Re: numeri algebrici


Arturo Magidin wrote:
In article <1146854962.594736.43280@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Randy Poe <poespam-trap@xxxxxxxxx> wrote:

Arturo Magidin wrote:
In article <1146852193.360406.222250@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
erg <mario.principato@xxxxxxxxxxxxx> wrote:
qualcuno mi sa dare una dimostrazione del seguente fatto :
il campo dei numeri algebrici =E8 algebricamente chiuso.

vorrei sapere come dimostrarlo.

Don't know if you know enough English, but:

If a is algebraic over the field of algebraic numbers, then there
exists a polynomial with algebraic number coefficients such that a
is a root. Let b_1,...,b_n be the coefficients. Then K=Q[b_1,...,b_n]
is finite over Q, and by assumption K(a) is finite over K, so K(a)
is finite over Q, hence a is algebraic over Q, hence a is in the
field of algebraic numbers.

I don't know enough about the notation to follow this.

OP was asking for a proof that the algebraic numbers
are algebraically closed, which I assume means closed
under addition and multiplication (question 1)?

It's possible I mistranslated what he wrote; but "algebraically
closed" means that every nonconstant polynomial has a root;

He definitely said "algebraically closed". So I'm mistranslating
the ENGLISH.

And after I posted I realized he couldn't have been
asking what I thought, since he said "the field of
algebraic numbers".

equivalently, that every root of a nonconstant polynomial with
coefficients in the field is already in the field.

OK. That's another result I've heard quoted but would
be interested in the proof, so I'm still interested in
the translation of your proof.

If he meant "is closed under the algebraic operations", which is what
you are asking here, then of course I did not address that. That would
require you to show that a+b is algebraic when a and b are algebraic,
and that ab is algebraic when a and b are.

What
does Q[b_1,...,b_n] mean and how does this proof establish
closure (question 2)?

Q[b_1,...,b_n] is the smallest subring of some ring containing Q which
contains b1, ..., bn. The ring of all polynomial expressions in
b_1,...,b_n with coefficients in Q. It is equal to the smallest
subfield of C that contains Q and contains b_1,..,b_n.

Of course, what I wrote does not address closure under addition and
under multiplication: it addresses algebraic closure. That any number
which is algebraic over the algebraic numbers is itself an algebraic
number.

And it seems to me that I've seen
a proof that a+b and a*b are algebraic by explicit construction
of polynomials that have a+b and a*b as roots. Could
you give a hint about that construction (question 3) and
is the above actually equivalent to that (question 4)?

No, the above is not equivalent to that; it addresses a completely
different question.

OK. Sorry.


As to the constructions, it's been outlined before. Here is an
explicit construction from Niven, Zuckerman, Montgomery, "Introduction
to the Theory of Numbers" fifth Edition, Theorem 9.12

Suppose a satisfies

a^m + r_{m-1} a^{m-1} + ... + r_1 a + r_0 = 0

and b satisfies

b^n + s_{n-1} b^{n-1} + ... + s_1 b + s_0 = 0.

Let k = n*m. Define the complex number c1,...,ck to be the numbers
a^i*b^j, 0<=i<m, 0<=j<n, in some order. For each c, you have either

ac = a^{u+1}b^v = some c_w if u+1<m or
ac = a^{u+1}b^v = -(r_{m-1}a^{m-1}+...+r_1 a + r_0)b^t if u=m-1.

In either case, we have rational constants h_{j,1},...,h_{j,k} such
that

ac_j = h_{j,1}c_1 + ... + h_{j,k}c_k.

There are also rational constants w_{j,1},..., w_{j,k} such that

bc_j = w_{j,1}c_1 + ... + w_{j,k}c_k.

So

(a+b)c_j = (h_{j,1}+w_{j,1})c_1 + ... + (h_{j,k}+w_{j,k})c_k

These latter equations can be thought of as a system of homogeneous
linear equations in c_1,...,c_k with coefficients in Q; the c_i are
not all zero, and the characteristic polynomial will have a+b as an
eigenvalue, hence is a root of the characteristic polynomial.

Ah. This is one reason why I asked. I saw someone give
a linear algebra-based proof and I was trying to
remember the more elementary proof, unsuccessfully.
I didn't realize they're the same proof.

We have

ab*c_j = a(w_{j,1}c_1 + ... + w_{j,k}c_k)
= w_{j,1}(ac_1) + ... + w_{j,k}(ac_k)

so substituting into each ac_i we get again equations for ab*c_j,
leading to a matrix whose characteristic polynomial has ab as a root.

Thank you.

Long ago I attempted to learn number theory by
signing up for a course taught from the book by
Takashi Ono, but I found I didn't have the vocabulary
to understand the class discussions and eventually
dropped the course. Perhaps it's time for another look at
Ono to see if I can learn this on my own.

- Randy

.

Relevant Pages

  • Re: numeri algebrici
    ... il campo dei numeri algebrici =E8 algebricamente chiuso. ... closed" means that every nonconstant polynomial has a root; ... of polynomials that have a+b and a*b as roots. ... explicit construction from Niven, Zuckerman, Montgomery, "Introduction ...
    (sci.math)
  • Re: numeri algebrici
    ... il campo dei numeri algebrici =E8 algebricamente chiuso. ... Don't know if you know enough English, ... a proof that a+b and a*b are algebraic by explicit construction ... of polynomials that have a+b and a*b as roots. ...
    (sci.math)