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We know that the set $\mathbb{A}$ of algebraic numbers is a field. But there is a constructive proof of this statement? I.e. : given a sum (or a product) of numbers of the form $\sqrt[n]{q}$ with $ q \in \mathbb{Q}$ we can find a polynomial in $x$, with rational coefficients, that has as solution this sum (or product)? I've find that such a proof can be given using the resultant of the polynomials that have as roots the given numbers, but, as far as I know, the resultant is a number, so how can we find a polynomial?

I would appreciate an example: e.g. how to find the rational polynomial that has root $ \sqrt{2}+\sqrt[3]{3}$.

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  • $\begingroup$ If $r_1,...,r_n$ are all the conjugates of $r_1$ and $s_1,...,s_m$ are the conjugates of $s_1$ consider $\prod_{i,j}(x-r_i-s_j)$, $\prod_{i,j}(x-r_is_j)$ and $\prod_i(x-1/r_i)$. $\endgroup$ – Pp.. Feb 2 '15 at 20:46
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Hint $ $ Let $\,a=\sqrt2,\,b=\sqrt[3]3.\,$ $\Bbb Q(a,b)\cong \Bbb Q\langle1,b,b^2, a,ab,ab^2\rangle$ as a vector space. $\,x\mapsto (a\!+\!b)\,x\,$ is a linear map on this vector space. Compute its matrix, then apply Cayley-Hamilton, to compute a (characteristic) polynomial having $\,a\!+\!b\,$ as root (e.g. by computing a determinant).

Remark $\ $ Such "determinant tricks" generalize to arbitrary rings. More efficiently (and precisely), one can use Grobner bases to do the elimination (e.g. via ideal-contraction), or resultants.

Mechanically, we can use resultants to eliminate $\,z,y\,$ from $\,x = y+z,\ x^2=2,\ y^3 = 3.$ First eliminating $\,z\,$ we have ${\rm res}(x\!-\!y\!-\!z,z^2\!-\!2,z) = (x\!-\!y)^2\!-\!2,\,$ then eliminating $\,y\,$ we have $\,{\rm res}((x\!-\!y)^2\!-\!2,y^2\!-\!3,y) = x^6\!-\!6x^4\!-\!6x^3\!+\!12x^2\!-\!36x\!+\!1\,$ (e.g. computed via Alpha)

Alternatively we can contract the ideal $\,(x\!-\!y\!-\!z,z^2\!-\!2,y^3\!-\!3)\,$ to $\,\Bbb Q[x],\,$ using Grobner bases.

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  • $\begingroup$ whaw! I have a lot of things to learn to understand this answer! So.. nothing to do with the resultant? $\endgroup$ – Emilio Novati Feb 2 '15 at 21:14
  • $\begingroup$ @Emilio Have you studied linear algebra / vector spaces? $\endgroup$ – Bill Dubuque Feb 2 '15 at 21:15
  • $\begingroup$ Yes, but I've to learn about Grobner bases and ideal-contraction. $\endgroup$ – Emilio Novati Feb 2 '15 at 21:29
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The response of @BillDubuque is probably the best general method, but in this case, I’d start with the minimal $\Bbb Q$-polynomial for $\root3\of3$, namely $X^3-3$, and pass from this to $a$ polynomial for $\root3\of3+\sqrt2$, namely $f(X)=(X-\sqrt2)^3-3$, I’ll let you multiply it out; and then multiply $f$ by its $\Bbb Q(\sqrt2)$-conjugate, that is replace $\sqrt2$ by $-\sqrt2$ wherever it appears in $f$. You obviously get a $\Bbb Q$-polynomial, of the right degree, namely $6$, of which your number is a root.

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  • $\begingroup$ Also this answer don't use resultants. So why en.wikipedia.org/wiki/Algebraic_number say that ''this fact can be demonstrated using the resultant''?. $\endgroup$ – Emilio Novati Feb 2 '15 at 21:36
  • $\begingroup$ @Emilio I added a remark to my answer showing how to do the elimination with resultants (it is essentially a more mechanical form of the above elimination). I presumed from your query that you sought a general method, which is why I mentioned the determinant trick. $\endgroup$ – Bill Dubuque Feb 2 '15 at 22:01
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    $\begingroup$ @Bill: Thank you very much. I've a lot of work to do from your answer. If I well understand we can find the rational polynomial also if a root is not writable by means of radicals, as a root of some polynomial of degree $n>4$. $\endgroup$ – Emilio Novati Feb 3 '15 at 10:30
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    $\begingroup$ Right, @EmilioNovati: the method I used is very special. But quick, when it works! $\endgroup$ – Lubin Feb 3 '15 at 13:01

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