From: Robin Chapman
Subject: Re: Eisenstein's Criterion a kind of converse
Date: Thu, 18 Nov 1999 14:57:27 GMT
Newsgroups: sci.math
Keywords: Some variant of Eisenstein's Criterion always sufficient? (no)
In article <942922983.777958@ns.idl.com.au>,
"Dilby" wrote:
> I'm assuming that the converse of Eisenstein's Criterion for irreducible
> polynomials is false.... but wondering whether it is possible to prove the
> converse for an irreducible polynomial f(x) with an appropriate
> substitution x= y-a where a is based on some condition on the co-efficeints
> of f(x).
>
Let a be a root of f. Then if f is Eisenstein at p, then p is
totally ramified in the extension Q(a) of Q. There are examples
of number fields no prime in which is totally ramified. One example
is Q(sqrt(2), sqrt(5)). If we take the minimal polynomial of a primitive
element of this field, say a = sqrt(2) + sqrt(5) yielding
f(x) = x^4 - 14x^2 + 9 (I think) then no massaging of f(x) will give
ann Eisenstein polynomial.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'"
Greg Egan, _Teranesia_
Sent via Deja.com http://www.deja.com/
Before you buy.
==============================================================================
From: Kurt Foster
Subject: Re: Eisenstein's Criterion a kind of converse
Date: Thu, 18 Nov 1999 16:38:10 GMT
Newsgroups: sci.math
In <942922983.777958@ns.idl.com.au>, Dilby said:
. I'm assuming that the converse of Eisenstein's Criterion for irreducible
. polynomials is false.... but wondering whether it is possible to prove
. the converse for an irreducible polynomial f(x) with an appropriate
. substitution x= y-a where a is based on some condition on the
. co-efficeints of f(x).
Good question! The answer is in general "no". As a specific example, I
give f(x) = x^4 - x^2 + 1, the irreducible polynomial satisfied by the
primitive 12th roots of unity. This f(x) is not Eisenstein with respect
to any prime p, and you can't get an Eisenstein polynomial with respect to
any prime p FROM it, by any "reasonable" transformation of the roots.
If you start with a *quadratic* polynomial, though, there are Eisenstein
polynomials to be found via "reasonable" transformations.
As an elementary exercise, you might like to show that, if your chosen
f(x) is Eisenstein with respect to the prime p, but isn't monic [the
coefficient a_n of the highest-degree term x^n is not 1], then you can
obtain a monic polynomial which is also Eieenstein with respect to the
prime p, using the transformation y = a_n * x, or x = y/a_n.
I don't know of a really good "elementary" explanation of why the answer
is in general "no", but here's an "at the speed of light" thumbnail
sketch: Let f(x) be Eisenstein with respect to the prime p, and r a root
of the equation f(x) = 0. If you perform the transformation x = a + b*y,
where a and b are rational and b is non-zero, then f(x) = g(y) where g is
a polynomial with rational coefficients. The root s of g(y) = 0
corresponding to r satisfies a + b*s = r. Therefore, the field Q(s)
obtained by adjoining s to the rational field Q, is the SAME field as
Q(r), obtained by adjoining r to Q; Q(r) = Q(s) = K, say. The "integral
elements" of K (those which satisfy a polynomial with integer coefficients
*and* have leading coefficient 1) form a ring R, called the ring of
algebraic integers of K. There is a well defined notion of factorization
of "ideals" of R into "prime ideals" of R (never mind the details for
now!).
It turns out that, if there is some element of K (and therefore of R, as
the above elementary exercise shows) which satisfies a polynomial which
is Eisenstein with respect to the prime p, then the ideal pR factors into
a very special way - it is the n-th power of a prime ideal; pR = P^n.
But this is a something that depends on the field K and its ring of
integers R, and not upon which particular polynomial is used to describe
them. And for some fields K, there simply is NO prime p for which pR
factors in this special way. Consequently, for such K, it is absolutely
guaranteed that, whatever polynomial f(x) you take, with a root r for
which Q(r) = K, f(x) will NOT be an Eisenstein polynomial. [That is how I
came up with my example - I know that for the field K = Q(z), z a
primitive 12th root of unity, R the ring of integers in K, there is NO
prime p for which pR = P^4.]