Subjects: Field Theory
Links: Field Extensions, Separable Field Extensions, Characteristic of a Ring, Splitting Fields and Normal Field Extensions

If $K$ is a field, let $L$ be the splitting field of the polynomial $x^n-1\in K[x]$. If $K=\mathbb{Q}$, the polynomial is separable, but if $K$ has characteristic $p>0$, and $n=p^{r}q$ with $p\nmid q$, then $$x^n-1 = (x^q-1)^{p^r}\in K[x], $$and the splitting field of $x^q-1$ is the same as the splitting field of $x^n-1$. With this in mind, we can focus on the splitting fields of $x^n-1$ with $p\nmid n$. Let suppose that the characteristic of $K$ doesn't divide $n$ and $L$ be the splitting field of $x^n-1$. The $n$th roots of unity form a finite group, ${\mu }_n\le L^{\ast }$, this must be a cyclic group of order $n$. A generator of ${\mu }_n$ is called a primitive $n$th root of unity. If $\omega$ is a primitive $n$th root of unity, then $L=K(\omega )$. We call the fields $L=K(\omega )$ is called a cyclotomic field.

We see that every primitive $n$th root of unity can be used to generate $L$, if we consiser the all of the primitive roots of unity, the polynomial that they define is $$\Phi_n(x) :=\prod_{\omega}(x-\omega)\in L[x],$$where the product is over all the primitive $n$th roots of unity, and this polynomial is known as the $n$th cyclotomic polynomial.

Prop: Let $K$ be a field and suppose that the characteristic of $K$ doesn't divide $n$. If $K_0$ is its prime field of $K$, then ${\mathrm{\Phi }}_n(x)\in K_0[x]$.

Th: Let $K$ be a field and suppose that the characteristic of $K$ doesn't divide $n$. Let ${\mathrm{\Phi }}_n$ the $n$th cyclotomic polynomial over $K$, and let $L$ be the splitting field of ${\mathrm{\Phi }}_n$ over $K$, then

• The extensions $L/K$ is finite and Galois, and $L=K(\omega )$, where $\omega$ is an primitive $n$th root of unity.
• The group $\text{Gal}(L/K)$ is abelian, and is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$.
• The polynomial ${\mathrm{\Phi }}_n$ is irreducible over $K$ iff $\text{Gal}(L/K)\cong (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$.

Prop: Let $p$ be a prime and $n$ be a positive integer such that $p\nmid n$.

• If $\omega$ is a primitive $n$th root of unity, then $[{\mathbb{F}}_p(\omega ):{\mathbb{F}}_p]=e$, where $e$ is the order of $p$ in $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$.
• ${\mathrm{\Phi }}_n$ is irreducible over ${\mathbb{F}}_p$ iff $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ is a cyclic group with $p$ as a generator.

The proposition above tells us that the there must be a primitive root modulo $n$, so must be of the form $q^k$, $2q^k$, $2$ and $4$.

Cyclotmic Extensions over $\mathbb{Q}$

Def: Let ${\mu }_n$ denote the group of $n$th roots of unity over $\mathbb{Q}$.

Def: A generator of the cyclic group ${\mu }_n$ is called a primitive $n$th roots of unity. Let ${\zeta }_n$ be the a primitive $n$th root of unity.

Obs: Suppose $m$ and $n$ are relatively prime positive integers. Then ${\zeta }_m{\zeta }_n$ is a primitive $mn$th root of unity. If ${\zeta }_n$ is a primitive $n$th root of unity, and $d$ is divisor of $n$, then ${\zeta }_n^d$ is a primitive $(n/d)$th root of unity.

Def: We define the $n$th cyclotomic polynomial ${\mathrm{\Phi }}_n(x)$ to be the polynomial whose roots are the primitive $n$th roots of unity $$\Phi_n(x) := \prod_{\zeta \text{ primitive }\in \mu_n}(x-\zeta) = \prod_{\substack{1 \le a <n \ (a, n) = 1}} (x-\zeta_n^a),$$which has degree $\phi (n)$, where $\phi (n)$ is the Euler Totient Function.

