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I am trying to understand the proof of the fact that The set of all algebraic numbers is a countable set. The proof I am reading is from "Theory of sets" by E.Kamke.The proof is quite similar to the proofs in some other website (Link 1) (Link 2). It starts from defining the height $h$ of a polynomial:

$$h = n + a_n + |a_{n-1}| + ... + |a_1| + |a_0|$$

Or in the second website

$$h = n + \sum_{i=0}^n |a_i| $$

What I dont understand is that , why do we also need $n$ ?

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    $\begingroup$ You need the $n$ to limit the power of the polynomial. If we didn't have the $n$ and we defined height as $\sum |a_i|$ then $3x^2 + 2x + 1$ and $3x^3 + 2x^2 + 1$ and $x^{97} + 2x^{56} + x^{43} + 2$ would all have height $6$ and there would be an infinite number of polynomials with height $6$. The proof relies and being a countable union of finite sets. (Although a countable union of countable sets is countable but that would not be a direct construction) $\endgroup$
    – fleablood
    Jun 8, 2021 at 16:00

1 Answer 1

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You want a finite number of polynomials, and so a finite number of algebraic numbers, to correspond to each height. This then makes it possible to list the algebraic numbers by height and thus show that there are a countable number of them.

If you did not have the $n$ term then the following would all have height $3$:

$$x-2=0$$ $$x^2-2=0$$ $$x^3-2=0$$ $$x^4-2=0$$ $$x^5-2=0$$ $$\cdots$$

so there would be an infinite number their roots and so of distinct algebraic numbers of height $3$ and you would not reach those of height $4$ in your list.

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  • $\begingroup$ thanks for your response. I have another question (Hope you dont mind) , In the proof of E.Kamke , it is said that "We may suppose , moreover , without loss of generality , that $a_n$ > 0 ".But why? Is it really necessary? $\endgroup$ Jun 9, 2021 at 7:51
  • $\begingroup$ If $a_n=0$ then you really have a polynomial of lower degree. If $a_n<0$ then you can multiply all the $a_i$ by $-1$ and you get the same roots. So you lose nothing by assuming $a_n>0$ $\endgroup$
    – Henry
    Jun 9, 2021 at 8:10
  • $\begingroup$ But , is it unnecessary ? $\endgroup$ Jun 9, 2021 at 8:12
  • $\begingroup$ It is probably convenient rather than necessary. If it halves the length of the proof then it is probably a sensible approach $\endgroup$
    – Henry
    Jun 9, 2021 at 8:15

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