Given an extension $K = \mathbb{Q}(\alpha)$ by a cubic polynomial $g(x)\in \mathbb{Q}[x]$ (not necessarily Galois extension) is there a criterion for a cubic polynomial $f(x) \in K[x]$ to be reducible (f is actually defined over $\mathbb{Q}$)? I have tried assuming that $\beta$ is a root of $f$ then writing $\beta=a_{0}+a_{1}\alpha+a_{2}\alpha^2$ where $a_{i}\in \mathbb{Q}$ and substituting $\beta$ back into $f$ and equating the result to zero but I get a horrible equation to try and solve. (Essentially I want to find out when $\mathbb{Q}(\alpha) \cong \mathbb{Q}(\beta)$, where $f(x)=x^3+6ax^2+9a^2x+27(3b-n)^2+4a^3$ and $g(x)=x^3-12ax^2+36a^2x+27(3b-n)^2+4a^3$ with $a,b,n \in \mathbb{Z}$).
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$\begingroup$ One condition is that the two polynomials have the same discriminant. Is that a manageable equation for your f, g? $\endgroup$– AsvinCommented May 9, 2021 at 19:11
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$\begingroup$ Two cubics can have the same discriminant but generate non-isomorphic fields though? $\endgroup$– H UCommented May 9, 2021 at 19:13
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4$\begingroup$ Do you want $f(x)$ to be irreducible over $\mathbf Q$? A necessary condition for two irreducible polynomials to each have a root that generates isomorphic field extensions of $\mathbf Q$ is that the discriminants of the polynomials have a square ratio (need not be equal). The condition is not sufficient. Irreducible cubics can have equal discriminant while their roots generate nonisomorphic extensions. $\endgroup$– KConradCommented May 9, 2021 at 19:29
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$\begingroup$ Hi @KConrad! Yes I am assuming that f is irreducible over $\mathbb{Q}$ and I think your answer will solve my problem so thank you! $\endgroup$– H UCommented May 9, 2021 at 19:36
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$\begingroup$ Sorry, yes. I should have said exactly what Conrad did say. $\endgroup$– AsvinCommented May 9, 2021 at 22:37
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