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It's not too hard to show that the splitting field of $f(x) = x^3 - 2 \in \mathbb{Q}[x]$ over $\mathbb{Q}$ is $\mathbb{Q}(2^{1/3}, \omega)$ where $\omega$ is a nontrivial third root of unity (a little working shows that it can also be written as $\mathbb{Q}(2^{1/3}, \sqrt{-3})$). However, as one expects of a degree 6 extension, the minimal polynomial is also of degree 6: $x^6 + 9x^4 - 4x^3 + 27x^2 + 36x + 31$. Is there any relation between this higher degree minimal polynomial and the third degree polynomial we started with, considering they both generate the same splitting field? This is meant to be a general question, the above is just an example.

EDIT: A clarification and generalisation of the question. Apologies for the earlier lack of clarity, I see that I made too many assumptions and such.

Given a field $K$ and a polynomial $f(x)\in K[x]$, supposing we can find a field extension $K(\alpha_1, ..., \alpha_n) \cong K[x]/(f)$, is there any relation between $f$ and the minimal polynomial of some linear combination of the $\alpha_i$ (that involves all $\alpha_i$.

The reason I ask this is that I would expect there to be something relating them as both the original polynomial and the resulting minimal polynomial will result in the aforementioned field extension.

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  • $\begingroup$ You wrote : "the minimal polynomials is also..." . What "the minimal polynomial" are you talking about?? $\endgroup$ – DonAntonio Sep 13 '16 at 8:48
  • $\begingroup$ I'm sorry, I'm afraid I'm not sure what you're asking. Minimal polynomials are unique, are they not? $\endgroup$ – rwmak Sep 13 '16 at 9:07
  • $\begingroup$ Would you like to explain us how you got the 6th degree polynomial? $\endgroup$ – Stefan4024 Sep 13 '16 at 9:16
  • $\begingroup$ @rwmak "Minimal polynomial" in fields extensions is always wrt some element. You give there a sixtic and call it "minimal polynomial": of what element and how is this related to the whole question? $\endgroup$ – DonAntonio Sep 13 '16 at 9:21
  • $\begingroup$ Ah, apologies. I skipped a step there. It's the minimal polynomial of the element $\sqrt{-3} + 2^{1/3}$. I worked it out by setting $\beta = \sqrt{-3} + 2^{1/3}$ and taking exponents so only rational numbers were left as coefficients. So the question then is this: given a field $K$ and a function $f(x)\in K[x]$, supposing we can find a field extension $K(\alpha_1, ..., \alpha_n) \cong K[x]/(f)$, is there any relation between $f$ and the minimal polynomial of some linear combination of the $\alpha_i$ (that involves all $\alpha_i$.). I'll edit my main question to reflect this clarification. $\endgroup$ – rwmak Sep 13 '16 at 10:55
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If $K$ is a field of characteristic zero, if $\alpha_1,\dots,\alpha_n$ are algebraic over $K$, if there is no proper subset $S$ of $\{\,\alpha_1,\dots,\alpha_n\,\}$ such that $K(S)=K(\alpha_1,\dots,\alpha_n)$, then there is some linear combination $\alpha=\sum c_i\alpha_i$ such that $K(\alpha_1,\dots,\alpha_n)=K(\alpha)=K[x]/(f)$, where $f$ is the minimal polynomial for $\alpha$.

In the other direction, if you start with $f$ irreducible over $K$, then there will be lots of ways to find $\alpha_1,\dots,\alpha_n$ such that $K(\alpha_1,\dots,\alpha_n)=K[x]/(f)$, and there's no reason to think there's a linear combination $\alpha=\sum c_i\alpha_i$ such that the minimal polynomial of $\alpha$ is related to $f$ (although there's a certain amount of wiggle room here, since it's not entirely clear what you mean by "any relation").

Consider this example. Let $K$ be the rationals, let $f$ be the minimal polynomial for $\sqrt2+\sqrt6$, then $K(\sqrt2,\sqrt3)=K[x]/(f)$, but there is no rational linear combination of $\sqrt2$ and $\sqrt3$ with minimal polynomial $f$.

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  • $\begingroup$ Sorry, 'any relation' is indeed vague, but I was really looking for an answer like yours that shows there is nothing 'intuitive' or 'obvious' linking them as such. Thank you for the concise answer. $\endgroup$ – rwmak Sep 13 '16 at 13:23

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