Abelian extensions

Overview

In these notes, we assume $F$ is a field of characteristic zero.

Abelian extension of a field

If $E$ is a finite extension of a field $F$, then $E$ is an abelian extension if $E$ is a Galois extension (i.e., both normal and separable) and the Galois group $\text{Gal}(E/F)$ is an abelian group.

Likewise, if $f\in F[x]$ is a nonconstant polynomial, then we say the Galois group of $f$ is abelian if the Galois group of its splitting field over $F$ is abelian.

Importantly, since every subgroup of an abelian group is a normal subgroup, it follows from (Theorem) Fundamental theorem of Galois theory that every subgroup of $\text{Gal}(E/F)$ is normal, and hence every intermediate field $K$ such that $F\le K\le E$ is a Galois extension of $F$ (as well as itself an abelian extension). In particular, for all $\alpha \in E$, the simple extension $F(\alpha )$ is normal, and hence the minimal polynomial $\text{irr}(\alpha ,F)$ splits into linear factors in $F(\alpha )[x]$; in other words, once one root of $\text{irr}(\alpha ,F)$ has been adjoined to $F$, all of its other roots are expressible in terms of $\alpha$.

The two main examples of abelian extensions are cyclotomic extensions—roughly, those obtained using roots of unity—and $n$th root extensions—those for which the Galois group is isomorphic to a subgroup of $\mathbb{Z}/n\mathbb{Z}$ (equivalently, have order dividing $n$).

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Cyclotomic extensions

Proposition

Let $E$ be a splitting field of $x^n-1$ over $F$ and let ${\mu }_n(E)$ denote the set of roots of $x^n-1$ in $E$. Given a generator $\zeta$ for the cyclic group ${\mu }_n(E)\le E^{\ast }$, we have $E=F(\zeta )$ and $\text{Gal}(E/F)$ is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. In particular, $\text{Gal}(E/F)$ is abelian.

Proof from Modern Algebra II. By the definition of a splitting field, $f=x^n-1$ factors into linear factors in $E$. We first show that all of these factors are distinct: since $F$ has characteristic zero, the roots of $Df=n{x}^{-1}$ are exactly $0$. But $0$ is not a root of $f$, so $f,Df$ do not share a root and by the criteria in Formal derivatives on polynomial rings, $f$ does not have multiple roots. Thus, the set ${\mu }_n(E)$ of $n$th roots of unity is a subgroup of $E^{\ast }$ with order $n$.

We also know that ${\mu }_n(E)\le E^{\ast }$ is a finite subgroup, hence cyclic by (Theorem) Existence of a primitive root. Then for any generator $\zeta \in {\mu }_n(E)$, we know that $F(\zeta )$ is a subfield of $E$ containing all the roots of $f=x^n-1$. But by the definition of a splitting field again, $E$ is generated over $F$ but the roots of $f$, so $E$ is the minimal such subfield and hence $F(\zeta )=E$. This proves the first part of the claim.

Next, we show that if $\zeta$ is a generator for ${\mu }_n(E)$, then $\sigma (\zeta )$ is also a generator for all $\sigma \in \text{Gal}(E/F)$. Let $\sigma \in \text{Gal}(E/F)$, so $\sigma (\zeta )$ is also a root of $x^n-1$ and hence $\sigma (\zeta )={\zeta }^i$ for some $i$. Then setting $d=\text{gcd}(i,n)$, we must have $({\zeta }^i{)}^{n/d}=1$, and hence ${\zeta }^i$ is a root of $f(x^{1/d})=x^{n/d}-1$. But this implies that $\zeta ={\sigma }^{-1}({\zeta }^i)$ is also a root of $x^{n/d}-1$, and since $\zeta$ has order $n$ in ${\mu }_n(E)$, this means $n/d\ge n$. Thus $d=\gcd (i,n)=1$ and $i$ is relatively prime to $n$.

To identify $\text{Gal}(E/F)$ with a subgroup of $(\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$, define the map $\phi :\text{Gal}(E/F)\to (\mathbb{Z}/n\mathbb{Z}{)}^{\ast },\sigma (\zeta )\mapsto {\zeta }^{\phi (\sigma )}$

where $\phi (\sigma )$ is an integer $i$ which we can take $\mathrm{mod}n$. The fact that $\text{gcd}(i,n)=1$ means that $\phi (\sigma )=i$ is a well-defined element of $(\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. It is straightforward to check that $\phi$ is a homomorphism, i.e., that $\phi (\sigma \circ \sigma ’)=\phi (\sigma )\phi (\sigma ’)$, using the fact that $(\sigma \circ \sigma ’)(\zeta )={\zeta }^{\phi (\sigma )\phi (\sigma ’)}$ for all $\sigma ,\sigma ’\in \text{Gal}(E/F)$. Finally, $\phi$ is injective since $\phi (\sigma )=1⟹\sigma (\zeta )=\zeta ⟹\sigma ={\text{id}}_E,$

and thus $\ker \phi =\{1\}$ is trivial. Thus, $\phi$ embeds $\text{Gal}(E/F)$ as a subgroup, as claimed. $\square$

Primitive $n$th root of unity

Let $E$ is a splitting field of $x^n-1$ and ${\mu }_n(E)\le E^{\ast }$ be the set of roots of $x^n-1$ in $E$. A primitive $n$th root of unity is a generator of ${\mu }_n(E)$.

