Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $a,b,c$)

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

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).

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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