Galois Theory

• Reduce to a group theory problem: $[E:F] = [G:H]$, despite the fact that $E/F$ is not necessarily Galois. This is because we can count in towers: $$\begin{align*} [K:F] = [K:E][E:F] &\implies [G:1] = [K:E][H:1] \\ &\implies {\sharp}G = [K:E] {\sharp}H \\ &\implies [G:H] = {{\sharp}G \over {\sharp}H} = [K:E] .\end{align*}$$

• Essential fact: if $P \in {\operatorname{Syl}}_p(G)$, we can use that $P \subseteq N_G(P) \subset H$ and so $P\in {\operatorname{Syl}}_p(H)$ as well.

• Now use that $N_G(P) \subseteq H$, and do Sylow theory for $P$ in both $G$ and $H$:

  • Sylow 3 on $G$ yields $n_p(G) = [G: N_G(P)] \equiv 1 \operatorname{mod}p$.
  • Sylow 3 on $H$ yields $n_p(H) = [G: N_H(P)] \equiv 1 \operatorname{mod}p$.

• Claim: $N_H(P) = N_G(P)$.

  • We have $N_H(P) \subseteq N_G(P)$ since $H \subseteq G$, so $hPh^{-1}= P$ remains true regarding either $h\in H$ or $h\in G$.
  • For $N_G(P) \subseteq N_H(P)$, use that $N_G(P) \subseteq H$ and so $gPg^{-1}= P$ implies $g\in H$, so $g\in N_H(P)$.

• Now morally one might want to apply an isomorphism theorem: $$\begin{align*} {G/ N_G(P) \over H/N_H(P)}= {G/ N_H(P) \over H/N_H(P)}\cong {G\over H} ,\end{align*}$$ but we don’t have normality. However, we can still get away with the corresponding counting argument if everything is finite: $$\begin{align*} {[G: N_G(P)] \over [H:N_H(P)] }= {[G: N_H(P)] \over [H:N_H(P)] }= {{\sharp}G / {\sharp}N_H(P) \over {\sharp}H / {\sharp}N_H(P)} = {{\sharp}G \over {\sharp}H} = [G: H] .\end{align*}$$

• We have an equation of the form $n_p(G)/n_p(H) = m$, and we want to show $m\equiv 1 \operatorname{mod}p$. So write $$\begin{align*} {n_p(G) \over n_p(H) } = m \implies m n_p(H) &= n_p(G) \\ \implies m n_p(H) &\equiv n_p(G) \operatorname{mod}p \\ \implies m\cdot 1 &\equiv 1 \operatorname{mod}p \\ \implies m &\equiv 1 \operatorname{mod}p .\end{align*}$$

• Some notation: let $\alpha_k \coloneqq\zeta_n^k + \zeta_n^{-k}$.

• Let $m(x)$ be the minimal polynomial of $\alpha_1 \coloneqq\zeta_n + \zeta_n^{-1}$. Note that $\alpha_1 \in {\mathbf{Q}}(\zeta_n)$.

• Use that ${ \mathsf{Gal}}({\mathbf{Q}}(\zeta_n)/{\mathbf{Q}}) \cong C_n^{\times}$, consisting of maps $\sigma_k: \zeta \mapsto \zeta^k$ for $\gcd(k, n) = 1$, of which there are $\phi(n)$ many.

• Galois transitively permutes the roots of irreducible polynomials, so the roots of $m$ are precisely the Galois conjugates of $\alpha$, i.e. the Galois orbit of $\alpha$, so we can just compute it. For illustrative purposes, suppose $n$ is prime, then $$\begin{align*} \sigma_1(\zeta_n + \zeta_n^{-1}) &= \zeta_n + \zeta_n^{-1}=\alpha_1 \\ \sigma_2(\zeta_n + \zeta_n^{-1}) &= \zeta_n^2 + \zeta_n^{-2} = \alpha_2 \\ \sigma_3(\zeta_n + \zeta_n^{-1}) &= \zeta_n^3 + \zeta_n^{-3} = \alpha_3 \\ \vdots&\\ \sigma_{n-1}(\zeta_n + \zeta_n^{-1}) &= \zeta_n^{n-1} + \zeta_n^{-(n-1)} = \zeta_n^{-1} + \zeta_n^{1} = \alpha_1 \\ \sigma_{n-2}(\zeta_n + \zeta_n^{-1}) &= \zeta_n^{n-2} + \zeta_n^{-(n-2)} = \zeta_n^{-2} + \zeta_n^{2} = \alpha_2 \\ \sigma_{n-3}(\zeta_n + \zeta_n^{-1}) &= \zeta_n^{n-3} + \zeta_n^{-(n-3)} = \zeta_n^{-3} + \zeta_n^{3} = \alpha_3 ,\end{align*}$$ where we’ve used that $\zeta^{k} = \zeta^{k\operatorname{mod}n}$. From this, we see that $\sigma_{k}(\alpha_1)=\sigma_{n-k}(\alpha_1)$ and we pick up $(n-1)/2$ distinct conjugates.

