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So, the field of complex algebraic numbers is algebraically closed, meaning that every root of a polynomial with algebraic coefficients is also algebraic.

But algebraic numbers are defined as roots of polynomials with integer coefficients.

So does that mean that every polynomial with algebraic coefficients can be transformed into a polynomial with integer coefficients that has the same roots? Or is it a system of integer polynomials with each corresponding to a different root of the initial polynomial?

I changed the title of the question a little so it makes more sense.

Servaes
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Yuriy S
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2 Answers2

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No; the polynomial $x-\sqrt{2}$ has algebraic coefficients and unique root $\sqrt{2}$. But any polynomial with integer coefficients that has $\sqrt{2}$ as a root also has $-\sqrt{2}$ as a root.

To the edited question; yes. You already note two key facts:

  1. The field of algebraic numbers is algebraically closed.
  2. Algebraic numbers are roots of polynomials with integer coefficients.

The first means that every root of a polynomial with algebraic coefficients is itself an algebraic number. The second means that it is a root of a polynomial with integer coefficients.

So given a polynomial $f$ with algebraic coefficients, for every root $\alpha$ of $f$ there exists a polynomial $f_{\alpha}$ with integer coefficients and with root $\alpha$. The product of all these polynomials for all roots $\alpha$ of $f$ then is a polynomial with integer coefficients, with all roots of $f$ as its roots.

Servaes
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  • I wasn't very clear in the title, I don't mean 'exactly one polynomial with exactly the same roots'. – Yuriy S Sep 10 '15 at 21:31
  • Thank you. Are there general algorithms developed for finding these integer polynomials for any particular class of algebraic polynomials? I do not need this information for any practical purpose, but I find this subject very interesting. – Yuriy S Sep 10 '15 at 21:43
  • I believe there are some fairly efficient algorithms, but things get pretty bad pretty quickly as the degrees of the algebraic numbers involved increase. Abstractly, one way to go about it is to adjoin all coefficients and roots of $f$ to $\Bbb{Q}$, and then finding a primitive element for this extension, and its minimal polynomial. This is all a lot of linear algebra, but I don't know of particularly efficient ways to do all this. Seems like a nice new question :) – Servaes Sep 10 '15 at 21:51
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To go into a bit more detail than Servaes has, you can start with your original polynomial $f(X)=\sum_na_nX^n$, with the coefficients algebraic. Then for any automorphism $\sigma$ of $\Bbb C$ (there are uncountably many such), you can take $f^\sigma$, by which I mean for you to apply $\sigma$ to all the coefficients of $f$. It’s crucial that there are actually only finitely many different $f^\sigma$ that occur by this process, and that their product has its coefficients in $\Bbb Q$. Then multiply this $\prod f^\sigma$ by an integer sufficient to kill all the denominators, and you’ll have a $\Bbb Z$-polynomial that has all the roots you were originally interested in.

Lubin
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