7 Cyclic extensions

Definition 7.1 A field extension $L/K$ is called a cyclic extension of degree $n$ if it is Galois and has Galois group $\Gal(L/K)\cong\mathbb{Z}/n$.

In the case where $K$ contains a primitive $n$-th root of unity, we can completely classify these. This is the first step in understanding abelian extensions and is (part of) a description called Kummer Theory.

To warm up, let’s first consider the case where $n=p$ is prime. Suppose a field $K$ contains a primitive $p$-th root of unity $\zeta_p$. (Recall this can’t happen if $\ch(K)=p$.) Given $a\in K$ which is not a $p$-th power of an element of $K$, define $L=K(\theta)$ where $\theta$ is a root of $x^p-a\in K[x]$. Then clearly $[L:K]\leq p$. Also, $x^p-a$ splits completely in $L$ as the roots are $\theta,\,\zeta_p\theta,\,...,\,\zeta_p^{p-1}\theta$. As these are distinct, the extension $L/K$ is Galois. Define an automorphism $\sigma\in\Gal(L/K)$ by $\sigma(\theta)=\zeta_p\theta$. Then $\sigma^k(\theta)=\zeta_p^k\,\theta$ and we have $p$ distinct $K$-automorphisms $\sigma^k$ for $0\leq k\leq p-1$. Thus $[L:K]=|\Gal(L/K)|\geq p$ and we deduce $\Gal(L/K)$ has order $p$. In particular, it must be cyclic.

Constructing cyclic extensions of arbitrary degree $n$ is not much more difficult, it’s just that $n$ might have other non-trivial divisors.

Theorem 7.1 Suppose $K$ contains a primitive $n$-th root of unity and $a\in K^\times$ is not a $d$-th power of an element of $K$ for any divisor $d\mid n$ with $d>1$. Then $L=K(\sqrt[n]{a})$ is a cyclic extension of $K$ of degree $n$. In particular, $x^n-a$ is irreducible in $K[x]$.

Proof. Let $\theta\in L$ be a given root of $x^n-a$ and $\zeta_n\in K$ be a primitive $n$-th root of unity. The set of roots of $x^n-a$ is exactly $\zeta_n^k\theta$ for $0\leq k\leq n-1$. Since these are distinct and contained in $L=K(\theta)$, the extension $L/K$ is normal and separable, hence Galois with $[L:K]=n$.

Given any $\sigma\in G=\Gal(L/K)$, the element $\sigma(\theta)\in L$ is also a root of $x^n-a$, hence $\sigma(\theta)=\zeta_n^{k_\sigma}\theta$ for some $0\leq k_\sigma\leq n-1$. We remark that $\sigma(\zeta_n)=\zeta_n$ since $\zeta_n\in K$.

Consider the map $\pi:G=\Gal(L/K)\longrightarrow\mathbb{Z}/n$ defined by $\sigma\mapsto k_\sigma\bmod n$. We need to check that $\pi$ is an isomorphism of groups. Suppose we are given $\sigma_1,\sigma_2\in G$ where $\sigma_1(\theta)=\theta\zeta_n^{k_1}$ and $\sigma_2(\theta)=\theta\zeta_n^{k_2}$. Then \[\sigma_1\sigma_2(\theta)=\sigma_1\left(\zeta_n^{k_2}\theta\right)=\zeta_n^{k_2}\sigma_1(\theta)=\zeta_n^{k_1+k_2}\theta\] and so $\pi(\sigma_1\sigma_2)=k_1+k_2\bmod n=\pi(\sigma_1)+\pi(\sigma_2).$ (Note we write the group $\mathbb{Z}/n$ additively.) Thus $\pi$ is a group homomorphism.

We next show $\pi$ is a surjection. The image of $\pi$ is a subgroup of $\mathbb{Z}/n$ so is cyclic. In fact, it must be of the form \[ \pi(G)=\{\,0,\,d,\,2d,\,...,\,(m-1)d\,\}\qquad\text{where $d\mid n$ and $md=n$}\] and we want to show that $d=1$.

Now, for any $\sigma\in G$, we see $k_\sigma$ is a multiple of $d$ and so $mk_\sigma$ is a multiple of $md=n$. That means \[\sigma(\theta^m)=\sigma(\theta)^m=\left(\zeta_n^{k_\sigma}\theta\right)^m=\zeta_n^{mk_\sigma}\theta^m=\theta^m.\] In particular, the element $\theta^m$ is fixed every $\sigma\in G$ and so $\theta^m\in L^G=K$, say $\theta^m=b\in K$. But now \[a=\theta^n=\theta^{md}=b^d\] and the conditions in the Theorem imply that $d=1$.

