I'm thinking about using the construction of splitting field by quotient out the irreducible polynomial every time to construct the splitting field of polynomial $\langle x^3-2\rangle$ over $\mathbb{Q}$, so we should quotient out by the irreducible polynomial $\langle x^3-2\rangle$ the first time, my question is how do we know it's isomorphic to $\mathbb{Q}(\sqrt[2]{3})$ but not $\mathbb{Q}$ adjoining other roots of the irreducible polynomial?


The polynomial $x^3-2$ has three distinct roots, one real and two complex. So $\mathbb{Q}(\sqrt[3]{2})$ cannot be the splitting field, because it's a subfield of the real numbers.

If $\omega$ is a complex non real number such that $\omega^3=1$, then you know that the roots of $x^3-2$ are $$ \sqrt[3]{2},\quad \omega\sqrt[3]{2},\quad\omega^2\sqrt[3]{2} $$ (because $\omega^2=\omega^{-1}=\bar{\omega}$ is the other complex cube root of $1$). Since $$ \omega=\frac{\omega\sqrt[3]{2}}{\sqrt[3]{2}} $$ belongs to the splitting field, you have that the splitting field contains $$ \mathbb{Q}(\sqrt[3]{2},\omega) $$ but, obviously, this field contains all roots of $x^3-2$ and so is the splitting field.

  • $\begingroup$ Thank you! I understand the part that the splitting field should be $\mathbb{Q}(\sqrt[3]{2},\omega)$, however, I'm just wondering what is Q\<x^3-2> isomorphic to? because it should be isomorphic to one of Q adjoining one of the three roots of the polynomial, but I don't know why is it isomorphic to Q adjoining $\sqrt{3}{2}$ rather than Q adjoining other two roots of the polynomial? Thanks so much! $\endgroup$ – Mark Nov 25 '15 at 16:41
  • $\begingroup$ @JoeyCheng If you add just one root (no matter which one), you get isomorphic fields, all isomorphic to $\mathbb{Q}[x]/\langle x^3-2\rangle$ $\endgroup$ – egreg Nov 25 '15 at 17:33

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