Looking at the polynomial $f\left(x\right)=x^{3}-2 $ over $\mathbb{Q}$, and with $ \rho=\sqrt[3]{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:

field extension diagram

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.

  • 1
    $\begingroup$ 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 $n > 1$). $\endgroup$
    – KCd
    Aug 14 at 11:10

3 Answers 3


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 $\{1,\rho,\rho^2,\zeta,\rho\zeta,\rho^2\zeta\}$


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