Finally, since $\pi$ is a surjective homomorphism between groups of the same order \[|\Gal(L/K)|=[L:K]=n=|\mathbb{Z}/n|\] we conclude it must be an isomorphism.

Example 7.1 $(a)\;$ Consider the extension $\mathbb{F}_5(\sqrt[4]{2})/\mathbb{F}_5$. Since $\mathbb{F}_5^{\,\times}$ is cyclic of order $4$, we know $\mathbb{F}_5$ contains a primitive $4$-th root of unity (either $2$ or $3$ works). Furthermore, since $2$ is not a square in $\mathbb{F}_5$, the above Theorem implies that $\mathbb{F}_5(\sqrt[4]{2})/\mathbb{F}_5$ is a cyclic extension of degree $4$. We also deduce that the polynomial $f(x)=x^4-2$ is irreducible over $\mathbb{F}_5$ without needing to check for quadratic factors.

In fact, as long as $K$ contains all $n$-th roots of unity, every cyclic extension can be written in the above form.

Theorem 7.2 Suppose $K$ contains a primitive $n$-th root of unity and $L/K$ is a cyclic extension of degree $n$. Then there is $a\in K^\times$ such that $L=K(\sqrt[n]{a})$.

Proof. Let $G=\Gal(L/K)=\{\id,\sigma,\sigma^2,...,\sigma^{n-1}\}$ be cyclic of order $n$ and suppose $K$ contains a primitive $n$-th root of unity $\zeta_n$. We would like to construct an element $\theta$ that behaves like $\sqrt[n]{a}$ under $G$, i.e. \[\theta \;\overset{\sigma}{\longmapsto}\; \zeta_n\,\theta \;\overset{\sigma}{\longmapsto}\; \zeta_n^2\,\theta \;\overset{\sigma}{\longmapsto}\; \cdots \;\overset{\sigma}{\longmapsto}\; \zeta_n^n\,\theta=\theta.\] Given any $b\in L$, define the Lagrange resolvent for $b$ by \[\theta_b=b+\zeta_n^{-1}\sigma(b)+\zeta_n^{-2}\sigma^2(b)+\cdots+\zeta_n^{-(n-2)}\sigma^{n-2}(b)+\zeta_n^{-(n-1)}\sigma^{n-1}(b).\] Notice that applying $\sigma$ to this, (remembering that $\sigma^n=\id$) gives \[\begin{align*} \sigma(\theta_b) &=\sigma(b)+\zeta_n^{-1}\sigma^2(b)+\zeta_n^{-2}\sigma^3(b)+\cdots+\zeta_n^{-(n-2)}\sigma^{n-1}(b)+\zeta_n^{-(n-1)}b \\ &=\zeta_nb +\sigma(b)+\zeta^{-1}\sigma^2(b)+\cdots+\zeta_n^{-(n-2)}\sigma^{n-1}(b) \\ &=\zeta_n\theta_b \end{align*}\]

It appears that we have succeeded. Unfortunately, there may be a disaster and it could be that $\theta_b=0$. That would certainly be the case if $b\in K$ since then $\theta_b=b\left(1+\zeta_n^{-1}+\zeta_n^{-2}+\cdots+\zeta_n^{-(n-1)}\right)=0$.

Fortunately, we have the following.

Fact (Artin’s Lemma): There is some $b\in L$ such that $\theta_b\neq 0$.

Given this holds, consider $a=\theta_b^n\neq 0$. Then \[\sigma(a)=\sigma(\theta_b)^n=(\zeta_nb)^n=b^n=a\] and therefore $a\in K$. We just need to verify that $L=K(\sqrt[n]{a})$. Clearly $K(\sqrt[n]{a})\subset L$ as $\theta_b\in L$. Furthermore, \[\begin{align*} \sigma^i\left(\sqrt[n]{a}\right)=\sigma^j\left(\sqrt[n]{a}\right) &\quad\iff\quad \sigma^i(\theta_b)=\sigma^j(\theta_b) \\ &\quad\iff\quad \zeta_n^i\theta_b=\zeta_n^j\theta_b \quad\iff\quad i\equiv j\bmod n \end{align*}\] so the $n$ distinct elements of $G$ act differently on $\sqrt[n]{a}$. This implies that $L=K(\sqrt[n]{a})$.

Remark. The Fact above essentially follows from the linear independence of automorphisms result (Dedekind’s Lemma) that we referred to in the proof of the Fundamental Theorem of Galois Theory.

