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Re: FLT Discussion: Simplifying
Posted:
Jan 19, 2001 5:11 AM
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jstevh@my-deja.com wrote:
>
>
> Hey, I've already seen the post where someone says you guys proved that
> AB = 0, when A = 0, or B = 0 by using x^2 + y^2 = 0.
>
> I concede that one could debate the question of whether or not there
> might exist some objects in an infinite ring that could be nonzero and
> multiply times each other to give 0. After all, it's trivally done in
> a finite ring.
>
> So, to me it's become a moot point. Maybe one of you will come back
> with some numbers like quarternions or something where that's not true,
> but hey, I don't mind that as I'd simply find that interesting, and
> worth thinking about whether or not I could work around that.
>
> So, bring it on.
>
> James Harris
Utterly trivial. Consider the ring Z[x], of polynomials in the
indeterminate x, with integral coefficients. Next, consider the
polynomial P(x) = x^2 - 3x + 2 = (x-2)(x-1). The ring I'll direct you to
is the quotient
R = Z[x] / < P(x) >
Where the angular brackets <> represent the ideal generated by the stuff
in parentheses. Here, that consists of Z[x]*P(x), the set of all
multiples of P(x). The quotient has as elements objects of the form F(x)
+ <P>, where F(x) is an element of Z[x]. Addition is done as follows:
(F + <P>) + (G + <P>) = (F+G) + <P>
and multiplication (using asterisk * to denote standard multiplication)
(F + <P>) * (G + <P>) = (F*G) + <P>
Since <P> is an additive subgroup of Z[x], addition of classes is
well-defined; since it is an ideal (and so closed under multiplication
by any element of Z[x]), multiplication of classes is also well-defined.
Associativity, commutativity, and the distributive property follow from
the same properties as they exist in the ring Z[x]. Therefore, R is a
ring.
R contains the integers, by the way. That's easy enough to prove.
Therefore the ring R is infinite.
Note, however, that in R, we have (x-2) + <P> != 0 , (x-1) + <P> != 0,
since neither (x-1) nor (x-2) is a multiple of P. However, ((x-2) +
<P>)((x-1) + <P>) = 0.
In a not too-uncommon shorthand, using an element of Z[x] as a name for
its equivalence class in R, under the canonical homomorphism, one would
write:
x-2 != 0 and x-1 != 0 in R
(x-2)(x-1) = 0 in R.
The moral of the story is that if you extend a ring (even a very nice
one, such as Z) by adding in roots of polynomials that already had roots
in the original ring, you will find yourself with a non-domain.
Dale.
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