• For $n$ not prime, the exact same argument runs through $\phi(n)$ values of $k$ for $\sigma_k$, and again yields $\sigma_{k}(\alpha_1) = \sigma_{\phi(n) - k}(\alpha_1)$. Matching them up appropriately yields $\phi(n)/2$ distinct roots.

The setup:

Link to Diagram

Part 1:

• $L/K$ is cyclic means $L/K$ is Galois and $G\coloneqq{ \mathsf{Gal}}(L/K) = C_n$ for some $n$.
• By the FTGT, setting $H \coloneqq{ \mathsf{Gal}}(L/E)$, we get $H {~\trianglelefteq~}G$ precisely because $E/K$ is normal, and ${ \mathsf{Gal}}(L/E) = G/H$.
• But then if $G$ is cyclic, $H \leq G$ must be cyclic, and $G/H$ is cyclic as well since writing $G = C_n = \left\langle{x}\right\rangle$, we have $G/H = \left\langle{xH}\right\rangle$.

Part 2:

• Letting $G\coloneqq{ \mathsf{Gal}}(L/K) = C_n$, by elementary group theory we have subgroups $H\coloneqq C_d \leq C_n$ for every $d$ dividing $n$.
  • A observation we’ll need: every subgroup is normal here since $G$ is abelian.
• By the fundamental theorem, taking the fixed field of $H \leq { \mathsf{Gal}}(L/K)$, we obtain some intermediate extension $E\coloneqq K^H$ fitting into a tower $L/E/K$.
• By the fundamental theorem, $[E: K] = [G:H] = n/d$, where we’ve used that $H{~\trianglelefteq~}G$.
• Letting $d$ range through divisors lets $n/d$ range through divisors, so we get extensions of every degree $d$ dividing $n$.

• $L/k$ is separable iff every element $\alpha$ is separable, i.e. the minimal polynomial $m(x)$ of $\alpha$ is a separable polynomial, i.e. $m(x)$ has no repeated roots in (say) the algebraic closure of $L$ (or just any splitting field of $m$).

• If $\operatorname{ch}k = p$, suppose toward a contradiction that $L/k$ is not separable. Then there is some $\alpha$ with an inseparable (and irreducible) minimal polynomial $f(x)\in k[x]$.

• Claim: since $f$ is inseparable and irreducible, $f(x) = g(x^p)$ for some $g\in k[x]$.

  • Note: write $g(x) \coloneqq\sum a_k x^k$, so that $f(x) = \sum a_k (x^p)^k = \sum a_k x^{pk}$.

• This is a contradiction, since it makes $f$ reducible by using the “Freshman’s dream”: $$\begin{align*} f(x) = \sum a_k x^{pk} = \qty{ \sum a_k^{1\over p} x^k}^p \coloneqq(h(x))^p .\end{align*}$$

• Proof of claim: in $\operatorname{ch}k = p, f$ inseparable $\implies f(x) = g(x^p)$.

  • Use that $f$ is inseparable iff $\gcd(f, f') \neq 1$, and since $f$ is irreducible this forces $f' \equiv 0$, so $ka_k = 0$ for all $k$.
  • Then $a_k\neq 0$ forces $p\divides k$, so $f(x) = a_0 + a_px^p + a_{2p}x^{2p} + \cdots$ and one takes $g(x) \coloneqq\sum a_{kp}x^{kp}$.

• A finite inseparable extension:

  • It’s a theorem that finite extensions of perfect fields are separable, so one needs a non-perfect field.
  • Take $L/k \coloneqq{ \mathbf{F} }_p(t^{1\over p}) / { \mathbf{F} }_p(t)$, which is a degree $p$ extension (although both fields are infinite are characteristic $p$).
  • Then the minimal polynomial of $t$ is $f(x) \coloneqq x^p - t \in { \mathbf{F} }_p(t)[x]$, where $f'(x) = px^p \equiv 0$ Alternatively, just note that $f$ factors as $f(x) = (x-t^{1\over p})^p$ in $L[x]$, which has multiple roots.

