# Understanding the field extension diagram of $x^3-2$

Looking at the polynomial $$f\left(x\right)=x^{3}-2$$ over $$\mathbb{Q}$$, and with $$\rho=\sqrt{2}, \zeta = \zeta_{3}$$ (3'rd unit root).

The roots of $$f$$ are $$\left(\rho,\rho\zeta_{3},\rho\zeta_{3}^{2}\right)$$, So $$\mathbb{E}=\mathbb{Q}\left(\rho,\rho\zeta_{3},\rho\zeta_{3}^{2}\right)=\mathbb{Q}\left(\rho,\zeta_{3}\right)$$ is the splitting field of $$f$$ over $$\mathbb{Q}$$.

Now I saw the following diagram regarding this problem, which made me quite confused: And I am not sure I understand it, and the numbers in it. It seems to me that

1. $$\left[\mathbb{Q}\left(\rho\right):\mathbb{Q}\right] = 3$$, as $$m_{\rho}^{\mathbb{Q}}\left(x\right)=x^{3}-2 =f$$, and its degree is $$3$$.
2. $$\left[\mathbb{Q}\left(\zeta\right):\mathbb{Q}\right] = 3$$ , as $$m_{\zeta}^{\mathbb{Q}}\left(x\right)=x^{3}-1$$, and its degree is $$3$$.

Since both are simple algebraic extensions, and those are indeed their minimal polynomials, am I wrong? So I would expect both numbers on the bottom arrows to be 3 (?).

It was also noted that $$6\mid\left[\mathbb{E}:\mathbb{Q}\right]=\deg m_{\rho}^{\mathbb{Q}\left(\zeta_{3}\right)}\cdot2\leq6$$ and I am not sure why this is true.

I saw this question but it didn't really help me.

• For $n > 1$, if $\zeta$ is an $n$th root of unity the extension $\mathbf Q(\zeta)/\mathbf Q$ never has degree $n$, because the polynomial $x^n - 1$ is never irreducible (it has at least the linear factor $x-1$, of smaller degree since $n > 1$).
– KCd
Aug 14 at 11:10

Note that the minimal polynomial of $$\zeta_3$$ is $$x^2+x+1$$, not $$x^3-1$$ (indeed $$x^3-1$$ is reducible as $$(x-1)(x^2+x+1)$$). This is where the $$2$$ comes from.

As for the equation $$6\mid[\mathbb E:\mathbb Q]=\text{deg } m_\rho^{\mathbb Q(\zeta_3)}\cdot 2\leq 6$$, we established that $$[\mathbb E:\mathbb Q]=6$$ with the diagram above and that $$\deg m_\rho=3$$, so substituting those in does indeed make the equation true. Do you have a specific concern surrounding this equation?

It's a non-trivial result of Gauss's that the $$n$$th cyclotomic polynomial is irreducible over $$\Bbb Q$$. It's by definition $$\Phi_n(x)=\prod (x-\zeta_i)$$ as $$\zeta_i$$ range over the $$\varphi (n)$$ primitive $$n$$th roots of unity.

In this case, $$n=3, \varphi (n)=2$$, and $$\Phi_3(x)=(x-e^{2\pi i/3})(x-e^{4\pi i/3})=x^2+x+1$$.

So the diagram is correct, and indeed $$[\Bbb Q(\zeta_3):\Bbb Q]=2$$.

In general, if $$\zeta_n$$ is a primitive $$n$$th root of unity, $$[\Bbb Q(\zeta_n):\Bbb Q]=\varphi (n)$$.

For the second part, we should have $$[E:\Bbb Q]=[E:\Bbb Q(\zeta_3)]\cdot [\Bbb Q(\zeta_3) :\Bbb Q]\le3\cdot 2=6$$, since we know $$[E:\Bbb Q(\zeta_3) ]\le \rm{deg}(m_\rho^\Bbb Q)=3$$ since $$\Bbb Q\subset\Bbb Q(\zeta_3)$$. Meanwhile $$6\mid[E:\Bbb Q]$$, because there are degree $$2,3$$ extensions in between.

(This is analogous to Lagrange's theorem in group theory.)

Thus $$[E:\Bbb Q]=6$$.

I would expect both numbers on the bottom arrows to be 3 (?)

Consider the extension $$\mathbb{Q}\left(\rho\right)|_{\mathbb{Q}}$$

Here minimal polynomial of $$\rho$$ over $$\Bbb{Q}$$ :$$m_{\rho}(x) =x^3-2$$

Hence $$\frac{\mathbb{Q}[x]}{\langle x^3-2\rangle}\cong\mathbb{Q}(\rho)$$

The basis of the vector space $$\mathbb{Q}(\rho)$$ over $$\Bbb{Q}$$ is $$\{1,\rho,\rho^2\}$$

$$[\mathbb{Q}\left(\rho\right):{\mathbb{Q}}]=\deg m_{\rho}(x)=3$$

Now consider the extension $$\mathbb{Q}\left(\zeta\right)|_{\mathbb{Q}}$$

$$1+\zeta+\zeta^2=0$$ implies the minimal polynomial of $$\zeta$$ over $$\Bbb{Q}$$ :$$m_{\zeta}(x) = x^2+x+1$$

Hence $$\frac{\mathbb{Q}[x]}{\langle x^2+x+1\rangle}\cong\mathbb{Q}(\zeta)$$

The basis of the vector space $$\mathbb{Q}(\zeta)$$ over $$\Bbb{Q}$$ is $$\{1,\zeta\}$$

$$[\mathbb{Q}\left(\zeta\right):{\mathbb{Q}}]=\deg m_{\zeta}(x)=2$$

Now let us consider the extension $$E|_{\Bbb{Q}(\zeta)}$$

This is also a simple extension which can be obtained by the field adjunction of $$\rho$$ to the field $$\Bbb{Q}(\zeta)$$.

Now the minimal polynomial of $$\rho$$ over the field $$\Bbb{Q}(\zeta)$$ is $$m_{\rho}^{\Bbb{Q}(\zeta) }(x) =x^3-2$$

Basis of the vector space $$E$$ over $${\Bbb{Q}(\zeta)}$$ has $$\{1,\rho,\rho^2\}$$

Hence $$[E:\Bbb{Q}(\zeta)]=3$$

Now by tower law of field extension , we have

\begin{align}[E:\Bbb{Q}]&=[Q(\zeta) :Q]\times[E:\Bbb{Q}(\zeta) ]\\&=2\times 3\\&=6\end{align}

Basis for the vector space $$E$$ over $$\Bbb{Q}$$ is $$\{1,\rho,\rho^2,\zeta,\rho\zeta,\rho^2\zeta\}$$