• Define a sample space \(\Omega = G \times G\), so \({\sharp}{\Omega} = ({\sharp}{G})^2\).

• Identify the event we want to analyze: \begin{align*} A \coloneqq\left\{{(g,h) \in G\times G {~\mathrel{\Big\vert}~}[g,h] = 1}\right\} \subseteq \Omega .\end{align*}

• Note that the slices are centralizers: \begin{align*} A_g \coloneqq\left\{{(g, h) \in \left\{{ g }\right\} \times G {~\mathrel{\Big\vert}~}[g, h] = 1}\right\} = Z(g) \implies A = \displaystyle\coprod_{g\in G} Z(g) .\end{align*}

• Set \(n\) be the number of conjugacy classes, note we want to show \(P(A) = n / {\left\lvert {G} \right\rvert}\).

• Let \(G\) act on itself by conjugation, which partitions \(G\) into conjugacy classes.

  • What are the orbits? \begin{align*} \mathcal{O}_g = \left\{{hgh^{-1}{~\mathrel{\Big\vert}~}h\in G}\right\} ,\end{align*} which is the conjugacy class of \(g\). In particular, the number of orbits is the number of conjugacy classes.

  • What are the fixed points? \begin{align*}X^g = \left\{{h\in G {~\mathrel{\Big\vert}~}hgh^{-1}= g}\right\},\end{align*} which are the elements of \(G\) that commute with \(g\), which is isomorphic to \(A_g\).

• Identifying centralizers with fixed points, \begin{align*} {\sharp}{A} = {\sharp}{\displaystyle\coprod_{g\in G} Z(g) } = \sum_{g\in G} {\sharp}{Z(g)} = \sum_{g\in G}{\sharp}{X^g} .\end{align*}

• Apply Burnside \begin{align*} {\sharp}{X / G} = \frac { 1 } { {\sharp}G } \sum _ { g \in G } {\sharp}X ^ { g } , \end{align*}

• Note \({\sharp}{X/G} = n\), i.e. the number of conjugacy classes is the number of orbits.

• Rearrange and use definition: \begin{align*} n \cdot {\sharp}{G} = \qty{{\sharp}{X/G} }\cdot {\sharp}{G} = \sum _ { g \in G } {\sharp}X ^ { g } \end{align*}

• Compute probability: \begin{align*} P(A) = {{\sharp}A \over {\sharp}\Omega} = \displaystyle\sum _{ g \in G } \frac{{\sharp}X ^ { g }}{ ( {\sharp}{G} )^2} = \frac{\qty{ {\sharp}{X/G}} \cdot {\sharp}{G}}{ ({\sharp}{G})^2} = \frac{n \cdot {\sharp}{G}}{( {\sharp}{G} )^2} = \frac n {{\sharp}G} .\end{align*}

Statement of the class equation: \begin{align*} {\left\lvert {G} \right\rvert} = Z(G) + \sum_{\substack{\text{One $x$ from each} \\ \text{conjugacy class}}}[G: Z(x)] \end{align*} where \(Z(x) = \left\{{g\in G {~\mathrel{\Big\vert}~}[g, x] = 1}\right\}\) is the centralizer of \(x\) in \(G\).

(DZG): I couldn’t convince myself that a previous proof using the class equation actually works. Instead, I’ll borrow the proof from this note

• Write the event as \(A = \displaystyle\coprod_{g\in G} \left\{{g}\right\} \times Z(g)\), then \begin{align*} P(A) = {{\sharp}A\over ({\sharp}G)^2} = {1\over ({\sharp}G)^2} \sum_{g\in G} {\sharp}Z(g) .\end{align*}
• Attempt to estimate the sum: pull out central elements \(g\in Z(G)\).
  • Note \(Z(g) = G\) for central \(g\), so \({\sharp}Z(g) = {\sharp}G\)
  • Note \begin{align*} g\not\in Z(G)\implies {\sharp}Z(g) \leq {1\over 2} {\sharp}G ,\end{align*} since \(Z(g) \leq G\) is a subgroup, and \begin{align*} [G:Z(g)] \neq 1 \implies [G: Z(g)] \geq 2 .\end{align*}
• Use these facts to calculate: \begin{align*} P(A) &= {1\over ({\sharp}G)^2 } \qty{ \sum_{g\in Z(g)} {\sharp}Z(g) + \sum_{g\not\in Z(g)} {\sharp}Z(g) } \\ &= {1\over ({\sharp}G)^2 } \qty{ \sum_{g\in Z(g)} {\sharp}G + \sum_{g\not\in Z(g)} {\sharp}Z(g) } \\ &= {1\over ({\sharp}G)^2 } \qty{ {\sharp}Z(G) \cdot {\sharp}G + \sum_{g\not\in Z(g)} {\sharp}Z(g) } \\ &\leq {1\over ({\sharp}G)^2 } \qty{ {\sharp}Z(G) \cdot {\sharp}G + \sum_{g\not\in Z(g)} {1\over 2} {\sharp}G } \\ &= {1\over ({\sharp}G)^2 } \qty{ {\sharp}Z(G) \cdot {\sharp}G + \qty{ \sum_{g\not\in Z(g)} {1\over 2} } \cdot {\sharp}G } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + \sum_{g\not\in Z(g)} {1\over 2} } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + {1\over 2} \sum_{g\not\in Z(g)} 1 } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + {1\over 2} {\sharp}(G \setminus Z(G) ) } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + {1\over 2} {\sharp}G - {1\over 2} {\sharp}Z(G) } \\ &= {1\over ({\sharp}G) } \qty{ {1\over 2} {\sharp}Z(G) + {1\over 2} {\sharp}G } \\ &= {1\over 2} \qty{1 + { {\sharp}Z(G) \over {\sharp}G }} \\ &= {1\over 2} \qty{1 + { 1 \over [G : Z(G)] }} .\end{align*}

• By definition, \(S\in {\operatorname{Syl}}_p(G) \iff S\) is a maximal \(p{\hbox{-}}\)subgroup: \(S<G\) is a \(p{\hbox{-}}\)group, so \({\sharp}S = p^k\) for some \(k\), \(S\) is a proper subgroup, and \(S\) is maximal in the sense that there are no proper \(p{\hbox{-}}\)subgroups \(S'\) with \(S \subseteq S' \subseteq G\).
• Since \({\sharp}H = p^i\), \(H\) is a \(p{\hbox{-}}\)subgroup of \(G\).
• If \(H\) is maximal, then by definition \(H\in {\operatorname{Syl}}_p(G)\)
• Otherwise, if \(H\) is not maximal, there exists an \(H' \supseteq H\) with \(H'\leq G\) a \(p{\hbox{-}}\)subgroup properly containing \(H\).
  • In this apply the same argument to \(H'\): this yields a proper superset containment at every stage, and since \(G\) is finite, there is no infinite ascending chain of proper supersets.
  • So this terminates in some maximal \(p{\hbox{-}}\)subgroup \(S\), i.e. a Sylow \(p{\hbox{-}}\)subgroup.
• So \(H \subseteq S\) for some \(S\in {\operatorname{Syl}}_p(G)\).
• All Sylows are conjugate, so for any \(S' \in {\operatorname{Syl}}_p(G)\) we can write \(S' = gSg^{-1}\) for some \(g\).
• Then using that \(H\) is normal, \(H \subseteq S \implies H = gHg^{-1}\subseteq gSg^{-1}\coloneqq S'\). So \(H\) is contained in every Sylow \(p{\hbox{-}}\)subgroup.