${\mathbf{Z}}/2{\mathbf{Z}}$?

${\mathbf{Q}}(2^{1\over p}, \zeta_p)$, which has degree $p(p-1)$ and is generated by the maps $$\begin{align*} \sqrt[p]{2} & \mapsto \sqrt[p]{2} \zeta^{a} \\ \zeta & \mapsto \zeta^{b} .\end{align*}$$

Part a:

• A finite extension $E/F$ is Galois if it is normal and separable:
  • Normal: every $f\in F[x]$ either has no roots in $E$ or all roots in $E$.
  • Separable: every element $e\in E$ has a separable minimal polynomial $m(x)$, i.e. $m$ has no repeated roots.

Part b:

• Note $f$ is irreducible by Eisenstein with $p=7$, and since ${\mathbf{Q}}$ is perfect, irreducible implies separable.

• Writing $L \coloneqq\operatorname{SF}(f)/{\mathbf{Q}}$, this is a Galois extension:

  • $L$ is separable: it is a finite extension of a perfect field, which is automatically separable.
  • $L$ is normal: $L$ is the splitting field of a separable polynomial, and thus normal.

• Since $f$ is degree 3, we have $G\coloneqq{ \mathsf{Gal}}(L/k) \leq S_3$, and since $G$ is a transitive subgroup the only possibilities are $$\begin{align*} G = S_3 \cong D_3, A_3 \cong C_3 .\end{align*}$$

• Factor $x^3 - 7 = (x-\omega)(x-\zeta_3\omega)(x-\zeta_3^2\omega)$ where $\omega \coloneqq 7^{1\over 3}$ and $\zeta_3$ is a primitive 3rd root of unity. Then $L = {\mathbf{Q}}(\zeta_3, \omega)$.

  • Aside: label the roots in this order, so $r_1 = \omega, r_2 = \zeta_3\omega, r_3 = \zeta_3^2\omega$.

• Write $\min_{\omega, {\mathbf{Q}}}(x) = x^3 - 7$ and let $L_0/{\mathbf{Q}}\coloneqq{\mathbf{Q}}(\omega)/{\mathbf{Q}}$ yields $[L_0: {\mathbf{Q}}] = 3$.

• Write $\min_{\zeta_3, {\mathbf{Q}}}(x) = (x^3-1)/(x-1) = x^2 + x + 1$, and note that this is still the minimal polynomial over $L_0$ since $L_0 \subseteq {\mathbf{R}}$ and $\zeta_3 \in {\mathbf{C}}\setminus{\mathbf{R}}$. So $[L:L_0] = 2$.

• Counting in towers, $$\begin{align*} [L:{\mathbf{Q}}] = [L:L_0][L_0: {\mathbf{Q}}] = (2)(3) = 6 .\end{align*}$$

• But ${\sharp}S_3 = 6$ and ${\sharp}A_3 = 3$, so $G = S_3$.

• Explicitly, since we can write $\operatorname{SF}(f) = {\mathbf{Q}}(\omega, \zeta_3)$, we can find explicit generators: $$\begin{align*} \sigma: &\begin{cases} \omega &\mapsto \omega \\ \zeta_3 &\mapsto \zeta_3\cdot \zeta_3. \end{cases} && \implies \sigma \sim (1,2,3) \\ \tau: &\begin{cases} \omega &\mapsto \omega \\ \zeta_3 &\mapsto \overline{\zeta_3}. \end{cases} && \implies \tau \sim (2, 3) .\end{align*}$$ So $G = \left\langle{\sigma, \tau {~\mathrel{\Big\vert}~}\sigma^3, \tau^2}\right\rangle$.

