# Splitting field and field extension $\mathbb Q(j,\sqrt[3]2)$

I have to clarify some stuffs which is not clear in my mind.

I denote $j=e^{\frac{2i\pi}{3}}$.

1) $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is a field extension of degree $3$ since $X^3-2$ is the minimal polynomial of $\sqrt[3]2$. The extension is separable (since $X^3-2$ is separable) but not normal (since $j\sqrt[3]2$ is a root conjugate that it's not in $\mathbb Q(\sqrt[3]2)$).

2) $\mathbb Q(j,\sqrt[3]2)/\mathbb Q$ is the splitting field of $X^3-2$ ? And thus $\mathbb Q(j,\sqrt[3]2)/\mathbb Q$ is a Galois extension since it's the splitting field of a separable polynomial.

3) $\mathbb Q(j,\sqrt[3]2)/\mathbb Q$ is an extension of degree $6$ since $[\mathbb Q(\sqrt[3]2):\mathbb Q]=3$ and since $j\notin \mathbb Q(\sqrt[3]2)$ and that it's minimal polynomial on $\mathbb Q(\sqrt[3]2)$ is $X^2+X+1$, we get $[\mathbb Q(j,\sqrt[3]2):\mathbb Q(\sqrt[3]2)]=2$ and thus $$[\mathbb Q(\sqrt[3]2,j):\mathbb Q]=[\mathbb Q(\sqrt[3]2,j):\mathbb Q(\sqrt[3]2)]\cdot [\mathbb Q(\sqrt[3]2):\mathbb Q]=2\cdot 3=6.$$

My questions are : What do you think about 2) ? And for 3) ?

• It's fine for me. – Bernard Nov 18 '15 at 21:36
• $2$ is correct. $3$ is correct in this example, but your argument may not work in general. It is not necessarily true that if $M/L/K$ and $x\in M\setminus L$, that the minimal polynomial of $x$ over $L$ is the same as its minimal polynomial over $K$. (Take $e^{i\pi/4}\in \mathbb Q(e^{i\pi/4})/\mathbb Q(i)/\mathbb Q$ for example) – Mathmo123 Nov 18 '15 at 21:38
• @Mathmo123: Thank you :-) For 3), the justification is simply that all roots of $X^2+X+1$ are complex, and since $\mathbb Q(\sqrt[3]2)\subset \mathbb R$, it must be irreducible over $\mathbb Q(\sqrt[3]2)$. Is it a correct justification ? For your example, $$e^{i\pi/4}=\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}\notin \mathbb Q(i)$$. Moreover, therefore $$[\mathbb Q(e^{i\pi/4}):\mathbb Q(i)]=2$$ since it is either $1,2$ or $4$. It can't be $1$ since $e^{i\pi/4}\notin \mathbb Q(i)$ and it can't be $4$, otherwise $\mathbb Q(i)=\mathbb Q$ which is impossible. Therefore it minimal polynomial has degree 2. – Rick Nov 18 '15 at 21:49
• So, now it's easy to find it :-) Do you think my argument are correct ? – Rick Nov 18 '15 at 21:50
• That's all completely correct in this case (so well done!). If your polynomial had degree bigger than 3, that argument might not work, as there are your polynomial could then be a product of irreducible quadratics. For example, $X^4+1$ has no real roots, but is reducible over $\mathbb Q(\sqrt 2)$. – Mathmo123 Nov 18 '15 at 22:50

To expand on what I said in the comments:

Your argument for $$3$$ (and also your argument in the comments) works in this particular case, because the degree of $$X^2 + X+1$$ is small.

Your claim that the minimal polynomial of $$j$$ over $$\mathbb Q(\sqrt[3]2)$$ is the same as the minimal polynomial of $$j$$ over $$\mathbb Q$$ because $$j\notin \mathbb Q$$ works in this case, but not in general. For example, $$e^{2i\pi/8}\notin \mathbb Q(i)$$, but its minimal polynomial over $$\mathbb Q(i)$$ is $$X^2-i$$, not $$X^4+1$$, its minimal polynomial over $$\mathbb Q$$.

So why does this argument work in this case? I'll write out a correct argument along your lines and highlight where we avoid the issue mentioned above.

Since $$j$$ is a root of $$f(X)=X^2+X+1$$, and $$f$$ has coefficients in $$\mathbb Q(\sqrt[3]2)$$, we know that $$[\mathbb Q(j,\sqrt[3]2):\mathbb Q(\sqrt[3]2)]\color{red}{\le}2.$$

Now the fact that $$j\notin \mathbb Q(\sqrt[3]2)$$ tells us that $$[\mathbb Q(j,\sqrt[3]2):\mathbb Q(\sqrt[3]2)]\ge 2$$and the rest of your argument now applies.

If we try this same argument for the example above, the fact that $$e^{2\pi/8}$$ is a root of $$X^4+1$$ tells us that $$[\mathbb Q(e^{2\pi i/8}):\mathbb Q(i)]\le 4$$and the fact that $$e^{2\pi i/8}\notin \mathbb Q(i)$$tells us that $$[\mathbb Q(e^{2\pi i/8}):\mathbb Q(i)]\ge2.$$These two facts aren't enough to deduce the value of $$[\mathbb Q(e^{2\pi i/8}):\mathbb Q(i)]$$.