• Claim: \(Z(H) \leq H\) works.
  • It is nontrivial since \(H\) is a \(p{\hbox{-}}\)group and \(p{\hbox{-}}\)groups have nontrivial centers
  • It is abelian since \(Z(Z(H)) = Z(H)\).
  • \({\sharp}Z(H) = p^\ell\) for some \(\ell \leq i\) by Lagrange
• It thus remains to show that \(Z(H) {~\trianglelefteq~}G\).
• Use that \(Z(H) \operatorname{ch}H\) and use transitivity of characteristic to conclude \(Z(H) {~\trianglelefteq~}H\).
• That \(Z(H) \operatorname{ch}H\): let \(\psi \in \mathop{\mathrm{Aut}}(H)\) and \(x=\psi(y)\in \psi(Z(H))\) so \(y\in Z(H)\), then for arbitrary \(h\in H\), \begin{align*} \psi(y)h &= \psi(y) (\psi \circ \psi^{-1})(h) \\ &= \psi( y \cdot \psi^{-1}(h) ) \\ &= \psi( \psi^{-1}(h) \cdot y ) && \text{since } \psi^{-1}(h)\in H, \, y\in Z(H) \\ &= h\psi(y) .\end{align*}
• That \(A \operatorname{ch}B {~\trianglelefteq~}C \implies A{~\trianglelefteq~}C\):
  • \(A\operatorname{ch}B\) iff \(A\) is fixed by every \(\psi\in \mathop{\mathrm{Aut}}(B)\)., WTS \(cAc^{-1}= A\) for all \(c\in C\).
  • Since \(B{~\trianglelefteq~}C\), the automorphism \(\psi({-}) \coloneqq c({-})c^{-1}\) descends to an element of \(\mathop{\mathrm{Aut}}(B)\).
  • Then \(\psi(A) = A\) since \(A\operatorname{ch}B\), so \(cAc^{-1}= A\) and \(A{~\trianglelefteq~}C\).

• Fix \(x\), then \(y\in {\mathrm{Orb}}(x) \implies g\cdot x = y\) for some \(g\), and \(x = g^{-1}\cdot y\).
• Then \begin{align*} h \in {\operatorname{Stab}}(x) &\iff h\cdot x = x && \text{by being in the stabilizer} \\ &\iff h\cdot (g^{-1}\cdot y) = g^{-1}\cdot y \\ &\iff (g h g^{-1}) \cdot y = y \\ &\iff ghg^{-1}\in G_y && \text{by definition}\\ &\iff h\in g ^{-1} {\operatorname{Stab}}(y) g ,\end{align*} so \({\operatorname{Stab}}(x) = g^{-1}{\operatorname{Stab}}(y) g\).

Let \(G\) act on its subgroups by conjugation,

• The orbit \(G\cdot H\) is the set of all subgroups conjugate to \(H\), and

• The stabilizer of \(H\) is \(G_H = N_G(H)\).

• By orbit-stabilizer, \begin{align*} G\cdot H = [G: G_H] = [G: N_G(H)] .\end{align*}

• Since \({\left\lvert {H} \right\rvert} = n\), and all of its conjugate also have order \(n\).

• Note that \begin{align*} H\leq N_G(H) \implies {\left\lvert {H} \right\rvert} \leq {\left\lvert {N_G(H)} \right\rvert} \implies {1\over {\left\lvert {N_G(H)} \right\rvert}} \leq {1\over {\left\lvert {H} \right\rvert}} ,\end{align*}

• Now strictly bound the size of the union by overcounting their intersections at the identity: \begin{align*} {\left\lvert {\displaystyle\bigcup_{g\in G}gHg^{-1}} \right\rvert} &< (\text{Number of Conjugates of } H) \cdot (\text{Size of each conjugate}) \\ & \text{strictly overcounts since they intersect in at least the identity} \\ &= [G: N_G(H)] {\left\lvert {H} \right\rvert} \\ &= {{\left\lvert {G} \right\rvert} \over {\left\lvert {N_G(H)} \right\rvert}} {\left\lvert {H} \right\rvert} \\ & \text{since $G$ is finite} \\ &\leq {{\left\lvert {G} \right\rvert} \over {\left\lvert {H} \right\rvert}} {\left\lvert {H} \right\rvert} \\ &= {\left\lvert {G} \right\rvert} .\end{align*}

• Let \(G\curvearrowright X\) transitively where \({\left\lvert {X} \right\rvert} \geq 2\).
• An action is transitive iff there is only one orbit, so \({\left\lvert {X/G} \right\rvert} = 1\).
• Apply Burnside’s Lemma \begin{align*} 1 = {\left\lvert {X/G} \right\rvert} = \frac{1}{{\left\lvert {G} \right\rvert}} \sum_{g\in G} {\left\lvert { \mathrm{Fix} (g)} \right\rvert} \implies {\left\lvert {G} \right\rvert} = \sum_{g\in G} {\left\lvert { \mathrm{Fix} (g)} \right\rvert} = \mathrm{Fix} (e) + \sum_{\substack{g\in G \\ g\neq e}} {\left\lvert { \mathrm{Fix} (g)} \right\rvert} \end{align*}
• Note that \(\mathrm{Fix} (e) = X\), since the identity must fix every element, so \({\left\lvert { \mathrm{Fix} (e)} \right\rvert} \geq 2\).
• If \({\left\lvert { \mathrm{Fix} (g)} \right\rvert} > 0\) for all \(g\neq e\), the remaining term is at least \({\left\lvert {G} \right\rvert} -1\). But then the right-hand side yields is at least \(2 + ({\left\lvert {G} \right\rvert} -1) = {\left\lvert {G} \right\rvert} + 1\), contradicting the equality.
• So not every \({\left\lvert { \mathrm{Fix} (g)} \right\rvert} > 0\), and \({\left\lvert { \mathrm{Fix} (g) } \right\rvert} = 0\) for some \(g\), which says \(g\) has no fixed points in \(X\).

• Every \(p{\hbox{-}}\)subgroup is contained in some Sylow \(p{\hbox{-}}\)subgroup, so \(P \subseteq S_p^i\) for some \(S_p^i \in \mathrm{Syl}_p(G)\).

• \(P {~\trianglelefteq~}G \iff gPg^{-1}= P\) for all \(g\in G\).

• Let \(S_p^j\) be any other Sylow \(p{\hbox{-}}\)subgroup,

• Since Sylow \(p{\hbox{-}}\)subgroups are all conjugate \(gS_p^i g^{-1}= S_p^j\) for some \(g\in G\).

• Then \begin{align*} P = gPg^{-1}\subseteq gS_p^i g^{-1}= S_p^j .\end{align*}

• If \(P\) is not contained in \(M\), then \(M < MP\) is a proper subgroup

• By maximality of \(M\), \(MP = G\)

• Note that \(M\cap P \leq P\) and \({\left\lvert {P} \right\rvert} = p^c\) implies \({\left\lvert {M\cap P} \right\rvert} = p^a\) for some \(a\leq c\) by Lagrange

• Then write \begin{align*} G = MP &\iff {\left\lvert {G} \right\rvert} = \frac{{\left\lvert {M} \right\rvert} {\left\lvert {P} \right\rvert}}{{\left\lvert {M\cap P} \right\rvert}} \\ \\ &\iff { {\left\lvert {G} \right\rvert} \over {\left\lvert {M} \right\rvert}} = {{\left\lvert {P} \right\rvert} \over {\left\lvert {M\cap P} \right\rvert}} = {p^c \over p^a} = p^{c-a} \coloneqq p^b \end{align*}

  where \(a\leq c \implies 0 \leq c-b \leq c\) so \(0\leq b \leq c\).