Obs: Let $n$ be a positive integer, then $$x^n-1 = \prod_{d \mid n}\Phi_d(x).$$This is equivalent to $$n = \sum_{d\mid n}\phi(d) $$
Lemma: The cyclotomic polynomial ${\mathrm{\Phi }}_n(x)$ is a monic polynomial in $\mathbb{Z}[x]$ of degree $\phi (n)$.

Prop: The cyclotomic polynomial ${\mathrm{\Phi }}_n$ is an irreducible monic polynomial in $\mathbb{Z}[x]$ of degree $\phi (n)$.

Prop: If $n$ is an odd integer, then ${\mathrm{\Phi }}_{2n}(x)={\mathrm{\Phi }}_n(-x)$.

Def: The field $\mathbb{Q}({\zeta }_n)/\mathbb{Q}$ is called the cyclotomic field of $n$th roots of unity.

Cor: The degree over $\mathbb{Q}$ of the cyclotomic field of $n$th roots of unity is $\phi (n)$: $$[\Bbb Q(\zeta_n) : \Bbb Q] = \varphi(n). $$
Prop: If $K/\mathbb{Q}$ is a finite field extension, then $K$ has only a finite number of roots of unity.

Prop: We can use the Möbius inversion formula to get the following formula: $$\Phi_n (x):= \prod_{d \mid n }(x^d-1)^{\mu(n/d)}.$$
Prop: Let $\ell$ be a prime, and let ${\mathrm{\Phi }}_{\ell }(x)=(x^{\ell }-1)/(x-1)\in \mathbb{Z}[x]$. We want to determine the factorisation of ${\mathrm{\Phi }}_{\ell }(x)$ modulo $p$ for any prime $p$. Let $\zeta$ denote any fixed primive $\ell$th root of unity.

• If $p=\ell$, then ${\mathrm{\Phi }}_{\ell }(x)=(x-1{)}^{\ell -1}\in (\mathbb{Z}/\ell \mathbb{Z})[x]$.
• Suppose $p\ne \ell$, and let $f$ denote the order of $p\mathrm{mod}\ell$. The smallest power of $p$ with $p^f\equiv 1(\mathrm{mod}\ell )$. We know that ${\mathbb{F}}_{p^n}$ is cyclic, then $n=f$ is the smallest power $p^n$ of $p$ with $\zeta \in {\mathbb{F}}_{p^n}$. This means that the minimal polynomial of $\zeta$ over $(\mathbb{Z}/p\mathbb{Z})[x]$ has degree $f$.
• We see that $(\mathbb{Z}/p\mathbb{Z})(\zeta )=(\mathbb{Z}/p\mathbb{Z})({\zeta }^a)$ for any $a$ that is not divisible by $\ell$. Finally, we see that ${\mathrm{\Phi }}_{\ell }(x)$ is the product of $(\ell -1)/f$ distinct irreducible polynomials of degree $f$in $(\mathbb{Z}/p\mathbb{Z})[x]$.

We can get a weakened version of Dirichlet's Theorem on Primes in Arithmetic Progressions, but without the machinery that are Dirichlet L-functions.

Lemma: Given any monic polynomial $p(x)\in \mathbb{Z}[x]$ of degree at least one there are infinitely many distinct prime divisors of the integers $${p(n) \mid n \in \Bbb N^+}.$$

Lemma: Let $p$ be a prime and $m$ an integer such that $p\nmid m$. If $a$ is an integer such that ${\mathrm{\Phi }}_m(a)\equiv 0(\mathrm{mod}p)$, then $(a,p)=1$, and the order of $a$ in $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$ is $m$.

