How Can We Prove the Countability of Algebraic Numbers of Fixed Degree?

[rbzima]

I posted this in the Homework/Coursework section, but I really don't consider it that at all because I'm working through this text on my own, and I'm a little stuck on this problem.

Fix n \in N, 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. (For each m \in N, consider the polynomials a_nx^n + a_n_-_1x^n^-^1 + ... + a_1x + a_0 that satisfy \left|a_n\right| + \left|a_n_-_1\right| + ... + \left|a_1\right| + \left|a_0\right| \leq m.)

By the way, this only deals with real roots. Complex roots are simply negligible.

So, I know a few things, but bringing the big picture together is really messing me up here. For example, I know that the sum of the absolute value of the coefficients for quadratic equations only has a certain number of solutions. So, whatever I elect m to be, there will always be a finite number of solutions. Also, the number of quadratics with coefficients is less than or equal to m: this is also finite. When we multiply this fact times the number of roots, we have the number of roots of a quadratic whose absolute value sums to some value less than or equal to m.

The big problem I have is trying to generalize this statement for all A_n. If anyone has any suggestions, this would be most helpful!

 

[mathman]

Your description for the quadratic essentially applies for any fixed m, for the specified n. There are only n+1 integer coeficients, so there are only a finite number of possibilites for the sum of abs. values bounded by m. Each polynomial has at most n roots - you can include the complex. Therefore for fixed m the total number of roots for all equations is finite.

You now have a countable sum of finite numbers when you let m -> inf.