• We have

• \(n_3 \divides 5\cdot 7, \quad n_3 \cong 1 \operatorname{mod}3 \implies n_3 \in \left\{{1, 5, 7, 35}\right\} \setminus \left\{{5, 35}\right\}\)

• \(n_5 \divides 3\cdot 7, \quad n_5 \cong 1 \operatorname{mod}5 \implies n_5 \in \left\{{1, 3, 7, 21}\right\}\setminus \left\{{3, 7}\right\}\)

• \(n_7 \divides 3\cdot 5, \quad n_7 \cong 1 \operatorname{mod}7 \implies n_7 \in \left\{{1, 3, 5, 15}\right\}\setminus\left\{{3, 5}\right\}\)

• Thus \begin{align*} n_3 \in \left\{{1, 7}\right\} \quad n_5 \in \left\{{1, 21}\right\} \quad n_7 \in \left\{{1, 15}\right\} .\end{align*}

• Toward a contradiction, if \(n_5\neq 1\) and \(n_7 \neq 1\), then \begin{align*} {\left\lvert {{\operatorname{Syl}}(5) \cup{\operatorname{Syl}}(7)} \right\rvert} = (5-1)n_5 + (7-1)n_7 + 1 &= 4(21) + 6(15) = 174 > 105 \text{ elements} \end{align*} using the fact that Sylow \(p{\hbox{-}}\)subgroups for distinct primes \(p\) intersect trivially (?).

• By (a), either \(Q\) or \(R\) is normal.
• Thus \(QR \leq G\) is a subgroup, and it has order \({\left\lvert {Q} \right\rvert} \cdot {\left\lvert {R} \right\rvert} = 5\cdot 7 = 35\).
• By the \(pqr\) theorem, since \(5\) does not divide \(7-1=6\), \(QR\) is cyclic.

• We want to show \(Q, R{~\trianglelefteq~}G\), so we proceed by showing \(\textbf{not }\qty{n_5 = 21 \text{ or } n_7 = 15}\), which is equivalent to \(\qty{n_5 = 1 \text{ and } n_7 = 1}\) by the previous restrictions.

• Note that we can write \begin{align*} G = \left\{{\text{elements of order } n}\right\} {\textstyle\coprod}\left\{{\text{elements of order not } n}\right\} .\end{align*} for any \(n\), so we count for \(n=5, 7\):

  • Elements in \(QR\) of order not equal to 5: \({\left\lvert {QR - Q\left\{{\operatorname{id}}\right\} + \left\{{\operatorname{id}}\right\}} \right\rvert} = 35 - 5 + 1 = 31\)
  • Elements in \(QR\) of order not equal to 7: \({\left\lvert {QR - \left\{{\operatorname{id}}\right\}R + \left\{{\operatorname{id}}\right\}} \right\rvert} = 35 - 7 + 1 = 29\)

• Since \(QR \leq G\), we have

  • Elements in \(G\) of order not equal to 5 \(\geq 31\).
  • Elements in \(G\) of order not equal to 7 \(\geq 29\).

• Now both cases lead to contradictions:

  • \(n_5 = 21\): \begin{align*} {\left\lvert {G} \right\rvert} &= {\left\lvert {\left\{{\text{elements of order } 5}\right\} {\textstyle\coprod}\left\{{\text{elements of order not } 5}\right\}} \right\rvert} \\ &\geq n_5(5-1) + 31 = 21(4) + 31 = 115 > 105 = {\left\lvert {G} \right\rvert} .\end{align*}

  • \(n_7 = 15\): \begin{align*} {\left\lvert {G} \right\rvert} &= {\left\lvert {\left\{{\text{elements of order } 7}\right\} {\textstyle\coprod}\left\{{\text{elements of order not } 7}\right\}} \right\rvert} \\ &\geq n_7(7-1) + 29 = 15(6) + 29 = 119 > 105 = {\left\lvert {G} \right\rvert} .\end{align*}

Suppose \(P\) is normal and recall \({\left\lvert {P} \right\rvert} = 3, {\left\lvert {Q} \right\rvert} = 5, {\left\lvert {R} \right\rvert} = 7\).

• \(P\cap QR = \left\{{e}\right\}\) since \((3, 35) = 1\)
• \(R\cap PQ = \left\{{e}\right\}\) since \((5, 21) = 1\)
• \(Q\cap RP = \left\{{e}\right\}\) since \((7, 15) = 1\)

We also have \(PQR = G\) since \({\left\lvert {PQR} \right\rvert} = {\left\lvert {G} \right\rvert}\) (???).

We thus have an internal direct product \begin{align*} G \cong P\times Q \times R \cong {\mathbb{Z}}_3 \times{\mathbb{Z}}_5 \times{\mathbb{Z}}_7 \cong {\mathbb{Z}}_{105} .\end{align*} by the Chinese Remainder Theorem, which is cyclic.

Strategy: get \(p\) to divide \({\left\lvert {Z(G)} \right\rvert}\).

• Apply the class equation: \begin{align*} {\left\lvert {G} \right\rvert} = {\left\lvert {Z(G)} \right\rvert} + \sum [G: C_G(x_i)] .\end{align*}

• Since \(C_G(x_i) \leq G\) and \({\left\lvert {G} \right\rvert} = p^k\), by Lagrange \({\left\lvert {C_G(x_i)} \right\rvert} = p^\ell\) for some \(0\leq \ell \leq k\).

• Since \({\left\lvert {G} \right\rvert} = p^k\) for some \(k\) and \(Z(G), C_G(x_i) \leq G\) are subgroups, their orders are powers of \(p\).

• Use \begin{align*}[G: C_G(x_i)] = 1 \iff C_G(x_i) = G \iff \left\{{g\in G{~\mathrel{\Big\vert}~}gx_ig^{-1}= x_i}\right\} = G \iff x_i \in Z(G).\end{align*}

  • Thus every index appearing in the sum is greater than 1, and thus equal to \(p^{\ell_i}\) for some \(1\leq \ell_i \leq k\)
  • So \(p\) divides every term in the sum

• Rearrange \begin{align*} {\left\lvert {G} \right\rvert} - \sum [G: C_G(x_i)] = {\left\lvert {Z(G)} \right\rvert} .\end{align*}

• \(p\) divides both terms on the LHS, so must divide the RHS, so \({\left\lvert {Z(G)} \right\rvert} \geq p\).

Strategy: examine \({\left\lvert {G/Z(G)} \right\rvert}\) by cases.

• \(1\): Then \(G = Z(G)\) and \(G\) is abelian.
• \(p\): Then \(G/Z(G)\) is cyclic so \(G\) is abelian
• \(p^2\): Not possible, since \({\left\lvert {Z(G)} \right\rvert} > 1\) by (a).

