Re: JSH "problem"

In article <gkief802799@xxxxxxxxxxxxxxxxx>,
David Bernier <david250@xxxxxxxxxxxx> wrote:
*** T. Winter wrote:
In article <Xns9B8FB298F6530tim111one@xxxxxxxxxxxxxxx> Tim Peters <tim.one@xxxxxxxxxxx> writes:
...
> So what is the object ring? You have to ask James ;-)

No need to ask James. What it comes down to is that given a quadratic with
two, non-integral roots, one of the roots is in the object ring and the other
is not. This must be the case for every quadratic. The problem, obviously,
but James does not see it, is that such a ring does not exist. That is, it
is not possible to make a consistent choice for every pair of roots of the
quadratics.

Let's see: x^2 + px + q = 0 (field = Q, p, q in Z).

The roots of such a quadratic would be integral. By "non-integral",
*** means that the quadratic must be a constant multiple of a
non-monic, primitive, irreducible quadratic.

Then x = (-p +/- sqrt(p^2 - 4q))/2 .

Assume a = (-p + sqrt(p^2 - 4q))/2 [whichever square root ...]

then a* ( - p - sqrt(p^2 - 4q))/2 = (p^2 - (p^2 - 4q))/4 = q.

So (-p - sqrt(p^2 - 4q))/2 = q/a .
Both roots are algebraic integers,

Indeed: and therefore they are integral. *** said "non-integral", not
"non-integer".

and the extension
field Q(a) contains q/a, which is an algebraic integer.
So the ring of integers of Q(a) contains the other root,
namely q/a .

Now consider the case of tx^2 + px + q, with t,p,q in Z, gcd(t,p,q)=1,
and t=/= 1,-1.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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