# Is this element quadratic in this field?

Can someone give me an idea how to prove :

a) that the minimal polynomial of $\sqrt[3]{2}\in F$ (where $F$ is the splitting field of $x^3-2$ over $\mathbb{Q}$) over the field $\mathbb{Q}(r\sqrt[3]{2})$ has degree 2 (hence the name of the title), where $r=e^{\frac{2\pi i}{3}}$ (notice that the roots of $x^3-2$ are $\sqrt[3]{2},r\sqrt[3]{2},r^2 \sqrt[3]{2}$);

b) that $\sqrt[3]{2}\not \in \mathbb{Q}(r\sqrt[3]{2})$ ?

What I know until now for a): I know that $\mathbb{Q}(r\sqrt[3]{2})(\sqrt[3]{2})=F$, since from $r\sqrt[3]{2},\sqrt[3]{2}$ I can recover $r^22^{\frac{1}{3}}$ (and by having all roots of the polynomial, the above must coincide with the splitting field); I think the polynomial I'm looking for is $x^2+\sqrt[3]{4}$, but I don't know how to prove that $\sqrt[3]{4} \in \mathbb{Q}(r\sqrt[3]{2})$, since this time I have no clue how to recover $\sqrt[3]{4}$ only from $r\sqrt[3]{2}$. For b) I think this must have something to do with the fact, that $\mathbb{Q}(r\sqrt[3]{2})=\{a+ b\cdot r\sqrt[3]{2}+ c \cdot r^2 \sqrt[3]{4}|a,b,c\in \mathbb{Q}\}$, since $(r\sqrt[3]{2})^3=1\cdot 2=2$, but I can't figure out how to turn this observation into a precise argument (because one could argue that it still might be possible to obtain $\sqrt[3]{2}$ if one only had suitable $a,b,c$)

-

Note that $\sqrt[3]{2}$ satisfies the polynomial $x^3-2$, which has coefficients in $\mathbb{Q}$; therefore, the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}(r\sqrt[3]{2})$ must divide the polynomial $x^3-2$.

Since $x^3 -2 = (x-\sqrt[3]{2})(x-r\sqrt[3]{2})(x-r^2\sqrt[3]{2})$, then the minimal polynomial must divide $$(x-\sqrt[3]{2})(x-r^2\sqrt[3]{2}) = x^2 - (1+r^2)\sqrt[3]{2}x+ r^2\sqrt[3]{4}.$$ Now simply note that since $r$ satisfies $x^2+x+1$, then $r^2 + 1 = -r$, so the minimal polynomial must divide $$x^2 + r\sqrt[3]{2}x + (r\sqrt[3]{2})^2.\qquad\qquad\qquad\qquad(1)$$ So the only remaining question is whether this is irreducible.

This can be seen from the fact that $\mathbb{Q}(r\sqrt[3]{2})$ is of degree $3$ over $\mathbb{Q}$, but the splitting field is of degree $6$; you know that $\sqrt[3]{2}$ must be quadratic over $\mathbb{Q}(\sqrt[3]{2})$.

Alternatively, if $\sqrt[3]{2}\in \mathbb{Q}(r\sqrt[3]{2})$, then you would necessarily have $\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(r\sqrt[3]{2})$ (since the former is contained in the latter, and they have the same degree over $\mathbb{Q}$). But $\mathbb{Q}(\sqrt[3]{2})\subseteq \mathbb{R}$, and $\mathbb{Q}(r\sqrt[3]{2})$ is not contained in $\mathbb{R}$, so this is impossible.

Thus, the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}(r\sqrt[3]{2})$ is (1).

-

I like this answer, but maybe it's not along the lines of what you'd like. We can see that this splitting field has degree $6$ over $\mathbb{Q}$ pretty easily by noting that $\mathbb{Q} (\sqrt[3] 2)$ has degree $3$ over the rationals and is entirely real, and $\mathbb{Q} (\sqrt[3] 2, r)$ is the splitting field.

Then you can us the tower law (or whatever you call the rule that states that if $L \subset K \subset F$, then $[F:L] = [F:K][K:L]$) to see that the remaining extension is quadratic. (This does assume part b however).

For part (b), maybe I'd like to be a bit witty. Suppose it was in that field for a moment. Then we can divide by it, and so we see that $r$ is also in our field. But then we have the entire splitting field, which we know is of degree 6. Is that a degree 6 extension? No - it's not.

-