• By Sylow

  • \(n_5 \divides 7^2,\quad n_5\cong 1\operatorname{mod}5 \implies n_5\in\left\{{1, 7, 49}\right\}\setminus\left\{{7, 49}\right\} = \left\{{1}\right\} \implies n_5 = 1\)
  • \(n_7 \divides 5^2, \quad n_7 \cong 1 \operatorname{mod}7 \implies n_7 \in \left\{{1, 5, 25}\right\}\setminus\left\{{5, 25}\right\} =\left\{{1}\right\} \implies n_7 = 1\)

• By recognition of direct products, \(G = S_5 \times S_7\)

  • By above, \(S_5, S_7{~\trianglelefteq~}G\)
  • Check \(S_5\cap S_7 = \left\{{e}\right\}\) since they have coprime order.
  • Check \(S_5S_7 = G\) since \({\left\lvert {S_5 S_7} \right\rvert} = 5^2 7^2 = {\left\lvert {G} \right\rvert}\)

• By (b), \(S_5, S_7\) are abelian since they are groups of order \(p^2\)

• The direct product of abelian groups is abelian.

• \({\mathbb{Z}}_{5^2} \times{\mathbb{Z}}_{7^2}\)
• \({\mathbb{Z}}_{5}^2 \times{\mathbb{Z}}_{7^2}\)
• \({\mathbb{Z}}_{5^2} \times{\mathbb{Z}}_{7}^2\)
• \({\mathbb{Z}}_{5}^2 \times{\mathbb{Z}}_{7}^2\)

• The important assumption is \(a(x)\not\in R[x]\), we’ll assume \(R\) is a UFD and try to contradict this.
• Write \(f(x) = a(x)b(x)\in F[x]\), then if \(R\) is a UFD we have \(r,s\in F\) such that \(f(x) = ra(x) sb(x) \in R[x]\).
• Since \(a(x), b(x)\) are monic and \(f=ab\), \(f\) is monic, and by the factorization in \(R[x]\) we have \(rs=1\). So \(r,s\in R^{\times}\).
• Then using that \(ra(x)\in R[x]\), we have \(r^{-1}ra(x) = a(x)\in R[x]\). \(\contradiction\)

• Set \(R = {\mathbb{Z}}[2\sqrt 2], F = {\mathbb{Q}}[2\sqrt 2]\).
• Let \(p(x) \coloneqq x^2-2 \in R[x]\) which splits as \(p(x) = (x+ \sqrt{2} )(x - \sqrt{2} ) \coloneqq a(x) b(x) \in F[x]\).
• Note neither \(a(x), b(x)\) are in \(R[x]\).
  • Explicitly, every monic linear \(p\in R[x]\) is of the form \(x + 2t\sqrt{2}\) with \(t\in {\mathbb{Z}}\), and \(\pm \sqrt{2} \neq 2t\sqrt{2}\) for any \(t\).
• So we have \(p(x) \in R[x]\) splitting as \(p=ab\) in \(F[x]\) with \(a\not \in R[x]\), so part (a) applies.

Define the set of proper ideals \begin{align*} S = \left\{{J {~\mathrel{\Big\vert}~}I \subseteq J < R}\right\} ,\end{align*}

which is a poset under set inclusion.

Given a chain \(J_1 \subseteq \cdots\), there is an upper bound \(J \coloneqq\cup J_i\), so Zorn’s lemma applies.

\(\implies\):

• We will show that \(x\in J(R) \implies 1+x \in R^{\times}\), from which the result follows by letting \(x=rx\).

• Let \(x\in J(R)\), so it is in every maximal ideal, and suppose toward a contradiction that \(1+x\) is not a unit.

• Then consider \(I = \left\langle{1+x}\right\rangle {~\trianglelefteq~}R\). Since \(1+x\) is not a unit, we can’t write \(s(1+x) = 1\) for any \(s\in R\), and so \(1 \not\in I\) and \(I\neq R\)

• So \(I < R\) is proper and thus contained in some maximal proper ideal \(\mathfrak{m} < R\) by part (1), and so we have \(1+x \in \mathfrak{m}\). Since \(x\in J(R)\), \(x\in \mathfrak{m}\) as well.

• But then \((1+x) - x = 1 \in \mathfrak{m}\) which forces \(\mathfrak{m} = R\).

\(\impliedby\)

• Fix \(x\in R\), and suppose \(1+rx\) is a unit for all \(r\in R\).

• Suppose towards a contradiction that there is a maximal ideal \(\mathfrak{m}\) such that \(x\not \in \mathfrak{m}\) and thus \(x\not\in J(R)\).

• Consider \begin{align*} M' \coloneqq\left\{{rx + m {~\mathrel{\Big\vert}~}r\in R,~ m\in M}\right\} .\end{align*}

• Since \(\mathfrak{m}\) was maximal, \(\mathfrak{m} \subsetneq M'\) and so \(M' = R\).

• So every element in \(R\) can be written as \(rx + m\) for some \(r\in R, m\in M\). But \(1\in R\), so we have \begin{align*} 1 = rx + m .\end{align*}

• So let \(s = -r\) and write \(1 = sx - m\), and so \(m = 1 + sx\).

• Since \(s\in R\) by assumption \(1+sx\) is a unit and thus \(m \in \mathfrak{m}\) is a unit, a contradiction.

• So \(x\in \mathfrak{m}\) for every \(\mathfrak{m}\) and thus \(x\in J(R)\).

\(\mathfrak N(R) \subseteq J(R)\):

• Use the fact \(x\in \mathfrak N(R) \implies x^n = 0 \implies 1 + rx\) is a unit \(\iff x\in J(R)\) by (b): \begin{align*} \sum_{k=1}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1- (-x)} = (1+x)^{-1} .\end{align*}

\(J(R) \subseteq \mathfrak N(R)\):

• Let \(x \in J(R) \setminus \mathfrak N(R)\).

• Since \(R\) is finite, \(x^m = x\) for some \(m > 0\).

• Without loss of generality, we can suppose \(x^2 = x\) by replacing \(x^m\) with \(x^{2m}\).

• If \(1-x\) is not a unit, then \(\left\langle{1-x}\right\rangle\) is a nontrivial proper ideal, which by (a) is contained in some maximal ideal \({\mathfrak{m}}\). But then \(x\in {\mathfrak{m}}\) and \(1-x \in {\mathfrak{m}}\implies x + (1-x) = 1 \in {\mathfrak{m}}\), a contradiction.

• So \(1-x\) is a unit, so let \(u = (1-x)^{-1}\).

• Then \begin{align*} (1-x)x &= x - x^2 = x - x = 0 \\ &\implies u (1-x)x = x = 0 \\ &\implies x=0 .\end{align*}

• Suppose \(c\coloneqq f(a)\neq 0\), noting that \(c\) may not be positive.
• By continuity, pick \({\varepsilon}\) small enough so that \({\left\lvert {x-a} \right\rvert}<{\varepsilon}\implies {\left\lvert {f(x) - f(a)} \right\rvert} < c/2\). Since we’re on the interval, we have \(f(x) \in (f(a) - c/2, f(a) + c/2) = (c/2, 3c/2)\) which is a ball of radius \(c/2\) about \(c\), which thus doesn’t intersect \(0\).
• So \(f(x) \neq 0\) on this ball, and \(g \coloneqq f^2 > 0\) on it. Note that ideals are closed under products, so \(g\in I\)
• Moreover \(f^2(x) \geq 0\) since squares are non-negative, so \(g\geq 0\) on \([0, 1]\).