Remark

• Since (Theorem) Existence of a primitive root implies that ${\mu }_n(E)$ is cyclic of order $n$, there are exactly $\phi (n)$ generators of ${\mu }_n(E)$, where $\phi$ is the Euler totient function. Thus, there are exactly $\phi (n)$ primitive $n$th roots of unity.
• The cyclotomic polynomial ${\Phi }_n={\prod }_{\zeta \in {\mu }_n(E),\zeta \text{ primitive}}(x-\zeta )$ therefore splits into $\phi (n)$ linear factors, showing that $\mathrm{deg}{\Phi }_n=\phi (n)$.
• The proof above shows that $\text{Gal}(E/F)$ permutes the primitive $n$th roots of unity. In particular, the roots of ${\Phi }_n$ are permuted by $\text{Gal}(E/F)$, so it follows from the definition of ${\Phi }_n$ that $\sigma ({\Phi }_n)={\Phi }_n,\sigma \in \text{Gal}(E/F).$
• Following the proof, since $\text{Gal}(E/F)$ is identified with a subgroup of $(\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$, by (Theorem) Lagrange we know that the order of $\text{Gal}(E/F)$ will divide $\phi (n)$ in general. However, we have $\text{Gal}(E/F)\cong (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }⟺\#(\text{Gal}(E/F))=\phi (n)⟺{\Phi }_n\text{ irreducible in }F[x].$

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$n$th root extensions

Proposition

Let $n>0$ and suppose $x^n-1$ has $n$ distinct roots in $F$, i.e., that $\#({\mu }_n(F))=n$. Let $a\ne 0\in F$ and $E$ be a splitting field for $x^n-a$. If $\alpha =\sqrt[n]{a}$ is a root of $x^n-a$ in $E$, then every root of $x^n-a$ is of the form $\zeta \alpha$ for some $\zeta \in {\mu }_n(F)$, i.e., a root of $x^n-1$. In this case,

$$x^n-a=\prod _{\zeta \in {\mu }_n(F)}(x-\zeta \alpha ),$$

and $E=F(\alpha )=F(\sqrt[n]{a})$. Moreover, $\text{Gal}(E/F)$ is abelian and isomorphic to a subgroup of ${\mu }_n(F)\cong \mathbb{Z}/n\mathbb{Z}$.

In this situation, we have $x^n-a\text{ irreducible}⟺[E:F]=n⟺\#(\text{Gal}(E/F))=n⟺\text{Gal}(E/F)\cong \mathbb{Z}/n\mathbb{Z}.$

Proof from Modern Algebra II. We first show that the splitting field is of the form $E=F(\alpha )$, where $\alpha =\sqrt[n]{a}$ is a root of $x^n-a$ in $E$. Since ${\mu }_n(F)\subseteq F\subseteq E$, if root $\alpha \in E$ is a root of $x^n-a$, then so is $\zeta \alpha$ for all $n$th roots of unity $\zeta \in {\mu }_n(F)$. Then there are at least $n$ distinct linear factors $(x-\zeta \alpha )$, and hence

$$\prod _{\zeta \in {\mu }_n(F)}(x-\zeta \alpha )$$

divides $x^n-a$ in $E[x]$. However, since both sides are monic of degree $n$, they must be exactly equal. This shows that $x^n-a$ splits into linear factors in $F(\alpha )$, so $F(\alpha )$ must be a splitting field for $x^n-a$ and thus $F(\alpha )\cong E$.

We now show the inclusion of $\text{Gal}(E/F)$ into ${\mu }_n(F)\cong \mathbb{Z}/n\mathbb{Z}$. Recall that the Galois group for the splitting field $E$ acts transitively on the roots of $x^n-a$. Thus for any $\sigma \in \text{Gal}(E/F)$, the image $\sigma (\alpha )$ is just another root of $x^n-a$ and hence $\sigma (\alpha )=\zeta (\alpha )$ for some uniquely specified $\zeta \in {\mu }_n(F)$ (since $\alpha$ assume nonzero). Then the map

$$\phi :\text{Gal}(E/F)\to {\mu }_n(F),\sigma \mapsto \sigma (\alpha )/\alpha$$

sending each $\sigma$ to the unique $\zeta \in {\mu }_n(F)$ such that $\sigma (\alpha )=\phi (\alpha )\cdot \alpha$, is well-defined. We can check that $\phi$ is a homomorphism using the fact that for all $\sigma ,\sigma ’\in \text{Gal}(E/F)$, we have

$$\sigma (\phi ({\sigma }^{\prime })\cdot \alpha )=\phi ({\sigma }^{\prime })\sigma (\alpha ),$$

since $\phi (\sigma ’)\in {\mu }_n(F)\subseteq F$ and hence is fixed. Finally, $\phi$ is injective since $E=F(\alpha )$ implies that

$$\sigma \in \ker \phi ⟺\sigma (\alpha )=\alpha ⟺\sigma ={\text{id}}_E,$$

and we conclude that $\text{Gal}(E/F)$ is isomorphic to a subgroup of ${\mu }_n(F)\cong \mathbb{Z}/n\mathbb{Z}$. $\square$

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Code snippets

(\mathbb Z/n \mathbb Z)^*