Part c:

• Note that the subgroup lattice for $S_3$ looks like the following:

• Note that we can identify
  • $\tau = (2,3)$ which fixes $r_1$
  • $\sigma \tau = (1,2)$ which fixes $r_3$
  • $\sigma^2\tau = (1, 3)$ which fixes $r_2$
  • $\sigma = (1,2,3)$, for which we need to calculate the fixed field. Using that $\sigma(\omega) =\zeta\omega$ and $\sigma(\zeta)=\zeta$, supposing $\sigma(\alpha) = \alpha$ we have $$\begin{align*} \sigma(\alpha) &\coloneqq\sigma(a + b\zeta_3 + c\zeta_3^2 + d\omega + e\zeta_3\omega + f\zeta_3^2\omega) \\ &= a + b\zeta_3 + c\zeta_3^2 + d\zeta_3\omega + e\zeta_3^2\omega + f\omega \\ \implies \alpha &= a + b\zeta_3 + c\zeta_3^2 + t_1(\omega + \zeta_3\omega + \zeta_3^2\omega) \\ \implies \alpha &= a + b\zeta_3 + c\zeta_3^2 + t_1\omega (1 + \zeta_3+ \zeta_3^2) \\ \implies \alpha &= a + b\zeta_3 + c\zeta_3^2 ,\end{align*}$$ using the general fact that $\sum_{k=0}^{n-1}\zeta_n^k = 0$. So the fixed field is ${\mathbf{Q}}(1, \zeta, \zeta^2) = {\mathbf{Q}}(\zeta)$.
• We thus get the following lattice correspondence:

Link to Diagram

• First consider $g(z) \coloneqq z^2 + 4z + 64$. Applying the quadratic formula yields $$\begin{align*} z = {-4 \pm \sqrt{16 - 64} \over 2} = -2 \pm {1\over 2}\sqrt{ -15 \cdot 16 } = -2 \pm 2i \sqrt{15} .\end{align*}$$

• Substituting $z=x^2$ yields the splitting field of $f$ as $L\coloneqq{\mathbf{Q}}(\pm \sqrt{ -2 \pm 2i\sqrt{15}})$.

  • Note that this factorization shows that $f$ is irreducible over ${\mathbf{Q}}$, since the two quadratic factors have irrational coefficients and none of the roots are real.
  • Irreducible implies separable over a perfect field, so $L/{\mathbf{Q}}$ is a separable extension.
  • $L$ is the splitting field of a separable polynomial and thus normal, making $L$ Galois.

• In this form, it’s not clear what the degree $[L:{\mathbf{Q}}]$ is, so we can find a better basis by rewriting the roots of $g$: $$\begin{align*} z = -2 \pm 2i\sqrt{15} = \qty{\sqrt{5}}^2 - \qty{\sqrt 3}^2 \pm 2i\sqrt{5}\sqrt{3} = (\sqrt 5 \pm i\sqrt{3})^2 ,\end{align*}$$ and so the roots of $f$ are $x = \pm \sqrt{5} \pm i\sqrt{3}$ and $L = {\mathbf{Q}}(\sqrt 5, i\sqrt 3)$.

• Counting in towers, $$\begin{align*} [L:{\mathbf{Q}}] = [{\mathbf{Q}}(\sqrt 5, i \sqrt{3} ) : {\mathbf{Q}}\sqrt{5} ][{\mathbf{Q}}\sqrt{5} : {\mathbf{Q}}] = (2)(2) = 4 ,\end{align*}$$ where we’ve used that $\min_{\sqrt 5, {\mathbf{Q}}}(x) = x^2-5$ and $\min_{i\sqrt 3, {\mathbf{Q}}}(x) = x^2 + 3$, which remains the minimal polynomial over ${\mathbf{Q}}(\sqrt 5) \subseteq {\mathbf{R}}$ since both roots are not real.

• So $G\coloneqq{ \mathsf{Gal}}(L/{\mathbf{Q}}) \leq S_4$ is a transitive subgroup of size 4, making it either $C_4$ or $C_2^2$.

• Label the roots: $$\begin{align*} r_1 &= \sqrt 5 + i\sqrt 3 \\ r_2 &= \sqrt{5} - i \sqrt{3} \\ r_3 &= - \sqrt 5 + i\sqrt 3 = -r_2 \\ r_4 &= -\sqrt{5} - i\sqrt{3} = -r_1 .\end{align*}$$

• We can start writing down automorphisms: $$\begin{align*} \sigma_1: \begin{cases} \sqrt 5 &\mapsto -\sqrt 5 \\ i\sqrt 3 &\mapsto i\sqrt 3 . \end{cases} && \sigma_1 \sim (1,3)(2,4) \\ \sigma_2 \begin{cases} \sqrt 5 &\mapsto \sqrt 5 \\ i\sqrt 3 &\mapsto -i\sqrt 3 . \end{cases} && \sigma_2 \sim (1, 2)(3, 4) .\end{align*}$$ Note that these define automorphisms because we’ve specified what happens to a basis and they send roots to other roots.