• By contrapositive, suppose \(V(I)= \emptyset\), we’ll show \(I\) contains a unit and thus \(I=R\).
• For each fixed \(x\in [0, 1]\), since \(V(I)\) is empty there is some \(f_x\) such that \(f_x(x) \neq 0\).
• By (a), there is some \(g_x\) with \(g_x(x) > 0\) on a neighborhood \(U_x\ni x\) and \(g_x \geq 0\) everywhere.
• Ranging over all \(x\) yields a collection \(\left\{{(g_x, U_x) {~\mathrel{\Big\vert}~}x\in [0, 1]}\right\}\) where \(\left\{{U_x}\right\}\rightrightarrows[0, 1]\).
• By compactness there is a finite subcover, yielding a finite collection \(\left\{{(g_k, U_k)}\right\}_{k=1}^n\) for some \(n\).
• Define the candidate unit as \begin{align*} G(x) \coloneqq{1\over \sum_{k=1}^n g_k(x)} .\end{align*}
• This is well-defined: fix an \(x\), then the denominator is zero at \(x\) iff \(g_k(x) = 0\) for all \(k\). But since the \(U_k\) form an open cover, \(x\in U_\ell\) for some \(\ell\), and \(g_\ell > 0\) on \(U_\ell\).
• Since ideals are closed under sums, \(H\coloneqq{1\over G} \coloneqq\sum g_k \in I\). But \(H\) is clearly a unit since \(HG = \operatorname{id}\).

• If \(I{~\trianglelefteq~}R\) is maximal, \(I\neq R\), and so by (b) we have \(V(I) \neq \emptyset\).
• So there is some \(x_0\in[0, 1]\) with \(f(x_0) = 0\) for all \(f\in I\).
• Define \({\mathfrak{m}}_{x_0} \coloneqq\left\{{f\in R {~\mathrel{\Big\vert}~}f(x_0) = 0}\right\}\), which is clearly an ideal.
  • This is a proper ideal, since constant nonzero functions are continuous and thus in \(R\), not not \({\mathfrak{m}}_{x_0}\).
• We thus have \(I \subseteq {\mathfrak{m}}_{x_0}\), and by maximality they are equal.

• Let \(\phi\) denote the map in question, it suffices to show that \(\phi\) is \(R{\hbox{-}}\)linear, i.e. \(\phi(s\mathbf{x} + \mathbf{y}) = s\phi(\mathbf{x}) + \phi(\mathbf{y})\): \begin{align*} \phi(s\mathbf{x} + \mathbf{y}) &= r(s\mathbf{x} + \mathbf{y}) \\ &= rs\mathbf{x} + r\mathbf{y} \\ &= s(r\mathbf{x}) + (r\mathbf{y}) \\ &= s\phi(\mathbf{x}) + \phi(\mathbf{y}) .\end{align*}

Let \(\phi_r(x) \coloneqq rx\) be the multiplication map.

• Let \(x\in \ker \phi_r \coloneqq\left\{{x\in R {~\mathrel{\Big\vert}~}rx = 0}\right\}\).

• Since \(R\) is commutative \(0 = rx = xr\), and so \(r\in \ker \phi_x\), so \(\ker \phi_x \neq 0\) and \(x\) is a zero divisor by definition.

• Let \(y\in \operatorname{im}\phi_r \coloneqq\left\{{y \coloneqq rx {~\mathrel{\Big\vert}~}x\in R}\right\}\), we want to show \(\ker \phi_y\) is nontrivial by producing some \(z\) such that \(yz=0\). Write \(y\coloneqq rx\) for some \(x\in R\).

• Since \(r\) is a zero divisor, we can produce some \(z\neq 0 \in \ker \phi_r\), so \(rz = 0\).

• Now using that \(R\) is commutative, we can compute \begin{align*} yz = (rx)z = (xr)z = x (rz) = x(0) = 0 ,\end{align*} so \(z\in \ker \phi_y\).

• Let \(Z \coloneqq\left\{{z_i}\right\}_{i=1}^n\) be the set of \(n\) zero divisors in \(R\).

• Let \(\phi_i\) be the \(n\) maps \(x \mapsto z_i x\), and let \(K_i = \ker \phi_i\) be the corresponding kernels.

• Fix an \(i\).

• By (b), \(K_i\) consists of zero divisors, so \begin{align*} {\left\lvert {K_i} \right\rvert} \leq n < \infty \quad \text{for each } i .\end{align*}

• Now consider \(R/K_i \coloneqq\left\{{r + K_i}\right\}\).

• By the first isomorphism theorem, \(R/K_i \cong \operatorname{im}\phi\), and by (b) every element in the image is a zero divisor, so \begin{align*} [R: K_i] = {\left\lvert {R/K_i} \right\rvert} = {\left\lvert {\operatorname{im}\phi_i} \right\rvert} \leq n .\end{align*}

• But then \begin{align*} {\left\lvert {R} \right\rvert} = [R:K_i]\cdot {\left\lvert {K_i} \right\rvert} \leq n\cdot n = n^2 .\end{align*}

• By (c), if there are exactly 2 zero divisors then \({\left\lvert {R} \right\rvert} \leq 4\). Since every element in a finite ring is either a unit or a zero divisor, and \({\left\lvert {R^{\times}} \right\rvert} \geq 2\) since \(\pm 1\) are always units, we must have \({\left\lvert {R} \right\rvert} = 4\).

• Since the characteristic of a ring must divide its size, we have \(\operatorname{ch}R = 2\) or \(4\).

• Using the hint, we see that only \({\mathbb{Z}}/(4)\) has characteristic 4, which has exactly 2 zero divisors given by \([0]_4\) and \([2]_4\).

• If \(R\) has characteristic 2, we can check the other 3 possibilities.

• We can write \({\mathbb{Z}}/(2)[t]/(t^2) = \left\{{a + bt {~\mathrel{\Big\vert}~}a,b\in {\mathbb{Z}}/(2)}\right\}\), and checking the multiplication table we have \begin{align*} \begin{array}{c|cccc} & 0 & 1 & t & 1+t \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & t & 1+t \\ t & 0 & t & \mathbf{0} & t \\ 1 + t & 0 & 1+t & t & 1 \\ \end{array} ,\end{align*}

  and so we find that \(t, 0\) are the zero divisors.

• In \({\mathbb{Z}}/(2)[t]/(t^2 - t)\), we can check that \(t^2 = t \implies t t^2 = t^2 \implies t(t^2 + 1) = 0 \implies t(t+1) = 0\), so both \(t\) and \(t+1\) are zero divisors, along with zero, so this is not a possibility.

• Similarly, in \({\mathbb{Z}}/(2)[t]/(t^2 + t + 1)\), we can check the bottom-right corner of the multiplication table to find \begin{align*} \left[\begin{array}{c|cc} & t & 1 +t \\ \hline t & 1+t & 1 \\ t & 1 & t \\ \end{array}\right] ,\end{align*}

  and so this ring only has one zero divisor.

• Thus the only possibilities are: \begin{align*} R &\cong {\mathbb{Z}}/(4) \\ R &\cong {\mathbb{Z}}/(2)[t] / (t^2) .\end{align*}

• Maximal: a proper ideal \(I{~\trianglelefteq~}R\), so \(I\neq R\), such that if \(J\supseteq I\) is any other ideal, \(J = R\).
• Existence of a maximal ideal: use that \(0\in \operatorname{Id}(R)\) always, so \(S\coloneqq\left\{{I\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}I\neq R}\right\}\) is a nonempty poset under subset inclusion. Applying the usual Zorn’s lemma argument produces a maximal element.

\(\impliedby\): By contrapositive: if \(r\in R\) is a unit and \({\mathfrak{m}}\) is maximal, then \(r\in {\mathfrak{m}}\implies {\mathfrak{m}}= R\), contradicting that \({\mathfrak{m}}\) is proper.

