Can I have two irreducible polynomials of different degree, having isomorphic splitting fields? The base field does not has to be perfect, . I mean if the base field is perfect, the extension is Galois, and therefore the result is trivial. But what happens when the polynomials are not separable? Thanks

  • $\begingroup$ Yeah! Im sorry! The polynomial is not necessarily separable (a field cannot be separable, just the extension), another way I guess is taht the original field is not necessarily perfect. $\endgroup$
    – Matt
    Dec 22, 2013 at 3:39

1 Answer 1


Take $f(X)=X^3-5$, whose splitting field over $\mathbb Q$ is of degree six, and for $g(X)$ take the minimal $\mathbb Q$-polynomial for $\lambda+\omega$, where $\lambda$ is a cube root of $5$ and $\omega$ is a primitive cube root of unity. Since $\lambda+\omega$ generates the whole splitting field of $f$, its minimal polynomial $g$ is of degree six. In fact, $g$ seems to be $36 + 18X - 9X^2 - 3X^3 + 6X^4 + 3X^5 + X^6$.


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