Tutorial 6 : Cyclotomic Extensions

We know that quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{(-1)^{\binom{p}{2}}p}$ . Hence $\mathbb{Q}(\zeta _7)$ contains $\mathbb{Q}(\sqrt{-7})$ and $\mathbb{Q}(\zeta _3)$ contains $\mathbb{Q}(\sqrt{-3})$ . If $\zeta _3 \;\in\;\mathbb{Q}(\zeta _7)$ ,then $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{-7})$ are subfields of $\mathbb{Q}(\zeta _7)$ . However $\mathbb{Q}(\zeta_p)$ contains a unique quadratic subfield. Hence $\zeta _3 \notin \mathbb{Q}(\zeta _7)$ and therefore $[\mathbb{Q}(\zeta _7,\zeta _3):\mathbb{Q}(\zeta _3)]=6$ by the diagram below.

$\displaystyle \xymatrix{ & \mathbb{Q}(\zeta_7,\zeta_3)=\mathbb{Q}(\zeta_{21}) \... ...eta_7) \ar@{-}[rd]^6 & & \mathbb{Q}(\zeta_3) \ar@{-}[ld]_2\\ & \mathbb{Q}& }$

2. Determine a primitive element of a subfield

of $E=\mathbb{Q}(\zeta _{13})$ so that $[K:\mathbb{Q}]=3.$

The Galois group $G(\mathbb{Q}(\zeta _{13})/\mathbb{Q}) \simeq U_{13}$ is a cyclic group of order

Hence it has a unique subgroup of order

for each divisor

. Hence there is a unique subfield

with $[K:\mathbb{Q}]=3.$ The powers of

mod

are given in the table below.

$\displaystyle \begin{tabular}{\vert l\vert l\vert l\vert l\vert l\vert l\vert l... ...2^i$ & 2 & 4 &8 & 3 & 6 & 12 & 11 & 9 & 5 & 10 & 7 & 1 \\ \hline \end{tabular}$

Hence $\mathbb{F}_{13}^{\times}=(2).$ Put $\omega=\zeta _{13}.$ The automorphism $\sigma(\omega)=\omega^2$ generates the Galois group $G=G(\mathbb{Q}(\omega)/\mathbb{Q}).$ Then $H=(\sigma^3)$ is a subgroup of

having order

Put $E=\mathbb{Q}(\omega).$ Then

and $[E^H:\mathbb{Q}]=3.$ Thus

as there is a unique intermediate subfield of degree

over $\mathbb{Q}$ for every divisor

Hence a primitive element of

$\displaystyle \beta_K$	$\displaystyle =$	$\displaystyle \sigma^3(\omega)+\sigma^6(\omega)+ \sigma^9(\omega)+\sigma^{12}(\omega)$
	$\displaystyle =$	$\displaystyle \omega^8+\omega^{12}+\omega^5+\omega$

3. Put $\zeta =\zeta _7.$ Determine the degrees of $\zeta +\zeta ^{5}$ and $\zeta +\zeta ^5+\zeta ^8$ over $\mathbb{Q}.$

The powers of

modulo

are :

$\displaystyle 3^1=3, \;3^2=2,\;3^3=6,\;3^4=4,\;3^5=5$ and $\displaystyle 3^6=1.$

Hence $<3>=\mathbb{F}_7^\times$ and the automorphism defined by $\sigma(\zeta )=\zeta ^3$ is a generator of the Galois group $G=G(\mathbb{Q}(\zeta)/\mathbb{Q})$ . Thus to find the degree of $\zeta +\zeta ^5$ , we find the orbit of $\zeta +\zeta ^5$ under the action of

$\displaystyle \zeta +\zeta ^5, \;\zeta ^3+\zeta ,\; \zeta ^2+\zeta ^3,\; \zeta ^6+\zeta ^2,\; \zeta ^4+\zeta ^6,\;\zeta ^5+\zeta ^4,\; \zeta +\zeta ^5.$