\(\implies\):

• Suppose \(a\) is not a unit, we’ll produce a maximal ideal containing it.
• Let \(I\coloneqq Ra\) be the principal ideal generated by \(a\), then \(Ra \neq R\) since \(a\) is not a unit.
• Take a poset \(S \coloneqq\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J\supseteq Ra, J\neq R}\right\}\) ordered by set inclusion.

  • Let \(C_*\) be a chain in \(S\), set \(\widehat{C} \coloneqq\cup C_i\). Then \(\widehat{C} \in S\):

    • \(\widehat{C} \neq R\), since if so it contains a unit, forcing some \(C_i\) to contain a unit and thus equal \(R\).
    • \(\widehat{C} \supseteq Ra\), since e.g. \(\widehat{C} \supseteq C_1 \supseteq Ra\).
    • \(\widehat{C}\) is an ideal since \(xy\in \widehat{C} \implies x\in C_i, y\in C_j\) and \(C_i \subseteq C_j\) without loss of generality, so \(xy\in C_j \subseteq \widehat{C}\).
• Then \(Ra \subseteq \widehat{C}\), some maximal ideal.

• Write \(I \coloneqq\operatorname{Ann}(u)\) for some \(u\), and toward a contradiction suppose \(ab\in I\) but \(a,b\not\in I\).
• Then \(abu=0\) but \(au\neq 0, bu\neq 0\).
• Since \(abu=0\), we have \(a\in \operatorname{Ann}(bu)\). Note that \(\operatorname{Ann}(bu) \supseteq\operatorname{Ann}(u)\), since \(su = 0\implies bsu=sbu=0\).
• We can’t have \(\operatorname{Ann}(bu) = R\), since if \(sbu=0\) for all \(s\in R\), so we could take \(s=1\) to get \(bu=0\) and \(b\in \operatorname{Ann}(u)\).
• By maximality, this forces \(\operatorname{Ann}(u) = \operatorname{Ann}(bu)\), so \(sbu = 0 \implies su=0\) for any \(s\in R\).
• Now take \(s=a\), and since \(abu=0\) we get \(au=0\) and \(a\in \operatorname{Ann}(u)\). \(\contradiction\)

• Let \(\mathfrak{p}\) be prime and \(x\in N\).
• Then \(x^k = 0 \in \mathfrak{p}\) for some \(k\), and thus \(x^k = x x^{k-1} \in \mathfrak p\).
• Since \(\mathfrak p\) is prime, inductively we obtain \(x\in\mathfrak p\).

• Let \(S = \left\{{r^k \mathrel{\Big|}k\in {\mathbb{N}}}\right\}\) be the set of positive powers of \(r\).

• Then \(S^2 \subseteq S\), since \(r^{k_1}r^{k_2} = r^{k_1+k_2}\) is also a positive power of \(r\), and \(0\not\in S\) since \(r\neq 0\) and \(r\not\in N\).

• By the corollary, \(\left\{{I {~\trianglelefteq~}R {~\mathrel{\Big\vert}~}I\cap S = \emptyset}\right\}\) has a maximal element \(\mathfrak{p}\).

• Since \(R\) is commutative, \(\mathfrak{p}\) is prime.

• Suppose \(R\) has a unique prime ideal \(\mathfrak{p}\).

• Suppose \(r\in R\) is not a unit, and toward a contradiction, suppose that \(r\) is also not nilpotent.

• Since \(r\) is not a unit, \(r\) is contained in some maximal (and thus prime) ideal, and thus \(r \in \mathfrak{p}\).

• Since \(r\not\in N\), by (b) there is a maximal ideal \(\mathfrak{m}\) that avoids all positive powers of \(r\). Since \(\mathfrak{m}\) is prime, we must have \(\mathfrak{m} = \mathfrak{p}\). But then \(r\not\in \mathfrak{p}\), a contradiction.

\(\implies\):

• Suppose \(\operatorname{Id}(R) = \left\{{\left\langle{0}\right\rangle, \left\langle{1}\right\rangle}\right\}\). Then for any nonzero \(r\in R\), the ideal \(\left\langle{r}\right\rangle = \left\langle{1}\right\rangle\) is the entire ring.
• In particular, \(1\in \left\langle{r}\right\rangle\), so we can write \(a = tr\) for some \(t\in R\).
• But then \(r\in R^{\times}\) with \(t\coloneqq r^{-1}\).

\(\impliedby\):

• Suppose \(R\) is a field and \(I\in \operatorname{Id}(R)\) is an ideal.
• If \(I \neq \left\langle{0}\right\rangle\) is proper and nontrivial, then \(I\) contains at least one nonzero element \(a\in I\).
• Since \(R\) is a field, \(a^{-1}\in R\), and \(aa^{-1}= 1\in I\) forces \(I = \left\langle{1}\right\rangle\).

• Let \(J{~\trianglelefteq~}R\) be a nonzero two-sided ideal, noting that \(R\) is noncommutative – the claim is that \(J\) contains \(I_n\), the \(n\times n\) identity matrix, and thus \(J = R\).
• Pick a nonzero element \(M\in I\), then \(M\) has a nonzero entry \(m{ij}\).
• Let \(A_{ij}\) be the matrix with a 1 in the \(i,j\) position and zeros elsewhere.
  • Left-multiplying \(A_{ij}M\) moves row \(j\) to row \(i\) and zeros out the rest of \(M\).
  • Right-multiplying \(MA_{ij}\) moves column \(i\) to column \(j\) and zeros out the rest.
  • So \(A_{ij} M A_{kl}\) moves entry \(j, k\) to \(i, l\) and zeros out the rest.
• So define \(B \coloneqq A_{i, i}MA_{j, i}\), which movies \(m_{ij}\) to the \(i,i\) position on the diagonal and has zeros elsewhere.
• Then \(m_{ij}^{-1}{\varepsilon}_{ii} \coloneqq A_{ii}\) is a matrix with \(1\) in the \(i, i\) spot for any \(i\). Since \(I\) is an ideal, we have \({\varepsilon}_{ii}\in I\) for every \(i\).
• We can write the identity \(I_n\) as \(\sum_{i=1}^n {\varepsilon}_{ii}\), so \(I_n \in I\) and \(I=R\).

• Existence: the claim is that \(2R \coloneqq\left\{{2y {~\mathrel{\Big\vert}~}y\in R}\right\}\) is a nontrivial two-sided ideal of \(R\), forcing \(2R = R\) by simpleness.
  • That \(2R \neq 0\) follows from condition (1): Provided \(y\neq 0\), we have \(2y\neq 0\), and so if \(R\neq 0\) then there exists some nonzero \(a\in R\), in which case \(2a\neq 0\) and \(2a\in 2R\).
  • That \(2R\) is a right ideal: clear, since \((2y)\cdot r = 2(yr)\in 2R\).
  • That \(2R\) is a left ideal: use that multiplication is distributive: \begin{align*} r\cdot 2y \coloneqq r(y+y) = ry + ry \coloneqq 2(ry) \in 2R .\end{align*}
• So \(2R = R\) by simpleness.
• Uniqueness:
  • Use the contrapositive of condition (1), so that \(2x = 0 \implies x=0\).
  • Suppose toward a contradiction that \(x=2y_1 = 2y_2\), then \begin{align*} 0 = x-x = 2y_1 - 2y_2 = 2(y_1 - y_2) \implies y_1 - y_2 = 0 \implies y_1 = y_2 .\end{align*}

• First we’ll show \(z=2(yz)\): \begin{align*} xy + xy &= x \\ \implies xy + xy - x &= 0 \\ \implies xyz + xyz - xz &= 0 \\ \implies x(yz + yz - z) &= 0 \\ \implies yz + yz - z &= 0 && \text{since } x\neq 0 \text{ and no zero divisors }\\ \implies 2(yz) &= z .\end{align*}

• Now we’ll show \(z=2(zy)\): \begin{align*} yz + yz &= z \\ \implies zyz + zyz &= zz \\ \implies zyz + zyz - zz &= 0 \\ \implies (zy + zy - z)z &= 0\\ \implies z=0 \text{ or } zy+zy-z &= 0 && \text{ no zero divisors } .\end{align*}

• Then if \(z=0\), we have \(yz = 0 = zy\) and we’re done.

