How to deduce whether $\sqrt{3}i \in \mathbb Q (u)$ or not, where u is a root of $x^3-2$

Question

How can one deduce whether $\sqrt{3}i \in \mathbb Q (u)$ or not, where u is a root of $x^3-2$ ?

$\mathbb Q (u) = \{c_0 +c_1u+c_2u^2 | c_i \in \mathbb Q\}$

Is it a trivial computation, or does this perhaps rely on heavy theory?

Edit: Observed that it is trivial if the root of $x^3-2$ is the real root. Would still like to know what the case is for the other roots.

Answer 1

First of all notice that $x^3-2$ is irreducable over $\mathbb Q$,

so $|\mathbb Q(u):\mathbb Q|=3$ and $m=\sqrt3i$ is a root of $x^2+3$ so $|\mathbb Q(m):\mathbb Q|=2$

Since $2$ does not divides $3$, $\mathbb Q(m)$ can not be subfield of $\mathbb Q(u)\implies m\notin \mathbb Q(u) $

Answer 2

You want to define $\sqrt{3} i$ in a condition that can be expressed internally to $\mathbb{Q}[u]$ namely: "is there an element in $\mathbb{Q}[u]$ that squares to $-3$?"

$$(c_0+c_1u+c_2u^2)^2=-3$$ $$c_0^2+2c_0c_1u+(2c_0c_2+{c_1}^2)u^2+4c_1c_2+2{c_2}^2u=-3$$

The only way of representing $-3$ is for the $\mathbb{Q}$ term coefficient to be $-3$ and the others to be zero:

$${c_0}^2+4c_1c_2=-3$$ $$2{c_2}^2+2c_0c_1=0$$ $${c_1}^2+2c_0c_2=0$$

And some mildly annoying algebra can show that this set of equations has no rational solution.

It's also worth noting that proving it in the real case proves it in the general case, so you don't need to consider the other roots separately. The field structure here "can't detect" the difference between different cube roots of $2$, since they all have the same algebraic relations to each other and the rationals.

Answer 3

If one is allowed to use some machinery, one can use a degree argument. The field $\mathbb{Q}(u)$ has degree $3$ over $\mathbb{Q}$, and $\mathbb{Q}(\sqrt{3}i)$ has degree $2$, and $2$ does not divide $3$.