Notes - Galois Theory HT25, Cyclotomic extensions

$\zeta$ has order precisely $n$ in the multiplicative group $K^\times$.

\[\mu_n(K) := \\{\zeta \in K : \zeta^n = 1\\}\]

It is a cyclic group of order $n$, and is generated by the primitive $n$-th roots of unity.

Consider a field $F$ of characteristic $p$. Then if $\zeta^p = 1$ for some $\zeta \in F$, it follows $\zeta^p - 1 = (\zeta - 1)^p = 0$ and hence $\zeta = 1$.

\[\Phi_n := \prod_{\zeta \in \mu_n(\mathbb C) :o(\zeta) = n} (t - \zeta) \in \mathbb C[t]\]

i.e. the monic polynomial whose roots are the primitive $n$-th roots of $1$ in $\mathbb C$.

\[\mathbb Q(\zeta_n) \subseteq \mathbb C\]

where $\zeta _ n = e^{\frac{2\pi i}{n} }$.

\[\deg \Phi_n = |(\mathbb Z/n\mathbb Z)^\times| = \phi(n)\]

where $\phi$ is Euler’s totient function.

$\mathbb Q(\zeta _ n)$ is the splitting field of $t^n - 1$ containing $\mathbb Q$, since $t^n - 1$ splits completely in $\mathbb Q(\zeta _ n)$ and is generated by its roots. Hence it is a Galois extension of $\mathbb Q$.

\[\chi_n : \Gamma_n \to (\mathbb Z / n \mathbb Z)^\times\]

where $\chi _ n(\sigma)$ for $\sigma \in \Gamma _ n$ is determined by

\[\sigma(\zeta_n) = \zeta_n^{\chi_n(\sigma)}\]

(i.e. each element of $\Gamma _ n$ is a permutation that shuffles the primitive roots of unity around. The $n$-th cyclotomic character tells you what power to raise $\zeta _ n$ in order to “mimic” the effect of the permutation).

Suppose that $\varepsilon \in \mathbb Q(\zeta _ n)$ and $o(\varepsilon) = n$ (i.e. it is a primitive $n$-th root). Consider some $\sigma \in \Gamma _ n$.

Then $o(\sigma(\varepsilon)) = n$. Why? $\sigma$ is some permutation that shuffles the primitive roots of unity around, so since $\varepsilon$ is a primitive root of unity, it must get mapped to another primitive root of unity (@todo, is this right? the notes are vague here).

Therefore the set of primitive $n$-th roots of unity is $\Gamma _ n$-stable, and so $\sigma$ permutes the linear factors of $\Phi _ n$ for all $\sigma \in \Gamma _ n$.::

But then $\Phi _ n \in \mathbb Q(\zeta _ n)^{\Gamma _ n}[t]$ by the theorem that says:

If:

• $H$ is a finite group of automorphisms of a field $L$
• $X \subseteq L$
• $X$ is $H$-stable Then the polynomial $f _ X := \prod _ {y \in X} (t - y) \in L[t]$ is always separable and has coefficients in $L^H$.

But then by the Galois correspondence,

\[\mathbb Q(\zeta_n)^{\Gamma_n} = \mathbb Q(\zeta_n)^{\text{Gal}(\mathbb Q(\zeta_n) / \mathbb Q)} \cong \mathbb Q\]

and so $\Phi _ n$ actually lives in $\mathbb Q[t]$.

\[\prod_{d \mid n, d \ne n} \Phi_d = t^n - 1\]

Write

\[\begin{aligned} h &= a_0 + a_1 t + \cdots + a_{m-1} t^{m-1} \\\\ f &= b_0 + b_1 t + \cdots + b_{n-1} t^{n-1} \\\\ k &= c_0 + c_1 t + \cdots + c_{m + n - 1} t^{m + n} \end{aligned}\]

where

\[\begin{aligned} a_0, \ldots, a_{m-1} &\in \mathbb Q \\\\ b_0, \ldots, b_{n-1} &\in \mathbb Z \\\\ c_0, \ldots, c_{n + m - 1} &\in \mathbb Z \end{aligned}\]

Then for $0 \le j \le m$, we have that

\[c_{n + j} = a_j b_n + a_{j+1} b_{n-1} + \cdots + a_{m-1} b_{n+j+1 - m} + a_m b_{n + j - m}\]

Since $h$ is monic, we have $a _ m = 1$.

Let $0 \le j < m$, and assume inductively that $a _ i \in \mathbb Z$ for $I > j$. Since $f$ is monic, $b _ n = 1$, and so

\[a_j = c_{n + j} - a_{k+1}b_{n-1} - \cdots - a_{m-1} b_{n+j + 1 - m} - a_m b_{n + j - m} \in \mathbb Z\]

as required. Hence $h \in \mathbb Z[t]$.

We have that:

• $\Phi _ n \in \mathbb Q[t]$
• If $k, f \in \mathbb Z[t]$ are monic polynomials, $h \in \mathbb Q[t]$, and $k = hf$ then $h \in \mathbb Z[t]$ also
• $\prod _ {d \mid n, d\ne n} \Phi _ d = t^n - 1$

Induct on $n$. Let $k = t^n - 1 \in \mathbb Z[t]$, $h \in \Phi _ n \in \mathbb Q[t]$ and $f = \prod _ {d \mid n, d\ne n} \Phi _ d = t^n - 1$.

By induction $f \in \mathbb Z[t]$ and $f$ is monic. Then $k = hf$, so we have $h = \Phi _ n \in \mathbb Z[t]$.

Suppose that $\Phi _ n$ is not irreducible over $\mathbb Q$. Then it is not irreducible over $\mathbb Z$ by Gauss’ lemma. So it suffices to show that $\Phi _ n$ is irreducible over $\mathbb Z$.

Assume, for a contradiction, that $\Phi _ n = fg$ for some monic $f, g \in \mathbb Z[t]$ of degree $\ge 1$. We can assume that $f$ is irreducible over $\mathbb Q$ and that $f(\zeta _ n) = 0$ (the first assumption comes from the definition of irreducibility, and for the second assumption, @todo).

Let $\varepsilon$ be a primitive $n$-th root of $1$, and let $p \not\mid n$ be a prime. We want to show that $f(\varepsilon^p) = 0$, since this can be used for a contradiction later.

To show this, assume for a contradiction that $f(\varepsilon^p) \ne 0$. Since $\varepsilon^p$ is still a primitive $n$-th root of unity, $\Phi _ n(\varepsilon^p) = 0$. So $g(\varepsilon^p) = 0$. Define

\[k(t) := g(t^p) \in \mathbb Z[t]\]

Then $k(\varepsilon) = g(\varepsilon^p) = 0$. Since $f$ is irreducible over $\mathbb Q$, it is equal to $m _ {\mathbb Q, \zeta _ n}$ and must hence divide $k$ in $\mathbb Q[t]$.

So $k = hf$ for some $h \in \mathbb Q[t]$.

Since $k$ and $f$ are both monic, the result that says

If $k, f \in \mathbb Z[t]$ are monic polynomials, $h \in \mathbb Q[t]$, and $k = hf$ then $h \in \mathbb Z[t]$ also.

implies that $h \in \mathbb Z[t]$. Reduce $k = hf$ modulo $p$ to obtain

\[\overline k (t) = \overline{g(t^p)} = \overline{g(t)}^p\]

and hence

\[\overline{hf} = \overline g^p\]

in $\mathbb F _ p[t]$.

Let $\overline q$ be any irreducible factor of $\overline f$ in $\mathbb F _ p[t]$. Then $\overline q$ divides $\overline g^p$ and therefore also $\overline g$.

But then $\overline q^2 \mid \overline f \overline g = \overline{fg} = \overline{\Phi _ n}$ and $\overline{\Phi _ n} \mid t^n - 1$.

Hence $q$ divides the formal derivative

\[D(t^n - 1) = nt^{n-1}\]

as well as $t^n - 1$.

Since $p \not\mid n$, it implies $q$ divides $1$, a contradiction. Hence $f(\varepsilon^p) = 0$ after all.

Now let $r$ be a positive integer coprime to $n$. Write

\[r = p_1 \cdots p_s\]

for primes $p _ 1, \ldots, p _ s$ coprime to $n$. By the above

\[f(\zeta_n) = 0 \implies f(\zeta_n^{p_1}) = 0\]

Since $\zeta _ n^{p _ 1}$ is still a primitive $n$-th root of unity, we can repeat the same argument to deduce that

\[f(\zeta_n^{p_1 p_2}) = 0\]

Continuing this, we see that

\[f(\zeta_n^r) = 0\]

So $f$ vanishes at all primitive $n$-th roots of unity and $\Phi _ n \mid f$. But then $\deg g = 0$, a contradiction.

Previous results say that it is a homomorphism.

$\chi _ n$ is injective: If $\chi _ n(\sigma) = 1$, then $\sigma(\zeta) = \zeta$ and so $\sigma = 1$.

$\chi _ n$ is surjective: By the result that says

For all $\alpha$ algebraic over $F$, we have $[F(\alpha) : F] = \deg m _ {F, \alpha}(t)$

it follows that:

\[|\Gamma_n| = [\mathbb Q(\zeta_n) : \mathbb Q] = \deg m_{\mathbb Q, \zeta_n}\]

Since $\Phi _ n$ is monic and irreducible over $\mathbb Q$ and since $\Phi _ n(\zeta) = 0$, it follows that

\[m_{\mathbb Q, \zeta_n} = \Phi_n\]

Therefore

\[\begin{aligned} |\text{Im}(\chi_n)| &= |\Gamma_n| \\\\ &= [\mathbb Q(\zeta_n) : \mathbb Q] \\\\ &= \deg m_{\mathbb Q, \zeta_n} \\\\ &= \deg \Phi_n \\\\ &= \phi(n) \\\\ &= |(\mathbb Z / n \mathbb Z)^\times| \end{aligned}\]

and hence $\chi _ n$ is surjective.

It follows that it is an isomorphism.

\[\text{Gal}(K / \mathbb Q)\]

is an abelian group.

• Consider $\text{Gal}(\mathbb Q(\zeta _ n) / \mathbb Q)$ where $\zeta _ n$ is a primitive $n$-th root of unity
• Consider the cyclotomic character $\chi _ n : \text{Gal}(\mathbb Q(\zeta _ n) / \mathbb Q) \to (\mathbb Z / n\mathbb Z)^\times$.
• This is an isomorphism, so $\text{Gal}(\mathbb Q(\zeta _ n) / \mathbb Q) \cong (\mathbb Z / n\mathbb Z)^\times$.

Let $K / \mathbb Q$ be an abelian Galois extension. Then $K$ embeds into $\mathbb Q(\zeta _ n)$ for some positive integer $n$.