• Otherwise, \(2(zy) = z\), and thus \begin{align*} 2(zy) = z = 2(yz) \implies 2(zy - yz) = 0 \implies zy-yz = 0 ,\end{align*} so \(zy=yz\).

• If \(1\in R\), \begin{align*} 2xy &= x \\ \implies 2xy-x &= 0 \\ \implies x(2y-1) &= 0 \\ \implies 2y-1 &= 0 && x\neq 0 \text{ and no zero divisors}\\ \implies 2y &= 1 .\end{align*}
• Now use \begin{align*} 1\cdot z &= z\cdot 1 \\ \implies (2y)z &= z(2y) \\ \implies (y+y)z &= z(y+y) \\ \implies yz+yz &= zy+zy \\ \implies 2(yz) &= 2(zy) \\ \implies 2(yz-zy) &= 0 \\ \implies yz-zy &= 0 \\ ,\end{align*} using condition (2).

\begin{align*} 2a = (2a)^2 = 4a^2 = 4a \implies 2a = 0 .\end{align*} Note that this implies \(x = -x\) for all \(x\in R\).

\begin{align*} a+b = (a+b)^2 &= a^2 + ab + ba + b^2 = a + ab + ba + b \\ &\implies ab + ba = 0 \\ &\implies ab = -ba \\ &\implies ab = ba \quad\text{by (a)} .\end{align*}

Note that \(g(x) = x^2 - 4x + 2\) has roots \(\beta = 2 \pm \sqrt{2}\), and so \(f\) has roots \begin{align*} \alpha_1 &= \sqrt{2 + \sqrt 2} \\ \alpha_2 &= \sqrt{2 - \sqrt 2} \\ \alpha_3 &= -\alpha_1 \\ \alpha_4 &= -\alpha_2 .\end{align*}

and splitting field \(K = {\mathbb{Q}}(\left\{{\alpha_i}\right\})\).

\(K\) is the splitting field of a separable polynomial and thus Galois over \({\mathbb{Q}}\). Moreover, Since \(f\) is irreducible by Eisenstein with \(p=2\), the Galois group is a transitive subgroup of \(S^4\), so the possibilities are:

• \(S_4\)
• \(A_4\)
• \(D_4\)
• \({\mathbb{Z}}/(2) \times{\mathbb{Z}}/(2)\)
• \({\mathbb{Z}}/(4)\)

We can note that \(g\) splits over \(L \coloneqq{\mathbb{Q}}(\sqrt 2)\), an extension of degree 2.

We can now note that \(\min(\alpha, L)\) is given by \(p(x) = x^2 - (2 + \sqrt 2)\), and so \([K: L] = 2\).

We then have \begin{align*} [K: {\mathbb{Q}}] = [K: L] [L : {\mathbb{Q}}] = (2)(2) = 4 .\end{align*}

This \({\left\lvert {{ \mathsf{Gal}} (K/{\mathbb{Q}})} \right\rvert} = 4\), which leaves only two possibilities:

• \({\mathbb{Z}}/(2) \times{\mathbb{Z}}/(2)\)
• \({\mathbb{Z}}/(4)\)

We can next check orders of elements. Take \begin{align*} \sigma &\in { \mathsf{Gal}} (K/{\mathbb{Q}}) \\ \alpha_1 &\mapsto \alpha_2 .\end{align*}

Computations show that

• \(\alpha_1^2 \alpha_2^2 = 2\), so \(\alpha_1 \alpha_2 = \sqrt 2\)
• \(\alpha_1^2 = 2 + \sqrt 2 \implies \sqrt 2 = \alpha_1^2 - 2\)

and thus \begin{align*} \sigma^2(\alpha_1) &= \sigma(\alpha_2) \\ &= \sigma\left(\frac{\sqrt 2}{\alpha_1}\right) \\ &= \frac{\sigma(\sqrt 2)}{\sigma(\alpha_1)} \\ &= \frac{\sigma(\alpha_1^2 - 2)}{\alpha_2} \\ &= \frac{\alpha_2^2 - 2}{\alpha_2} \\ &= \alpha_2 -2\alpha_2^{-1}\\ &= \alpha_2 - \frac{2\alpha_1}{\sqrt 2} \\ &= \alpha_2 -\alpha_1 \sqrt 2 \\ &\neq \alpha_1 ,\end{align*}

and so the order of \(\sigma\) is strictly greater than 2, and thus 4, and thus \({ \mathsf{Gal}} (K/{\mathbb{Q}}) = \left\{{\sigma^k {~\mathrel{\Big\vert}~}1\leq k \leq 4}\right\} \cong {\mathbb{Z}}/(4)\).

?? The subgroup of index 2 \(\left\langle{\sigma^2}\right\rangle\) corresponds to the field extension \(Q(\sqrt 2) / {\mathbb{Q}}\).

Letting \(\sigma \in K\) be arbitrary, noting that \([K: {\mathbb{Q}}]\) has a basis \(\left\{{1, \zeta, \zeta^2, \cdots, \zeta^{n-1}}\right\}\), it suffices to specify \(\sigma(\zeta)\) to fully determine the automorphism. (Since \(\sigma(\zeta^k) = \sigma(\zeta)^k\).)

In particular, \(\sigma(\zeta)\) satisfies the polynomial \(x^n - 1\), since \(\sigma(\zeta)^n = \sigma(\zeta^n) = \sigma(1) = 1\), which means \(\sigma(\zeta)\) is another root of unity and \(\sigma(\zeta) = \zeta^k\) for some \(1\leq k \leq n\).

Moreover, since \(o(\zeta) = n \in K^{\times}\), we must have \(o(\zeta^k) = n \in K^{\times}\) as well. Noting that \(\left\{{\zeta^i}\right\}\) forms a cyclic subgroup \(H\leq K^{\times}\), then \(o(\zeta^k) = n \iff (n, k) = 1\) (by general theory of cyclic groups).

Thus \(\theta\) is surjective.

\begin{align*} \tau_j \circ \tau_k (\zeta) =\tau_j(\zeta^k) = \zeta^{jk} \implies \tau_{jk} = \theta(jk) = \tau_j \circ \tau_k .\end{align*}

We have \(K \cong {\mathbb{Z}}_{20}^{\times}\) and \(\phi(20) = 8\), so \(K \cong {\mathbb{Z}}_8\), so we have the following subgroups and corresponding intermediate fields:

• \(0 \sim {\mathbb{Q}}(\zeta_{20})\)
• \({\mathbb{Z}}_2 \sim {\mathbb{Q}}(\omega_1)\)
• \({\mathbb{Z}}_4 \sim {\mathbb{Q}}(\omega_2)\)
• \({\mathbb{Z}}_8 \sim {\mathbb{Q}}\)

For some elements \(\omega_i\) which exist by the primitive element theorem.

