202306012006

Type :Note Tags : Number Theory

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Cyclotomic Fields

Note

Let $\omega =e^{2\pi i/m}$, where $m>1$ is an integer. Then the field $\mathbb{Q}(\omega )$ is called the $m^{th}$ cyclotomic field. It can be seen that:

• For odd $m$, the $m^{th}$ cyclotomic field is equal to the $2m^{th}$ one. ($e^{2\pi i/2m}=-e^{2\pi i(m+1)/2m}\in \mathbb{Q}[e^{2\pi i/m}]$).
• The field is a finite extension of $\mathbb{Q}$.

Theorem 1:

Note

All ${\omega }^k$, $1\le k\le m$, $(m,k)=1$, are exactly the conjugates of $\omega$.

Proof:

It is easy to see that every conjugate of $\omega$ is also a root of $x^m-1$ but not of $x^n-1$ for all $n<m$. But these are the numbers ${\omega }^k$ with $1\le k\le m,(m,k)=1$. Thus the set of conjugates is contained in the set $S=\{{\omega }^k:1\le k\le m,(m,k)=1\}$.

Consider the field homomorphism $\sigma :\mathbb{Q}(\omega )\to \mathbb{Q}(\omega )$ which takes $\omega$ to ${\omega }^k$ and fixes $\mathbb{Q}$ for some $1\le k\le m$, $(k,m)=1$. Now, $\omega \in \sigma (\mathbb{Q}(\omega ))$. Why? Because $(k,m)=1$, and so $um+vk=1$ for some $u,v\in \mathbb{Z}$. Thus, $\omega ={\omega }^{um+vk}=({\omega }^k{)}^v=\sigma ({\omega }^v)$. So $\sigma$ is a linear map from a finite dimensional $\mathbb{Q}$-vector space to itself, which is injective (since it’s a non zero field homomorphism).

This makes it a vector space isomorphism, and hence a field automorphism fixing $\mathbb{Q}$. Note that $\mathbb{Q}(\omega )$ is a galois extension of $\mathbb{Q}$, since it is the splitting field of $x^m-1$ over $\mathbb{Q}$. This makes $\sigma$ a member of $\mathrm{Gal}(\mathbb{Q}(\omega )/\mathbb{Q})$, implying $\phi (m)=|S|\le |\mathrm{Gal}(\mathbb{Q}(\omega )/\mathbb{Q})|=deg(\omega )\le \phi (m)$. This means $|\mathrm{Gal}(\mathbb{Q}(\omega )/\mathbb{Q})|=\phi (m)=deg(\omega )$, and so $S=$ the set of conjugates of $\omega$.

Corollary:

Note

$\mathbb{Q}(\omega )$ has degree $\phi (m)$ over $\mathbb{Q}$.

Corollary:

Note

$\mathrm{Gal}(\mathbb{Q}(\omega )/\mathbb{Q})$ is isomorphic to the multiplicative group of integers mod $m$. For each $k$, the corresponding automorphism sends $\omega$ to ${\omega }^k$ as in the proof.

Note:

By the fundamental theorem of galois theory, we find that subfields of $\mathbb{Q}(\omega )$ and subgroups of ${\mathbb{Z}}_m^{\ast }$ are in correspondence with each other. In particular, for $p^{th}$ cyclotomic fields, with $p$ an odd prime, there is a unique quadratic field (since ${\mathbb{Z}}_p^{\ast }$ is of order $p-1$ and cyclic). It turns out that this field is in fact $\mathbb{Q}[\sqrt[]{\pm p}]$. The proof follows from Theorem 1 in Gauss sums

Corollary:

Note

Let $\omega =e^{2\pi i/m}$. If $m$ is even, the only roots of $1$ in $\mathbb{Q}[\omega ]$ are the $m^{th}$ roots of unity. If $m$ is odd, the only ones are the $2m^{th}$ roots of unity.

Proof:

The second part follows from the first and the fact that $2m^{th}$ cyclotomic field is equal to the $m^{th}$ one, if $m$ is odd.

To prove the first part, suppose $\theta$ is a primitive $k^{th}$ root of unity in $\mathbb{Q}(\omega )$. Then there is a primitive $r^{th}$ root of unity in $\mathbb{Q}(\omega )$ where $r=lcm(k,m)$. Implying $\varphi (r)\le \varphi (m)$, but since $r=\frac{km}{gcd(k,m)}=k^{\prime }m$ where $gcd(k^{\prime },m)=1$, we get $\varphi (r)=\varphi (k^{\prime })\varphi (m)\le \varphi (m)⟹\varphi (k^{\prime })\le 1⟹k^{\prime }=1\text{or}2$. But $k^{\prime }$ can’t be 2 since it is coprime to $m$, thus $k^{\prime }=1$ giving $r=m$. Hence, $k\mid m$ and $\theta$ is an $m^{th}$ root of unity.

Corollary:

Note

The $m^{th}$ cyclotomic fields, for $m$ even, are all distinct, and pairwise non isomorphic.