When $n=p$ is prime, we can avoid this with a slightly different approach, choosing the primitive root of unity instead: for any $b\in L\setminus K$, there is some primitive $p$-th root of unity such that $\theta_b\neq 0$.

Indeed, suppose $b+\zeta^{-1}\sigma(b)+\cdots+\zeta^{-(p-1)}\sigma^{p-1}(b)=0$ for each primitive $p$-th root of unity $\zeta$. Adding over all the $p-1$ possible choices of $\zeta$, and using the fact that the sum the primitive $p$-th roots of unity equals $-1$, gives \[(p-1)b-\sigma(b)-\sigma^2(b)-\cdots-\sigma^{p-1}(b)=0.\] Also, the element $b+\sigma(b)+\sigma^2(b)+\cdots+\sigma^{p-1}(b)$ is fixed by $\langle\,\sigma\,\rangle$ so lies in $K$. By adding, we find $pb\in K$ and hence $b\in K$.

Example 7.2 Let’s use the above Theorems to construct the field $L=\mathbb{F}_{3^{16}}$ by finding an irreducible polynomial of degree $16$ in $\mathbb{F}_3[x]$. We know that $L/\mathbb{F}_3$ is a cyclic extension of degree $16$ from the Chapter on Finite Fields. However, $\mathbb{F}_3$ doesn’t contain a primitive $16$-th root of unity (it doesn’t even contain $16$ elements!) so we can’t directly apply Kummer Theory. Instead, we can build $L$ in stages - we first make an extension $K/\mathbb{F}_3$ which contains a suitable primitive $n$-th root of unity, then adjoin the $n$-th root of a suitable element of $\alpha\in K$ to get $L=K(\sqrt[n]{\alpha})$.

We know $\mathbb{F}_9=\mathbb{F}_3(\theta)$ where $\theta$ is a root of $x^2+1$. Then \[\mathbb{F}_9=\{a+b\theta\;\mid\;a,b\in\mathbb{F}_3\}.\] Notice $K=\mathbb{F}_9$ now contains a primitive $8$-th root of unity since $\mathbb{F}_9^{\,\times}\cong\mathbb{Z}/8$.
We know $L/K$, i.e. $\mathbb{F}_{9^8}/\mathbb{F}_9$ is a cyclic extension of degree $8$, so by the above Theorem there must exist an element $\alpha\in K=\mathbb{F}_9$ such that $L=K(\sqrt[8]{\alpha})$. To find $\alpha$, we need an element that is not a square or fourth power in $\mathbb{F}_9$ and finding $\alpha\in\mathbb{F}_9^{\,\times}$ of order exactly $8$ will work.
If we try $\alpha=\theta$, it goes wrong since $\theta^2=-1$ and $\theta^4=1$. Instead, try $\alpha=1+\theta$. Then \[(1+\theta)^2=\theta^2+2\theta+1=-\theta,\qquad (1+\theta)^4=\theta^2=-1,\qquad (1+\theta)^8=1\] so $\alpha=1+\theta$ has order exactly $8$ in $\mathbb{F}_9$.
Then $L=K(\sqrt[8]{\alpha})=\mathbb{F}_9(\sqrt[8]{1+\theta})=\mathbb{F}_3(\theta,\sqrt[8]{1+\theta})=\mathbb{F}_3(\eta)$ where $\eta^8=1+\theta$. Now $L/K$ is cyclic of degree $8$, $[L:K]=8$ and we must have $L=\mathbb{F}_{3^{16}}$ by uniqueness of finite fields: \[\mathbb{F}_3\;\subset\; K=\mathbb{F}_3(\theta)=\mathbb{F}_9\;\subset\; L=\mathbb{F}_3(\eta)=\mathbb{F}_{3^{16}} \] The extension $\mathbb{F}_3(\eta)/\mathbb{F}_3$ has degree $16$ so the minimal polynomial $f(x)$ of $\eta$ over $\mathbb{F}_3$ has degree $16$. We can find it easily by \[\eta^8=1+\theta \quad\implies\quad (\eta^8-1)^2=\theta^2=-1\qquad\implies\qquad \eta^{16}+\eta^8+2=0\] and so $f(x)=x^{16}+x^8+2\in\mathbb{F}_3[x]$ is irreducible.

Remark. The two Theorems above only apply when the base field contains a primitive $n$-th root of unity. There are other Galois extensions with cyclic Galois group, e.g. if $L=\mathbb{Q}(\cos(2\pi/7))$, then $L/\mathbb{Q}$ is cyclic of degree $3$ but is not of the form $L=\mathbb{Q}(\sqrt[3]{a})$ for any $a\in\mathbb{Q}$.