# Question about Rudin's PMA, Chapter 2 Exercise 2

I know that there are many questions related to this exercise that have been answered, and I have found a couple of different solutions to this exercise as well. But the solutions usually ignore the hint provided by Rudin, and even though some of them do deal with the hint, I am still not quite sure about what we should actually do with it. I wrote up my answer based on my understanding of those solutions used Rudin's hint. I want to know if this is correct. Thanks in advance for any clarification!

The exercise goes like this:

A complex number $$z$$ is said to be algebraic if there are integers $$a_0$$, $$\dots$$, $$a_n$$, not all zero, such that $$$$a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0.$$$$ Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $$N$$ there are only finitely many equations with $$$$n + |a_0| + |a_1| + \dots + |a_{n-1}| + |a_n| = N.$$$$

My attempt based on my understanding of those solutions used Rudin's hint:

Consider the set $$A$$ of all algebraic numbers, where for each $$z \in A$$ there are integers $$a_0$$, $$\dots$$, $$a_n$$, not all zero, such that $$$$a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0. \tag{1}$$$$ Now fix an $$N \in \mathbb{Z}^+$$, and consider the subset $$A_N$$ of $$A$$, where for each $$z \in A_N$$ there are integers $$a_0$$, $$\dots$$, $$a_n$$, not all zero, such that (1) holds, and that $$$$n + |a_0| + |a_1| + \dots + |a_{n-1}| + |a_n| = N. \tag{2}$$$$ Since, by the hint, there are only finitely many such equations as (2); and since, by the Fundamental Theorem of Algebra, for each $$(a_0, a_1, \dots, a_{n-1}, a_n)$$ satisfying (2) there are $$n$$ roots, and thus finitely many roots, for the polynomial $$$$a_0x^n + a_1x^{n-1} + \dots + a_{n-1}x + a_n = 0, \tag{3}$$$$ we have that the set $$A_N$$ is a finite union of finite sets and hence is finite. Since $$A = \bigcup_{N=2}^{\infty} A_N$$, we have $$A$$ is a countable union of finite sets and thus is countable, by the Corollary to Theorem 2.12.

Is my answer correct and rigorous? I really appreciate it!

• Why list equation (4) (mistakenly written with variable $x$) when it’s a duplicate of (2), which is in turn a duplicate of (1)? Can’t you make this more succinct? What you have is correct, otherwise, although it takes a second of thought to understand why every algebraic number is in some $A_N$. Jun 8, 2023 at 4:42

Let $$M = \{ (a_0,\ldots, a_n) \in \mathbb Z^{n+1} \mid (a_0,\ldots, a_n) \ne (0,\ldots,0) \}.$$ This is a countable set.
For $$(a_0,\ldots, a_n) \in M$$ consider the equation $$$$a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0 \tag{1}$$$$ and define $$A(a_0,\ldots, a_n) = \{ z \in \mathbb C \mid z \text{ satisfies } (1)\} ,$$ $$A = \{ z \in \mathbb C \mid \text{There exists } (a_0,\ldots, a_n) \in M \text{ such that } z \text{ satisfies } (1)\}.$$ Clearly $$A = \bigcup_{(a_0,\ldots, a_n) \in M } A(a_0,\ldots, a_n)$$. Since $$(1)$$ has at most $$n$$ distinct roots, $$A(a_0,\ldots, a_n)$$ is finite.
Thus $$A$$ is a countable union of finite sets and thus is countable.