2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '13 at 3:39
4
$\begingroup$

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 $36 + 18X - 9X^2 - 3X^3 + 6X^4 + 3X^5 + X^6$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.