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.]