Re: The Algebraic Set

On Jul 31, 2005 4:00 AM, William Elliot wrote:
> On Sun, 31 Jul 2005, Narcoleptic Insomniac wrote:
> > On Jul 31, 2005 1:47 AM, William Elliot wrote:
> > > On Sun, 31 Jul 2005, Narcoleptic Insomniac wrote:
> > >
> > > > A number z is said to be algebraic if there are
> > > > integers c_0,c_1,...,c_n, not all zero, such that
> > > > c_0 z^n + c_1 z^(n-1) +...+ c_(n-1) z + c_n = 0.
> > > >
> > > > Given that any n'th degree polynomial has n
> > > > (not necessarily distinct) roots, prove that the
> > > > set of algebraic numbers is countably infinite.
> > >
> > > For all n in N, P_n = { p in Z[x] | deg p = n }
> > > is countable.
> > > \/_n P_n is countable, ie there's only countably
> > > many polynomials.
> > >
> > > Now each polynomial has a finite number of roots
> > > but let's increase
> > > this to each polynomials has countably many
> > > roots. Thus over all an
> > > upper bound for the number of roots is countable
> > > times countable = countable.
> >
> > Yes, I definitely should have said something about
> > the cardinality of A_n (or similarly P_n).
> >
> > > > PROOF. Let A_n be the set of all distinct roots of
> > > > all the n'th degree polynomials with integral
> > > > coefficients. Then clearly {\/A_n : n =
> > > > positive integers} is the set of algebraic
> > > > numbers.
> > >
> > > No, \/{ A_n : n is positive integer }
> >
> > Oppps, I did mean the union of A_n over the
> > positive integers though.
> >
> > > You don't need to count distinct roots, only get an
> > > upper bound for the
> > > number of roots that's not too large.
> >
> > This is why you chose P_n instead of A_n correct,
> > since the A_n I used
> > follows directly and naturally from P_n?
> >
> No, A_n is more complex than P_n. However |A_n| <=
> |P_n| , and that's all
> that's needed. I chose P_n, as it's easier to
> 'count' and provides for a
> sufficiently 'small' upper bound.

Maybe I misunderstood something, why is the cardinality of A_n less than or equal to P_n?

If P_n is the set of n'th degree polynomials (with integral coefficients) and A_n is the set of roots of all n'th degree polynomials, wouldn't the cardinality of A_n be larger than or equal to the cardinality of P_n?

I mean, for each element of P_n (each polynomial of degree n) there are between 1 and n elements of A_n associated with that element of P_n. Granted both P_n and A_n become countably infinite as n --> oo, but to each element of P_n there is one or more elements of A_n. If not, I think I must have misunderstood the definition of P_n.

> We actually need to finish up with the observation
> countable <= #distinct roots <= |Z[x]| <= countable
> to conclude #distinct roots = #algebraics = countable.
>
> Thus the easy and necessary task of showing finite /=
> #distinct roots, ie countable <= #distinct roots.
.

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