Thus there are

conjugates of $\zeta +\zeta ^5$ and hence $[\mathbb{Q}(\zeta +\zeta ^5):\mathbb{Q}]=6$ . The orbit of $\zeta +\zeta ^5+\zeta ^8$ under the action of

is : $2\zeta +\zeta ^5,\;2\zeta ^3+\zeta ,\;2\zeta ^2+\zeta ^3,\;2\zeta ^6+\zeta ^2,\; 2\zeta ^4+\zeta ^6,\;2\zeta ^5+\zeta ^4,\;2\zeta +\zeta ^5.$ Hence $[\mathbb{Q}(2\zeta +\zeta ^5):\mathbb{Q}]=6$ .

4. Put $\zeta =\zeta _{11}$ and $\alpha=\zeta +\zeta ^3+\zeta ^4+\zeta ^5 +\zeta ^9.$ Show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=2.$

The powers of

modulo

are: $2^2=4;2^3=8,\;2^4=5,\;2^5=10,\;2^6=9,\;2^7=7,\;2^8=3,\;2^9=6,\;2^{10}=1$
Therefore $\sigma(\zeta)=\zeta^2$ is a generator for $G=G(\mathbb{Q}(\zeta)/\mathbb{Q})$ . The orbit of $\alpha$ is
$\alpha,\sigma(\alpha)=\zeta ^2+\zeta ^6+\zeta ^8+\zeta ^{10} +\zeta ^7,\sigma^2(\alpha)=\zeta ^4+\zeta +\zeta ^5+\zeta ^9 +\zeta ^3=\alpha$
Thus $\alpha$ has only two conjugates and therefore $[\mathbb{Q}(\alpha):\mathbb{Q}]=2.$

5. Let $\nu$ be a primitive element modulo

where

is a prime. Thus $\mathbb{F}_p^{\times}=(\nu).$ Let $\zeta=\zeta _p.$ Using the list $\{ \zeta ^{\nu^0}, \zeta ^{\nu^1}, \zeta ^{\nu^2}, \ldots, \zeta ^{\nu^{p-2}}\},$ show how to find the sum $\beta$ of powers of $\zeta$ which determines a subfield $\mathbb{Q}(\beta)$ of $\mathbb{Q}(\zeta)$ so that $[\mathbb{Q}(\beta):\mathbb{Q}]=d$ where $d \,\vert\, (p-1).$

The map $\sigma(\zeta )=\zeta ^{\nu}$ is a generator of

Then $H=(\sigma^d)$ is the unique subgroup of

of order

Let $E=\mathbb{Q}(\zeta ).$ Then $[E^H:\mathbb{Q}]=d.$ Hence a primitive element of

is $\beta=\sum_{i=1}^{p-1/d} \sigma^{di}(\zeta ).$

6. Suppose $A \in \mathbb{C}^{n \times n}$ and

for some integer $k \in \mathbb{N}.$ Show that

can be diagonalized. Prove that the matrix $A= \left[\begin{array}{ll} 1 & \alpha \\ 0 & 1 \end{array}\right]$ where $\alpha \in K$ and

is a field of chracteristic

satisfies

and cannot be diagonalized if $\alpha \neq 0.$

By the equation

we see that the minimal polynomial of

is a divisor of

which has distinct roots. Hence

can be diagonalized.

$\displaystyle A^i= \left[\begin{array}{ll} 1 & \alpha \\ 0 & 1 \end{array}\rig... ...t[\begin{array}{ll} 1 & i\alpha \\ 0 & 1 \end{array}\right] \Rightarrow A^p=I.$

The minimal polynomial of

divides

which has only one root namely 1. Hence if there exists $P \in GL_2(K)$ so that

$\displaystyle PAP^{-1}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \Rightarrow A=I.$

This is a contradiction.

