Proof that algebraic numbers are countable

[Mr Davis 97]

Homework Statement

Fix $n,m \in \mathbb{N}$. The set of polynomials of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$
satisfying $|a_n| + |a_{n-1}| + \cdots + |a_0| \le m$ is finite because there are only a finite
number of choices for each of the coefficients (given that they must be integers.)
If we let $A_{nm}$ be the set of all the roots of polynomials of this form, then because
each one of these polynomials has at most $n$ roots, the set $A_{nm}$ is finite. Thus
$A_n$, the set of algebraic numbers obtained as roots of any polynomial (with
integer coefficients) of degree $n$, can be written as a countable union of finite
sets: $\displaystyle A_n = \bigcup_{m=1}^{\infty}A_{nm}$. It follows that $A_n$ is countable. Since the countable union of countable sets is countable, $\displaystyle A = \bigcup_{n=1}^{\infty}A_{n}$, which is the set of all algebraic numbers, is countable.

Homework Equations

The Attempt at a Solution

I generally understand this proof, but I just have a few questions. In the first part of the proof, is it okay if $a_n = 0$? That is, do the roots contained in $A_{nm}$ necessarily have to come from an nth degree polynomial, or can they also come from polynomials of degree n or less? For example, is $A_{12}$ a subset of $A_{22}$?

If this is not the case, should we stipulate in the beginning that $a_n \ne 0$ so that the roots can actually be of an nth degree polynomial?

 

[andrewkirk]

Generally what is meant by saying that a polynomial has degree n is that the position of the highest-position, non-zero coefficient is n. Thus, by definition a n-th degree polynomial cannot have $a_n=0$. In order not to disrupt this, the degree -1 is awarded to the zero polynomial (although that doesn't affect this case, as only polynomials of positive degree are used in the definition of the algebraic numbers). So $A_{12}$ is generally not a subset of $A_{22}$.

 

[Mr Davis 97]

Generally what is meant by saying that a polynomial has degree n is that the position of the highest-position, non-zero coefficient is n. Thus, by definition a n-th degree polynomial cannot have $a_n=0$. In order not to disrupt this, the degree -1 is awarded to the zero polynomial (although that doesn't affect this case, as only polynomials of positive degree are used in the definition of the algebraic numbers). So $A_{12}$ is generally not a subset of $A_{22}$.

So is it implicit in the following statement that $a_n \ne 0$, even though $a_n=0$ could possibly satisfy the inequality stated?

The set of polynomials of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ satisfying $|a_n| + |a_{n-1}| + \cdots + |a_0| \le m$

 

[fresh_42]

This depends alone on whether you consider the $A_{n}$ as all roots of polynomials of degree exactly $n$ or of degree at most $n$. One is a graduation and the other one a filtration. The "exactly of degree $n$" makes probably more sense here to avoid multiple counts, which means $a_n \neq 0$.

 

[Mr Davis 97]

So to make the proof better, or at least more obvious, should I write at the beginning:

The set of polynomials of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ satisfying $|a_n| + |a_{n-1}| + \cdots + |a_0| \le m$, where $a_n \ne 0$ is finite because there are only a finite number of choices for each of the coefficients (given that they must be integers.)?

 

[fresh_42]

I would do so, because this way you count all roots of polynomials of degree $n$. To be exact you get an upper bound, as there are still double counts as for $(x-a)^n$, but each one has at most $n$ different roots. If $a_n=0$ would be allowed, then you count all roots of polynomials of degree lesser than $n$ again and again.

 

[andrewkirk]

I noticed that the proof doesn't state anywhere directly that the coefficients must be integers. Rather, it makes a comment in parentheses that implies the statement has already been made, which it hasn't.

Another way to write the first piece, taking care of that issue and maybe shortening the expression a bit, be:

For positive integers $m,n$, let $P_{nm}$ be the set of all polynomials of degree $n$ with integral coefficients, such that the sum of the absolute values of each polynomial's coefficients is less than $m$, and let $A_{nm}$ be the set of all roots of polynomials in $A_{nm}$.