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I did not find any material on this anywhere else, so I decided to ask this here.

Let $f(x),g(x)\in\mathbb{Z}[x]$ be a pair of monic irreducible polynomials. Is there any classification of when does there exist two roots of $f(x)$, and $g(x)$, say $\alpha$, and $\beta$ such that the field extensions $\mathbb{Q}[\alpha]$ and $\mathbb{Q}[\beta]$ are linearly disjoint?

I read several theorems regarding the classification of generic linear disjoint field extensions, however, I am interested in this special case, and was wondering whether there was more one could say in this situation.

A conjecture I just made up (edited): the extensions induced by $f(x),g(x)$ as above is not linearly disjoint for any choice of $\alpha$ and $\beta$ if and only if there exists some polynomials $h(x),F(x),G(x)$ such that $\deg(h) > 1$ and $f(x) = h(F(x))$ and similarly $g(x) = h(G(x))$. Is that true?

Thanks in advance!

kindasorta

kindasorta
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  • Linking to this thread for Pete Clark gives a lot of information. And links to his lecture notes. – Jyrki Lahtonen Dec 28 '19 at 22:20
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    Are you sure that your condition guarantees that $\Bbb{Q}(\alpha)$ and $\Bbb{Q}(\beta)$ intersect trivially? Consider the case $f(x)=x^4+x^3+x^2+x+1$, $g(x)=x^2-5$, when $\Bbb{Q}(\beta)\subset\Bbb{Q}(\alpha)$. I didn't look very hard, and I'm prepares to be wrong, but it seems unlikely to me that a suitable $h(x)$ exists. – Jyrki Lahtonen Dec 28 '19 at 22:34
  • Very interesting example, and elementary arguments show that for your example a suitable $h(x)$ does not exist. However, notice that the same thread you linked to shows that for all roots $\alpha$ of $f(x)$ for which $\mathbb{Q}(\beta)\not\subseteq \mathbb{Q}(\alpha)$, then $\mathbb{Q}(\beta)$ and $\mathbb{Q}(\alpha)$ are in fact linearly disjoint! – kindasorta Dec 28 '19 at 23:01
  • Haha, and the fact that one $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\beta)$ implies that all conjugate fields of $\mathbb{Q}(\alpha)$, or that all $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\beta)$, since $\mathbb{Q}(\beta)$ is Galois :) – kindasorta Dec 29 '19 at 14:55

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