• Checking that $\sigma_1^2 = \sigma_2^2 = \operatorname{id}$, this produces two distinct order 2 elements, forcing $G \cong C_2^2$ since $C_4$ only has one order 2 element. Explicitly, we have $$\begin{align*} C_2^2 \cong G = \left\langle{\tau_1, \tau_2}\right\rangle = \left\{{\operatorname{id}, \tau_1, \tau_2, \tau_1 \tau_2}\right\} = \left\{{\operatorname{id}, (1,3)(2,4), (1,2)(3,4), (1,4)(2,3) }\right\} ,\end{align*}$$ and the generic subgroup lattice looks like:

• Computing some fixed fields. Write $i \sqrt{3} = x, \sqrt{5} = y$, then elements in the splitting field are of the form $\alpha = 1 + ax + by + cxy$.

  • For $\sigma_1$, we have $x\mapsto -x$, so $$\begin{align*} \sigma_1(\alpha) = 1 - ax + by - cxy = \alpha \implies a=-a=0, c=-c=0 ,\end{align*}$$ so this preserves $1+by$, making the fixed field ${\mathbf{Q}}(1, y) = {\mathbf{Q}}(i \sqrt{3})$.

  • For $\sigma_2$, we have $y\mapsto -y$, so $$\begin{align*} \sigma_2(\alpha) = 1 +ax -by -cxy = \alpha \implies b=-b=0,c=-c=0 ,\end{align*}$$ preserving $1 + ax$ and making the fixed field ${\mathbf{Q}}(1, x) = {\mathbf{Q}}(\sqrt 5)$.

  • For $\sigma_1 \sigma_2$, we have $x\mapsto -x$ and $y\mapsto -y$, so $$\begin{align*} \sigma_1\sigma_2(\alpha) = 1 -ax -by +cxy = \alpha \implies a=-a=-, b=-b=0 ,\end{align*}$$ preserving $1 + cxy$ and yielding ${\mathbf{Q}}(xy) = {\mathbf{Q}}(i\sqrt 3 \sqrt 5)$.

• So the lattice correspondence we get here is

Link to Diagram

\todo[inline]{Finish (c)}

Let $K = {\mathbf{Q}}(\zeta)$. Then $K$ is the splitting field of $f(x) = x^n - 1$, which is irreducible over ${\mathbf{Q}}$, so $K/{\mathbf{Q}}$ is normal. We also have $f'(x) = nx^{n-1}$ and $\gcd(f, f') = 1$ since they can not share any roots.

Or equivalently, $f$ splits into distinct linear factors $f(x) = \prod_{k\leq n}(x-\zeta^k)$.

Since it is a Galois extension, ${\left\lvert {{ \mathsf{Gal}}(K/{\mathbf{Q}})} \right\rvert} = [K: {\mathbf{Q}}] = \phi(n)$ for the totient function.

We can now define maps $$\begin{align*} \tau_j: K &\to K \\ \zeta &\mapsto \zeta^j \end{align*}$$ and if we restrict to $j$ such that $\gcd(n, j) = 1$, this yields $\phi(n)$ maps. Noting that if $\zeta$ is a primitive root, then $(n, j) = 1$ implies that that $\zeta^j$ is also a primitive root, and hence another root of $\min(\zeta, {\mathbf{Q}})$, and so these are in fact automorphisms of $K$ that fix ${\mathbf{Q}}$ and thus elements of ${ \mathsf{Gal}}(K/{\mathbf{Q}})$.

So define a map $$\begin{align*} \theta: {\mathbf{Z}}_n^{\times}&\to K \\ [j]_n &\mapsto \tau_j .\end{align*}$$

from the multiplicative group of units to the Galois group.

The claim is that this is a surjective homomorphism, and since both groups are the same size, an isomorphism.

Let $L/K/F$.

