Let $\mathbb{K}$ be the splitting field of the polynomial $X^3 + X + 1$ over $\mathbb{Q}$

Question

Let $\mathbb{K}$ be the splitting field of the polynomial $X^3 + X + 1$ over $\mathbb{Q}$.

(a) Determine the elements of $\mathbb{K}$ whose cube is rational number.

(b) Does $\sqrt{-1} \in \mathbb{K}$?

Hint required will finish the proof

Answer 1

Hints:

• If $x^3=q$, $q$ rational, has a solution $z\in\Bbb{K}\setminus\Bbb{Q}$ then prove that $\Bbb{K}$ contains the splitting field of $x^3-q$ as a subfield. Such splitting fields are known to contain a certain quadratic algebraic integer.
• What's the discriminant of your polynomial? What do you know about discriminants of cubics and intermediate fields?

Spoiler follows:

If $m(x)=x^3-q$ has a non-rational zero in $\Bbb{K}$, then, $\Bbb{K}$ being a normal extension of $\Bbb{Q}$, it must contain the splitting field of $m(x)$. If $m(x)$ has no rational zeros it is irreducible. Hence its splitting field is $\Bbb{Q}(\root3\of q,e^{2\pi i/3})$ implying that $\sqrt{-3}\in \Bbb{K}$. On the other hand, the discriminant of the given cubic is $-31$. Galois theory then tells us that the only quadratic subfield of $\Bbb{K}$ is $\Bbb{Q}(\sqrt{-31})$. Therefore $\sqrt{-3}\notin K$ and $\sqrt{-1}\notin K$. The answers to the questions are thus: (a) the rational numbers, (b) no.