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Consider a polynomial in one variable $x$ with irrational coefficients which are algebraic, i.e., they have a defining polynomial. As an example, take $p(x) = (x-3)(x-\sqrt{2}) = x^2-(3+\sqrt{2})x+3\sqrt{x}$. In my setting, $\sqrt{2}$ could also be some real root $r$ not being a radical expression, but I always have a defining polynomial $q(y)$ for $r$ so that $q(r)=0$.

My question is, whether there is a polynomial $\tilde p(x)$ with rational coefficients such that $p(a) = 0 \Rightarrow \tilde p(a) = 0$ (so I only need an over-approximation of the roots of $p$).

As a side information, such polynomials occur if I start with polynomials in multiple variables and substitute variables by real roots found by another procedure.

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Yes, this follows immediately by transitivity of algebraic dependence. You have $\rm\:r\:$ algebraic over $\rm\:\mathbb Q\:$ and $\rm\:a\:$ algebraic over $\rm\:\mathbb Q(r)\:$ so by transitivity we infer that $\rm\:a\:$ is algebraic over $\rm\:\mathbb Q\:.\:$ One can explicitly construct a polynomial $\rm\:p(x)\in \mathbb Q[x]\:$ with $\rm\:p(a) = 0\:$ by employing various elimination techniques, e.g. using Grobner bases.

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Yes. The branch of mathematics that is of help here is Galois theory. Specifically, let $K'$ be the field generated over $\mathbb{Q}$ by all the coefficients of your polynomial, and let $K$ be it's so-called Galois closure. Let $G = Aut(K)$ be the group of $\mathbb{Q}$-automorphisms of $K$. The polynomial $P(x) = \prod_{\sigma \in G}(\sigma p)(x)$ has the desired properties. [Here $G$ acts on $K[X]$ by acting on the coefficients alone.]

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  • $\begingroup$ I work with polynomials which possibly have irrational roots not being representable by radicals, e.g., $q(y)=y^5 - 3y^4 + y^3 - y^2 + 2y - 2$ having one irrational real root which is not a radical expression. As this root could also occur as coefficient in another polynomial, I would have to construct the Galois closure of the algebraic extension of $\mathbb{Q}$ w.r.t. $q(y)$. I'm not an expert in Galois theoy, but isn't that a problem? $\endgroup$ – loup Jan 12 '12 at 13:32
  • $\begingroup$ Expressibility using radicals is not relevant here. As long as the coefficients are algebraic everything is fine. NB: as Bill Dubuque points out you don't even need full-fledged galois theory here (although this was the first that came to my mind). $\endgroup$ – Tom Bachmann Jan 12 '12 at 20:16

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