Let $G\coloneqq{ \mathsf{Gal}}(K/F)$, then it suffices to show that $G$ is always a solvable group, i.e. any group of order $n=2015$ is solvable. Factor $2015 = 5\cdot 13\cdot 31$ – this is a $pqr$ factorization, and in fact any group with exactly 3 prime factors (so $n$ is squarefree in particular) will be solvable. Let $p=5, q=13, r=31$ so that $p<q<r$. We aim to construct a composition series whose successive quotients are simple groups. Applying Sylow 3 yields

• $n_p \divides qr, n_p \equiv 1 \operatorname{mod}p \implies n_5 \divides 13\cdot 31 = 403$ and $n_5\equiv 1 \operatorname{mod}5$.
  • So $n_5 \in \left\{{1, 13, 31}\right\}$ by divisibility and imposing the congruence forces $n_5\in \left\{{1, 31}\right\}$ since $13\not\equiv 1 \operatorname{mod}5$.
• $n_q \divides pr, n_q \equiv 1 \operatorname{mod}q \implies n_{13} \divides 5\cdot 31 = 155$ and $n_{13}\equiv 1 \operatorname{mod}13$.
  • So $n_{13} \in \left\{{1,5,31}\right\}$ by divisibility and the congruence imposes $n_{13}\in \left\{{1}\right\}$ since $5,31 \not\equiv 1\operatorname{mod}13$. In particular, there is one Sylow 13-subgroup which is normal.
• $n_r \divides pq, n_r \equiv 1 \operatorname{mod}r \implies n_{31} \divides 5 \cdot 13 = 65$ and $n_{31}\equiv 1 \operatorname{mod}31$.
  • So $n_{31}\in \left\{{1,5,13}\right\}$ by divisibility and the congruence imposes $n_{31}\in \left\{{1}\right\}$ since $5,13 \not\equiv 1\operatorname{mod}31$. In particular, the one Sylow 31-subgroup is normal.

Let $H_{13}, H_{31}$ be the two normal subgroups of $G$. Taking the quotient $\tilde G \coloneqq G/H_{31}$ yields a group of order $5\cdot 13$, and a similar argument as above using the Sylow theorems shows that $\tilde G$ has a normal subgroup of order 13. By the subgroup correspondence theorem, this yields a normal subgroup $N_1{~\trianglelefteq~}G$ containing $H_{31}$ which has order $13\cdot 31$. So define $N_2 \coloneqq H_{31}$ to obtain $$\begin{align*} G \trianglerighteq N_1 \trianglerighteq N_2 \coloneqq H_{31} \trianglerighteq 0 .\end{align*}$$ Since the quotients $N_i/N_{i+1}$ have prime power order, they are cyclic and thus simple. We know $N_1$ is normal in $G$ since it came from extending a normal group in a quotient, and we know $N_2$ is normal in $N_1$ since it was normal in all of $G$. So $G$ is solvable.

• Suppose $a$ is not a $p$th power in $F$, then $f(x) \coloneqq x^p-a$ has no roots in $F$.
• Toward a contradiction, suppose $f$ is reducible in $F[x]$.
• In $\operatorname{SF}(f)$, since $\operatorname{ch}F = p$ we have $f(x) = (x-\zeta)^p$ for some $\zeta = a^{1\over p}$.
  • So if $f$ is reducible in $F[x]$, we have $f(x) = p_1(x) p_2(x)$ where $p(x) = (x-\zeta)^q\in F[x]$ for some $1\leq q < p$, since these are the only factors of $f$.
  • The claim is that $\zeta\in F$ as well, which is a contradiction since $\zeta$ is a $p$th root of $a$.
• We have $x^q-\zeta^q \in F[x]$, so $\zeta^q\in F$.
• We know $a = \zeta^p\in F$, and thus $\zeta^{d} = \zeta\in F$ for $d \coloneqq\gcd(p, n) = 1$. $\contradiction$
  • Why this is true: write $d = \gcd(p, n)$ in ${\mathbf{Z}}$ to obtain $d = tp + sn$ for some $t, s$.
  • Then $\zeta^d = \zeta^{tp+sn} = (\zeta^p)^t \cdot (\zeta^n)^s \in F$.

• WTS: $f(x) \coloneqq x^p - a\in { \mathbf{F} }[x]$ reducible $\implies f$ has a root in the base field ${ \mathbf{F} }$.

• Write $f(x) = g(x) h(x)$ and factor $f(x) = \prod_{i=1}^p (x- r_i) \in \operatorname{SF}(f)[x]$ where the $r_i$ are not necessarily distinct roots.

• WLOG, $g(x) = \prod_{i=1}^\ell (x-r_i)$ for some $1\leq \ell \leq p-1$, i.e. rearrange the factors so that $g$ is the first $\ell$ of them.

  • $\ell \neq 1, p$ since $f$ is reducible, making $g, h$ nonconstant.