Cor: Let $a\in \mathbb{Z}$. If $p$ is an odd prime that divides ${\mathrm{\Phi }}_m(a)$, then either $p$ divides $m$ or $p\equiv 1(\mathrm{mod}m)$.

Th: For any positive integer $m$ there are infinitely many primes $p$ with $p\equiv 1(\mathrm{mod}m)$.

Th: The Galois group of the cyclotomic field $\mathbb{Q}({\zeta }_n)$ of $n$th roots of unity is isomorphic to the multiplicative group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. The isomorphism is given explicitly by the map $a(\mathrm{mod}n)\mapsto {\sigma }_a$, where ${\sigma }_a({\zeta }_n)={\zeta }_n^a$.

Obs: A basis for $\mathbb{Q}({\zeta }_p)$ over $\mathbb{Q}$ is given by $$\zeta_p, \zeta_p^2, \dots, \zeta_p^{p-2},\zeta_p^{p-1}.$$The reason for choosing this basis is that any $\sigma$ in the Galois group $\text{Gal}(\mathbb{Q}({\zeta }_p)/\mathbb{Q})$ simply permutes these basis elements since these are precisely the primitive $p$ roots of unity. This where we need that $p$ is prime. Let $H$ be any subgroup of the Galois group of $\mathbb{Q}({\zeta }_p)$ over $\mathbb{Q}$ and let $$\alpha_H = \sum_{\sigma\in H} \sigma(\zeta_p), $$the sum of the conjugates by the elements in $H$. We can prove that the fixed field of $H$ is $\mathbb{Q}(\alpha )$.

Cor: Let $n=\prod _{k=0}^{\mathrm{\infty }}{p}_k^{{\alpha }_k}$ be the prime decomposition of the positive integer $n$ into distinct prime powers. Then the cyclotomic fields $\mathbb{Q}({\zeta }_{p_k^{{\alpha }_k}})$, $k\in \mathbb{N}$ intersect only in the field $\mathbb{Q}$ and their composite is the cyclotomic field $\mathbb{Q}({\zeta }_n)$. We have $$\text{Gal}(\Bbb Q(\zeta_n)/\Bbb Q) \cong \prod_{k = 0}^\infty \text{Gal}(\Bbb Q(\zeta_{p_k^{\alpha_k}})/\Bbb Q) $$which by the we get the following isomorphism, equivalent to the Chinese Remainder Theorem$$(\Bbb Z/n \Bbb Z)^\times \cong \prod_{k = 0}^\infty (\Bbb Z/p_k^{\alpha_k}\Bbb Z)^\times $$
Th: Let $G$ be any finite abelian group. Then there is a subfield $K$ of a cyclotomic field with $\text{Gal}(K/\mathbb{Q})\cong G$.

$(\ast )$ Kronecker-Weber Theorem: Let $K$ be a finite abelian extension of $\mathbb{Q}$. Then $K$ is contained in a cyclotomic extension of $\mathbb{Q}$.

Prop: Let ${\zeta }_n$ denote the primitive $n$th root of unity and let $K=\mathbb{Q}({\zeta }_n)$ be the associated cyclotomic field. We see that ${\text{Tr}}_{K/\mathbb{Q}}({\zeta }_n)=\mu (n),$ where $\mu$ is the Möbius function, and ${\text{Tr}}_{K/\mathbb{Q}}$ is the trace of an element.

Prop: The primitive $n$th roots of unity form a basis over $\mathbb{Q}$ for the cyclotomic field of $n$th roots of unity iff $n$ is squarefree.

Def: We see that complex conjugation restricts to the automorphism ${\sigma }_{-1}\in \text{Gal}(\mathbb{Q}({\zeta }_n)/\mathbb{Q})$ of the cyclotomic field of $n$th roots of unity. We see that $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$ is a subfield of the real elements of $\mathbb{Q}({\zeta }_n)$. We call this field the maximal real subfield of $\mathbb{Q}({\zeta }_n)$.