Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $n$ a positive number, and let $A_n$ be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree $n$. Using the fact that every polynomial has a finite number of roots, show that $A_n$ is countable.

Hint: For each positive number $m$, consider polynomials

$\sum_{i=0}^n a_i x^i$ that satisfy

$\sum | a_i | \le m$.

I'm having difficulty grasping the concepts and method to write the proof. Can someone please explain in simple terms?

Thank you.

share|improve this question
You have a fairly low acceptance rate; it's been nine days since you asked this question and you've gotten two good answers. You may want to think about accepting one of them. –  Rick Decker Oct 11 '12 at 1:05
Sorry, don't know fully how this website works yet. –  Alti Oct 12 '12 at 1:40

2 Answers 2

up vote 1 down vote accepted

Hint: Here is a hint along the same lines as in the post, but with a slight twist that makes things easier.

For any polynomial $P(x)$ with integer coefficients, let $C(P)$, the complexity of $P$, be the sum of the degree of $P$ and the sum of the absolute values of the coefficients of $P$. Instead of sum of the absolute values of the coefficients, one can just use the maximum absolute value of the coefficients, but we definitely want the degree as a component of the complexity.

For any $k$, there are only finitely many polynomials of complexity $k$, and these produce only finitely many algebraic numbers.

share|improve this answer
Thank you, but why is it countable and not finite? That's what I'm confused about. –  Alti Oct 1 '12 at 4:46
@Alti: Technically, finite sets are countable. But the number of algebraic numbers is actually countably infinite. For note that any natural number $n$ is algebraic, since it is a root of the polynomial equation $x-n=0$. So there certainly are infinitely many algebraic numbers. –  André Nicolas Oct 1 '12 at 4:50

There are countably many polynomials with rational coefficients and each of them has finite number of roots, so there has to countably many roots of such polynomials.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.