• Set $R_\ell \coloneqq\prod_{i=1}^\ell r_i$, which is the constant term in $g$, so $R_\ell \in { \mathbf{F} }$ since $g\in { \mathbf{F} }[x]$.

• Each $r_i$ is a root of $f$, so $r_i^p - a = 0$ for all $i$, so $r_i^p = a$.

• Trick: what is the $p$th power of $R_\ell$? $$\begin{align*} R_\ell^p &\coloneqq\qty{ \prod_{i=1}^\ell}^p \\ &= \prod_{i=1}^\ell r_i^p \\ &= \prod_{i=1}^\ell a \\ &= a^\ell ,\end{align*}$$ so $R_\ell^p = a^\ell$.

• Use Bezout: $\gcd(\ell, p) = 1$ since $p$ is prime, so write $tp + s\ell = 1$ for some $t,s\in {\mathbf{Z}}$

• Use this to build a root of $f$ that’s in ${ \mathbf{F} }$: write $$\begin{align*} a &= a^1\\ &= a^{tp + s\ell} \\ &= a^{tp} a^{s\ell} \\ &=a^{tp} (a^\ell)^s\\ &= a^{tp} (R_\ell^p)^s \\ &= (a^t R_\ell^s)^p \\ &\coloneqq\beta^p ,\end{align*}$$ so $a = \beta^p$.

  • Check $\beta\in { \mathbf{F} }$: use that $R_\ell \in { \mathbf{F} }$ since it was a constant term of a polynomial in ${ \mathbf{F} }[x]$, $a\in { \mathbf{F} }$ by assumption, and fields are closed under taking powers and products.

Consider $L \coloneqq{ \mathbf{F} }_p[x]/(x^{p^\ell} - x)$, this yields a field extension $L/{ \mathbf{F} }_p$ with $[L: { \mathbf{F} }_p] = \ell$ and so $L\cong { \mathbf{F} }_{p^\ell}$ is the splitting field of $x^{p^\ell} - x$. Note that $x^{p^\ell}-x$ is the product of all monic irreducible polynomials in ${ \mathbf{F} }_p[x]$ of degree dividing $\ell$, and since $\ell$ is prime, the only such polynomials are of degrees $1$ or $\ell$ – this follows because any such polynomial would generate an intermediate extension $L'$ with $L/L'/{ \mathbf{F} }_p$, and multiplicativity in towers yields $$\begin{align*} [L:{ \mathbf{F} }_p] = [L: L'] \cdot [L' : { \mathbf{F} }_p] = \ell ,\end{align*}$$ forcing either $[L: L'] = 1$ or $[L': { \mathbf{F} }_p] = 1$.

Let $P$ be the desired number of monic irreducible degree $\ell$ polynomials in ${ \mathbf{F} }_p[x]$. The idea is now to get a formula involving all monic irreducible (not necessarily degree $\ell$) polynomials and use it to solve for $P$. By the above observation, we have a factorization $$\begin{align*} x^{p^\ell} - x = \prod_{i\in I} f_i(x) = \prod_{i\in I_1} f_i(x) \cdot \prod_{i\in I_2} g_i(x) ,\end{align*}$$ where $I$ is the set of all monic irreducible polynomials, and the above observation shows $I = I_1{\textstyle\coprod}I_2$ where $I_1$ are those of degree 1 and $I_2$ are those of degree $\ell$. Taking degrees of both sides yields $$\begin{align*} p^\ell = \sum_{i\in I_1} \deg f_i(x) + \sum_{i\in I_2} \deg g_i(x) = \sum_{i\in I_1} 1 + P\ell = {\sharp}I_1 + P\ell ,\end{align*}$$ since each $\deg f_i(x) = 1$ and there are ${\sharp}I_1$ many of them, and $\deg g_i(x) = \ell$ and there are $P$ of them. Rearranging yields $$\begin{align*} P\ell = p^\ell - {\sharp}I_1 \implies P = \ell^{-1}\qty{p^\ell - {\sharp}I_1} ,\end{align*}$$ and so it suffices to determine ${\sharp}I_1$, the number of monic linear irreducible polynomials in ${ \mathbf{F} }_p[x]$. These are all of the form $x + a$ where $a\in { \mathbf{F} }_p$, and there are $p$ choices for $a$, so the final count is $$\begin{align*} P = \ell^{-1}\qty{p^\ell - p} .\end{align*}$$

Let $K = {\mathbf{Q}}(\zeta)$.