- #1

Mr Davis 97

- 1,462

- 44

## 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?

Last edited: