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 $\mathbb C$.

This is equivalent to $f$ being separable, and so we require

\[\text{gcd}(f, f') = 1 \iff \text{gcd}(x^n - 1, nx^{n-1}) = 1 \iff n \ne 0 \in F\]

$L/F$ is a cyclotomic field extension of $F$ if $L = F(\varepsilon)$ where $\varepsilon \in L$ and $\varepsilon$ is a primitive $n$-th root of unity.

If $\varepsilon = 1$, $L = F$ and the result is trivial. So we may assume that $\varepsilon \ne 1$.

Since $p$ is a prime, $\varepsilon$ generates a cyclic group of order $p$ in $L^\times$. Hence $t^p - 1$ splits completely in $L$ as

\[t^p - 1 = \prod^{p-1}_{i = 0} (t - \varepsilon^i)\]

So $L$ is a splitting field of $t^p - 1$ over $F$. Since $t^p - 1$ has no repeated roots, it is separable.

So $L/F$ is Galois.

To show commutativity, fix some $\sigma \in \text{Gal}(L/F)$. Since $\sigma(\varepsilon)^p = \sigma(1) = 1$, it follows $\sigma(\varepsilon) \in \{1, \varepsilon, \ldots, \varepsilon^{p-1}\}$, i.e. it is another root of unity. Hence $(\sigma \circ \tau)(\varepsilon) = (\tau \circ \sigma)(\varepsilon)$ because the addition in the powers of $\varepsilon$ is commutative.

Let $\chi : \text{Gal}(L/F) \to \mathbb N$ be the function such that

\[\sigma(\varepsilon) = \varepsilon^{\chi(\sigma)}\]

for any $\sigma$. For $\sigma, \tau \in \text{Gal}(L/F)$,

\[\begin{aligned} (\sigma \circ \tau)(\varepsilon) &= \sigma(\varepsilon^{\chi(\tau)}) \\\\ &= (\varepsilon^{\chi(\sigma)})^{\chi(\tau)} \\\\ &= \varepsilon^{\chi(\sigma) \chi(\tau)} \\\\ &= (\tau \circ \sigma)(\varepsilon) \end{aligned}\]

Since $L = F(\varepsilon)$, it follows that $\sigma \circ \tau = \tau \circ \sigma$ for all $\sigma, \tau$ and therefore $\text{Gal}(L/F)$ is abelian.

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

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

$\mathbb Q(\zeta _ n)$ 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 $\mathbb Q$.

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

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

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Quick proof: by definition

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

and hence $\deg \Phi _ n = \vert \lbrace \zeta \in \mu _ n(\mathbb C) :o(\zeta) = n \rbrace \vert = \vert \mathbb (Z / n\mathbb Z)^\times \vert $.

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
• $\prod _ {d \mid n} \Phi _ d = t^n - 1$

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

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

Inductive proof:

We may show it actually lies 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
• $\prod _ {d \mid n} \Phi _ d = t^n - 1$

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

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

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Proof by appealing to results on fixed fields:

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

Then $o(\sigma(\varepsilon)) = n$ (since $\sigma(\varepsilon)^k = \sigma(\varepsilon^k)$ as $\sigma$ is an automorphism).

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 $\mathbb Q[t]$.

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 $f(\zeta _ n) = 0$ (consider that any polynomial over $\mathbb Q$ can be factored into a product of irreducible polynomials; choose $f$ to be the factor that vanishes at $\zeta _ n$ and choose $g$ to be the product of the other factors).

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Let $\varepsilon$ be an arbitrary primitive $n$-th root of $1$ (not necessarily $\zeta _ n$), and let $p \not\mid n$ be a prime. The overall idea is this: show that $f(\varepsilon^p) = 0$, since then we may conclude that actually $f(\zeta _ n^r) = 0$ for any $r$ by factoring $r$ as a product of primes and repeatedly applying this result. This would imply $\Phi _ n = f$, a contradiction to the reducibility of $\Phi$.

To show $f(\varepsilon^p) = 0$, assume for a contradiction that $f(\varepsilon^p) \ne 0$. Since $\varepsilon^p$ is still a primitive $n$-th root of unity, $\Phi _ n(\varepsilon^p) = 0$ and hence $g(\varepsilon^p) = 0$. 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 $\mathbb Q[t]$ as $\mathbb Q(\zeta _ n) / \mathbb Q$ is Galois and so the minimal polynomial of $\zeta _ n$ must consist of conjugates of $\zeta _ n$, among which is $\varepsilon$.

Therefore $k = hf$ for some $h \in \mathbb Q[t]$. Since $k$ and $f$ are both monic with coefficients in $\mathbb Z[t]$, by 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.

we have that $h \in \mathbb Z[t]$. Hence it makes sense to 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 $\mathbb F _ p[t]$. Now 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 $\overline g$.

Since $\overline q$ divides both $f$ and $g$, it follows that $\overline q^2 \mid \overline f \cdot \overline g = \overline{fg} = \overline{\Phi _ n}$ and $\overline{\Phi _ n} \mid \overline{t^n - 1}$, so $\overline{q}^2 \mid \overline{t^n - 1}$. This would imply that $\overline{t^n - 1}$ has multiple roots, but

\[D(\overline{t^n - 1}) = \overline{nt^{n-1} } \ne 0\]

a contradiction. Thus we may conclude $f(\varepsilon^p) = 0$.

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Using this to derive a contradiction: We now wish to use this to show $f(\zeta _ n^r) = 0$ for all $r$ where $r$ is 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 for the rest of the primes in term to deduce that $f(\zeta _ n^{p _ 1 \cdots p _ n}) = 0$ as required.

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

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Note: There is a much shorter proof when $n$ is a prime, by rewriting $x^n - 1$ as $(1 + x + \cdots + x^{n-1})(x - 1)$, making a substitution, and applying Eisenstein’s criterion.

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

\[\begin{aligned} t^n - 1 &= \prod_{1 \le k \le n} (t - e^{2\pi i k/n}) \\\\ &= \prod_{d \mid n} \prod_{1 \le k \le n, \gcd(k,n) =d} (t - e^{2\pi ik/n}) \\\\ &= \prod_{d \mid n} \Phi_d \end{aligned}\]

\[\Phi_n(x) = \frac{x^n - 1}{\prod_{d \mid n, d < n} \Phi_d(n)}\]

and so you may compute $\Phi _ n$ if you know $\Phi _ d$ for each $d \mid n$ with $d < n$.

\[\Phi_{mp}(t) = \Phi_m(t^p)\]

Note that $\deg \Phi _ {mp}(t) = \phi(mp)$ and $\text{deg}(\Phi _ {m}(t^p)) = p\phi(m)$ where $\phi$ is Euler’s totient function. Since $p \mid m$, $m = p^k m’$ for some $k, m’ \in \mathbb Z$ such that $p \not\mid m’$. Then:

\[\begin{aligned} \phi(mp) &= \phi(p^k m' p) \\\\ &= \phi(p^{k+1} m') \\\\ &= \phi(p^{k+1}) \phi(m') \\\\ &= p(p-1)p^{k-1} \phi(m') &&(\star) \\\\ &= p \phi(p^k) \phi(m') \\\\ &= p\phi(p^k m') \\\\ &= p\phi(m) \end{aligned}\]

where $(\star)$ is justified by the result that $\phi(p^{k+1}) = p^{k}(p-1)$.

Recall that

\[x^n - 1 = \prod_{t \mid n} \Phi_{t, \mathbb Q}(x)\]

and by assumption $\overline{\Phi _ {n, \mathbb Q}(k)} = 0$ and $\overline{\Phi _ {d, \mathbb Q}(k)} = 0$, therefore $k$ is a repeated root of the polynomial $x^n - 1$ in $\mathbb F _ p$. Since this is only possible if $(X^n - 1, nX^{n-1}) \ne 1$, it follows that $n = 0 \pmod p$, i.e. $p \mid n$.

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

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).

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.

• 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 $n$.