close as desired to $f$.

Richard Pinch;	Queens' College, Cambridge