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From: Maury Barbato on 30 Jan 2010 03:00
Hello,
it's a well known fact that if the complex numbers
a_1, ..., a_n are algebraic independent over Q,
they are also algebraic independent over the field
of all algebraic numbers.
Is there some simple proof of this fact?
I haven't ever studied the theory of fields, so ...

Thank you very much for your help.
My Best Regards,
Maury Barbato
From: Gerry on 30 Jan 2010 17:59
On Jan 31, 5:00 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote:
> Hello,
> it's a well known fact that if the complex numbers
> a_1, ..., a_n are algebraic independent over Q,
> they are also algebraic independent over the field
> of all algebraic numbers.
> Is there some simple proof of this fact?
> I haven't ever studied the theory of fields, so ...

Suppose there's a polynomial P(x_1, ..., x_n) with algebraic coeffs
such that P(a_1, ..., a_n) = 0. Then you can multiply P by all of
its "conjugates" to get a polynomial Q(x_1, ..., x_n) with rational
coefficients and, since P is a factor of Q, you'll have
Q(a_1, ..., a_n) = 0, contradicting algebraic independence over Q.

It's a bit tricky to make "conjugates" more precise to someone
who hasn't studied fields, but here's a small example of what
I mean; if P(x_1, x_2) = (sqrt2) x_1 + (sqrt3) x_2, then its
conjugates are (sqrt2) x_1 - (sqrt3) x_2, -(sqrt2) x_1 + (sqrt3) x_2,
-(sqrt2) x_1 - (sqrt3) x_2, and (always) P itself. Multiply these
four polynomials together and you'll see all the square roots
go away.
--
GM
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