From: rgep@emu.pmms.cam.ac.uk (Richard Pinch)
Newsgroups: sci.math.research
Subject: Re: Hasse principle for polynomials?
Date: 12 Sep 1994 11:07:43 GMT
Summary: No. (X^2-2)(X^2-17)(X^2-34)
Keywords: Hasse principle. Local-to-global

In article <940909111428755-MTARUBA*Joerg.Winkelmann@RUBA.RZ.ruhr-uni-bochum.de> Joerg.Winkelmann@RUBA.RZ.ruhr-uni-bochum.de writes:
>My question is the following:
>
>[...] If a polynomial equation in one variable is everywhere
>locally solvable, does this imply the existence of a global solution?
>
No.  Cassels, Local fields, Chapter IV.3bis gives the example

	(X^2 - 2)(X^2 - 17)(X^2 - 34)

which has a root everywhere locally but not in Q.

Richard Pinch;		Queens' College, Cambridge

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From: gerry@macadam.mpce.mq.edu.au (Gerry Myerson)
Newsgroups: sci.math.research
Subject: Re: Hasse principle for polynomials?
Date: 11 Sep 1994 19:23:32 -0500

In article
<940909111428755-MTARUBA*Joerg.Winkelmann@RUBA.RZ.ruhr-uni-bochum.de>,
Joerg.Winkelmann@RUBA.RZ.ruhr-uni-bochum.de wrote:
> 
> In other words: If a polynomial equation in one variable is everywhere
> locally solvable, does this imply the existence of a global solution?

No. The equation (X^2 - 13)(X^2 - 17)(X^2 - 221) has a solution in every 
p-adic completion of the rationals, but not in the rationals. 

There's also an example with degree 5, but you can prove this never 
happens (over the rationals) with any smaller degree. 

Gerry Myerson (gerry@mpce.mq.edu.au)
Centre for Number Theory Research (E7A) 
Macquarie University, NSW 2109, Australia

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From: rgep@pmms.cam.ac.uk (Richard Pinch)
Newsgroups: sci.math.research
Subject: Re: Hasse principle for polynomials?
Date: 13 Sep 1994 16:05:34 GMT

In article <9409120022.AA26787@macadam.mpce.mq.edu.au>, gerry@macadam.mpce.mq.edu.au (Gerry Myerson) writes:
|> In article
|> <940909111428755-MTARUBA*Joerg.Winkelmann@RUBA.RZ.ruhr-uni-bochum.de>,
|> Joerg.Winkelmann@RUBA.RZ.ruhr-uni-bochum.de wrote:
|> > 
|> > In other words: If a polynomial equation in one variable is everywhere
|> > locally solvable, does this imply the existence of a global solution?
|> 
|> No. [...]
|> There's also an example with degree 5, but you can prove this never 
|> happens (over the rationals) with any smaller degree. 

Both the examples posted were rather obviously composite.  It is not
hard to show that an irreducible polynomial cannot be a counter-example.

Proof: Suppose f is irreducible over Q with Galois group G necessarily
transitive on the roots.  The Artin symbol of f at p (action of the mod p
Frobenius) has a fixed point since f has a p-adic root.  So by the 
Chebotarev density theorem, every conjugacy class in G has a fixed point
in its permutation representation on the roots.  The average number of
fixed points in G is thus strictly greater than 1 (the identity).  So
by the Burnside formula, the number of orbits > 1, contradicting G being
transitive.  []

So a quintic counter-example has to be a product of a quadratic and a
cubic, say (X^2 - d)g(X).  Now we want g to have a root whenever f doesn't,
and the easiest way of ensuring that is to make d the discriminant of g
up to a square.  So try something like g = (X^2 - 257)(X^3 - 5X - 3).

Richard Pinch;		Queens' College, Cambridge