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Ring of integers of Cyclotomic fields

Theorem 2:

Note

Let $\omega =e^{2\pi i/m}$, where $m=p^r,p$ a prime. Then $\overline{\mathbb{Z}}\cap \mathbb{Q}(\omega )=\mathbb{Z}[\omega ]$.

Lemma 1:

Note

For $m\ge 3$, $\mathbb{Z}[\omega ]=\mathbb{Z}[1-\omega ]$ and $\mathrm{disc}(\omega )=\mathrm{disc}(1-\omega )$.

Proof:

$\mathbb{Z}[\omega ]=\mathbb{Z}[1-\omega ]$ is clear.

$$\mathrm{disc}(\omega )=\prod _{1\le r<s\le n}({\alpha }_r-{\alpha }_s{)}^2=\prod _{1\le r<s\le n}(1-{\alpha }_r-(1-{\alpha }_s){)}^2=\mathrm{disc}(1-\omega )$$

Lemma 2:

Note

For $m=p^r$, ${\prod }_k(1-{\omega }^k)=p$ where the product is over all $1\le k\le m$ such that $(k,p)=1$

Proof:

Set $f(x)=\frac{x^{p^r}-1}{x^{p^{r-1}}-1}=1+x^{p^{r-1}}+\cdots +x^{(p-1)p^{r-1}}$. Then all ${\omega }^k$ ($k$ as above) are roots of $f$. Thus $f(x)={\prod }_k(x-{\omega }^k)$ since there are exactly $\varphi (p^r)$ values of $k$. Now set $x=1$.

Proof of Theorem:

By theorem 1 of Number Rings, every $\alpha \in R$ can be expressed in the form $$ \alpha = \frac{m_{1} + m_{2}(1-\omega) + \dots m_{n}(1-\omega)^{n-1}}{d}

where $n = \phi(p ^{r})$, and $d = \mathrm{disc}(\omega) = \mathrm{disc}(1-\omega)$. We know that $\mathrm{disc}(\omega) \mid m ^{\phi(m)}$ ([[Discriminant of an n-tuple]]/ related problems/ problem 2), thus $\mathrm{disc}(1-\omega)$ is a power of $p$. Now suppose there is an $\alpha \in R$ such that some $m_{i}$ is not divisible by $d$, then $R$ contains

\beta = \frac{m_{i}(1-\omega)^{i-1} + \dots + m_{n}(1-\omega)^{n-1}}{p}

where $p \nmid m_{i}$ (because, we can take $m_{i}$ to be the one with the least power of $p$, then cancel out all prime powers with $d$ until $p \nmid m_{i}$ and multiply everything by a suitable power of $p$ until only $p$ remains in the denominator, and then remove terms of the form $\mathbb{Z}(1-\omega)^{k}$). Lemma 2 gives $\frac{p}{(1-\omega)^{n}} \in \mathbb{Z}[\omega]$. Then $\frac{p}{(1-\omega)^{i}} \in \mathbb{Z}[\omega]$ and hence $\frac{\beta p}{(1-\omega)^{i}} \in R$. Implying $\frac{m_{i}}{(1-\omega)} \in R$. It follows that $N(1-\omega) | N(m_{i}) \implies p | m_{i}^{\phi(m)}$ a contradiction. Thus $R = \mathbb{Z}[1-\omega] = \mathbb{Z}[\omega]$. ### Theorem 3: > [!note] > > Let $K = \mathbb{Q}(\omega), \omega = e ^{2\pi i/m}, R = \overline{\mathbb{Z}}\cap K$. Then $R = \mathbb{Z}[\omega]$. ###### Proof: We know this holds for $m = p ^{r}$. We induct on the number of distinct prime factors. Let $m = p_{1} ^{r_{1}}p_{2}^{r_{2}}$. Let $\omega_{1} = e ^{2\pi i/p_{1}^{r_{1}}}$ and $\omega_{2} = e ^{2\pi i/p_{2}^{r_{2}}}$, $\omega = e ^{2\pi i/m}$ Then $\mathbb{Q}(\omega) = \mathbb{Q}(\omega_{1})\mathbb{Q}(\omega_{2})$ and $\mathbb{Z}[\omega] = \mathbb{Z}[\omega_{1}]\mathbb{Z}[\omega_{2}]$. Since $d = \mathrm{gcd}(\mathrm{disc}(\mathbb{Z}[\omega_{1}]),\mathrm{disc}(\mathbb{Z}[\omega_{2}])) = \mathrm{gcd}(p_{1}^{t_{1}},p_{2}^{t_{2}}) = 1$, therefore by corollary 1 to theorem 3 in [[Number Rings]], we get $\overline{\mathbb{Z}} \cap \mathbb{Q}(\omega) = \mathbb{Z}[\omega_{1}]\mathbb{Z}[\omega_{2}] = \mathbb{Z}[\omega]$. We can write down a similar argument for the induction step. --- # References [[Number Field]] [[Galois Extensions]] [[The fundamental theorem of Galois Theory]] [[Gauss sums]] [[Number Rings]] [[Discriminant of an n-tuple]]