7. Show that $n=\sum_{d\vert n} \phi(d)$ and deduce that $\phi(n)=\sum_{d\vert n} \mu(n/d)d.$

Consider the fractions

$\displaystyle \left\{\frac{1}{n},\frac{2}{n},\frac{3}{n}, \ldots, \frac{n-1}{n}, \frac{n}{n}\right\} .$

Reduce them to lowest terms. Then the fraction with denominator

where $d \mid n$ will be $\phi(d)$ in number. The total number of fractions is

. Thus $n=\sum_{d \mid n}\phi(d)$ .By M $\ddot{o}$ bius inversion $\phi(n)=\sum_{d \mid n}\mu(n/d)d$ .

8. Show that $% latex2html id marker 35501 $ \Phi_n(x)=\prod_{d\mid n}(x^{n/d}-1)^{\mu(d)}$$ .

Use the formula $% latex2html id marker 35503 $ x^n-1=\prod_{ \mid n} \Phi_d(x)$$ and apply the multiplicative version of the M $\ddot{o}$ bious inversion formula.

9. Show that $\Phi_n(x)=x^{\phi(n)}\Phi_n(1/x)$ and deduce that the coefficients of $\Phi_n(x)$ satisfy $a_k=a_{\phi(n)-k}$ for all $0 \leq k \leq \phi(n).$

We know that

is a root of $\Phi_n(x)$ if and only if

is a root of $\Phi_n(x).$ Verify that the product of the roots of $\Phi_n(x)$ is

Note that if

is a root of $\Phi_n(x)$ then $z^{\phi(n)}\Phi_n(1/z)=\Phi_n(z).$ This translates into the palindromic property of $\Phi_n(x).$

10. Let

be a prime number and

Show that

$\displaystyle \Phi_p(x)$	$\displaystyle =$	$\displaystyle x^{p-1}+x^{p-2}+\cdots+x^2+x+1$
$\displaystyle \Phi_{p^r}(x)$	$\displaystyle =$	$\displaystyle \Phi_p(x^{p^{r-1}})$
$\displaystyle \Phi_{2n}(x)$	$\displaystyle =$	$\displaystyle \Phi_n(-x),$ if n is odd
$\displaystyle \Phi_{np}(x)$	$\displaystyle =$	$\displaystyle \Phi_n(x^p)/\Phi_n(x)$
$\displaystyle \Phi_{p_1^{r_1} p_2^{r_2}\ldots p_m^{r_m}}(x)$	$\displaystyle =$	$\displaystyle \Phi_{p_1\ldots p_m} (x^{p_1^{r_1-1}p_2^{r_2-1}\ldots p_m^{r_m-1}}).$

$% latex2html id marker 35570 $ \Phi_p(x)=\prod_{d\vert p}(x^{p/d}-1)^{\mu(d)}= x^{p-1}+\cdots+ x+1$$
$% latex2html id marker 35572 $ \Phi_{p^r}(x)=\prod_{d\vert p^r}(x^{p^{r}/d}-1)^{\mu(d)}=(x^{p^r}-1) (x^{p^{r-1}}-1)^{-1}=\Phi_p(x^{p^{r-1}})$$ .
$% latex2html id marker 35574 $ \Phi_{2n}(x)=\prod_{d\vert n}(x^{\frac{2n}{d}}-1)^{\mu(d)} \prod_{d\vert n}(x^{\frac{2n}{2d}}-1)^{\mu(2d)}$$ $% latex2html id marker 35576 $ =\Phi_n(x^2)\prod_{d/n}(x^{n/d}-1)^{-\mu(d)}$$ $=\Phi_n(x^2)/\Phi_n(x)$ $% latex2html id marker 35580 $ =\prod_{d\vert n}\biggr(\frac{(x^{n/d})^2-1)}{x^{n/d}-1}\biggr)^{\mu(d)}$$ $% latex2html id marker 35582 $ =\prod_{d\vert n}(x^{n/d}+1)^{\mu(d)}=\Phi_n(-x)$$

As $% latex2html id marker 35584 $ \Phi_n(x)=\prod(x-z)$$ , where product is taken over all primitive nth roots of unity and $% latex2html id marker 35586 $ \Phi_n(-x)=\prod(x+z)$$ , therefore $\Phi_n(x)\Phi_n(-x)= \Phi_n(x^2)$ .