4
$\begingroup$

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.

$\endgroup$
1
  • 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

7
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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


$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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