False: Take \(L/K/F = {\mathbb{Q}}(\zeta_2, \sqrt[3] 2) \to {\mathbb{Q}}(\sqrt[3] 2) \to {\mathbb{Q}}\).

Then \(L/F\) is Galois, since it is the splitting field of \(x^3 - 2\) and \({\mathbb{Q}}\) has characteristic zero.

But \(K/F\) is not Galois, since it is not the splitting field of any irreducible polynomial.

True: If \(L/F\) is Galois, then \(L/K\) is normal and separable:

• \(L/K\) is normal, since if \(\sigma: L \hookrightarrow\overline K\) lifts the identity on \(K\) and fixes \(L\), i-t also lifts the identity on \(F\) and fixes \(L\) (and \(\overline K = \overline F\)).

• \(L/K\) is separable, since \(F[x] \subseteq K[x]\), and so if \(\alpha \in L\) where \(f(x) \coloneqq\min(\alpha, F)\) has no repeated factors, then \(f'(x) \coloneqq\min(\alpha, K)\) divides \(f\) and thus can not have repeated factors.

False: Use the fact that every quadratic extension is Galois, and take \(L/K/F = {\mathbb{Q}}(\sqrt[4] 2) \to {\mathbb{Q}}(\sqrt 2) \to {\mathbb{Q}}\).

Then each successive extension is quadratic (thus Galois) but \({\mathbb{Q}}(\sqrt[4] 2)\) is not the splitting field of any polynomial (noting that it does not split \(x^4 - 2\) completely.)

By the Galois correspondence, it suffices to show that the fixed field of \(H_1 \cap H_2\) is \(E_1 E_2\).

Let \(\sigma \in H_1 \cap H_2\); then \(\sigma \in \mathop{\mathrm{Aut}}(K)\) fixes both \(E_1\) and \(E_2\).

Not sure if this works – compositum is not literally product..?

Writing \(x \in E_1E_2\) as \(x=e_1 e_2\), we have \begin{align*} \sigma(x) = \sigma(e_1 e_2) = \sigma(e_1) \sigma(e_2) = e_1 e_2 =x, \end{align*}

so \(\sigma\) fixes \(E_1 E_2\).

That \(H_1 H_2 \subseteq G\) is clear, since if \(\sigma = \tau_1 \tau_2 \in H_1 H_2\), then each \(\tau_i\) is an automorphism of \(K\) that fixes \(E_i \supseteq {\mathbb{Q}}\), so each \(\tau_i\) fixes \({\mathbb{Q}}\) and thus \(\sigma\) fixes \({\mathbb{Q}}\).

By the Galois correspondence, the subgroup \(H_1H_2 \leq G\) will correspond to an intermediate field \(E\) such that \(K/E/{\mathbb{Q}}\) and \(E\) is the fixed field of \(H_1 H_2\).

But if \(\sigma \in H_1 H_2\), then \(\sigma = \tau_1 \tau_2\) where \(\tau_i\) is an automorphism of \(K\) that fixes \(E_i\), and so \begin{align*} \sigma(x) = x \iff \tau_1\tau_2(x) = x &\iff \tau_2(x) = x \\ &~\&~ \\ \tau_1(x) = x &\iff x \in E_1 \cap E_2 .\end{align*} .

• Since \({\left\lvert {F} \right\rvert} = q\) and \([E:F] = k\), we have \({\left\lvert {E} \right\rvert} = q^k\) and \({\left\lvert {E^{\times}} \right\rvert} = q^k-1\).

• Noting that \(\zeta \in E^{\times}\) we must have \(n = o(\zeta) \divides {\left\lvert {E^{\times}} \right\rvert} = q^k-1\) by Lagrange’s theorem.

• Rephrasing (a), we have \begin{align*} n \divides q^k-1 &\iff q^k-1 \cong 0 \operatorname{mod}n \\ &\iff q^k \cong 1 \operatorname{mod}n \\ &\iff m \coloneqq o(q) \divides k .\end{align*}

• Since \(m\divides k \iff k = \ell m\), (claim) there is an intermediate subfield \(M\) such that \begin{align*} E \leq M \leq F \quad k = [F:E] = [F:M] [M:E] = \ell m ,\end{align*}

  so \(M\) is a degree \(m\) extension of \(E\).

• Now consider \(M^{\times}\).

• By the argument in (a), \(n\) divides \(q^m - 1 = {\left\lvert {M^{\times}} \right\rvert}\), and \(M^{\times}\) is cyclic, so it contains a cyclic subgroup \(H\) of order \(n\).

• But then \(x\in H \implies p(x)\coloneqq x^n-1 = 0\), and since \(p(x)\) has at most \(n\) roots in a field.

• So \(H = \left\{{x \in M {~\mathrel{\Big\vert}~}x^n-1 = 0}\right\}\), i.e. \(H\) contains all solutions to \(x^n-1\) in \(E[x]\).

• But \(\zeta\) is one such solution, so \(\zeta \in H \subset M^{\times}\subset M\).

• Since \(F[\zeta]\) is the smallest field extension containing \(\zeta\), we must have \(F = M\), so \(\ell = 1\), and \(k = m\).

We can consider the quotient \(K = \displaystyle{\frac{{\mathbb{F}}_p[x]}{\left\langle{\pi(x)}\right\rangle}}\), which since \(\pi(x)\) is irreducible is an extension of \({\mathbb{F}}_p\) of degree \(d\) and thus a field of size \(p^d\) with a natural quotient map of rings \(\rho: {\mathbb{F}}_p[x] \to K\).

Since \(K^{\times}\) is a group of size \(p^d-1\), we know that for any \(y \in K^{\times}\), we have by Lagrange’s theorem that the order of \(y\) divides \(p^d-1\) and so \(y^{p^d} = y\).

So every element in \(K\) is a root of \(q(x) = x^{p^d}-x\).

Since \(\rho\) is a ring morphism, we have

\begin{align*} \rho(q(x)) = \rho(x^{p^d} - x) &= \rho(x)^{p^d} - \rho(x) = 0 \in K \\ &\iff q(x) \in \ker \rho \\ &\iff q(x) \in \left\langle{\pi(x)}\right\rangle \\ &\iff \pi(x) \divides q(x) = x^{p^d}-x ,\end{align*} where we’ve used that “to contain is to divide” in the last step.

• \(\zeta \coloneqq e^{2\pi i / 8}\) is a primitive \(8\)th root of unity
• The minimal polynomial of an \(n\)th root of unity is the \(n\)th cyclotomic polynomial \(\Phi_n\)
• The degree of the field extension is the degree of \(\Phi_8\), which is \begin{align*} \phi(8) = \phi(2^3) = 2^{3-1} \cdot (2-1) = 4 .\end{align*}
• So \([{\mathbb{Q}}(\zeta): {\mathbb{Q}}] = 4\).

• \({ \mathsf{Gal}} ({\mathbb{Q}}(\zeta)/{\mathbb{Q}}) \cong {\mathbb{Z}}/(8)^{\times}\cong {\mathbb{Z}}/(4)\) by general theory
• \({\mathbb{Z}}/(4)\) has exactly one subgroup of index 2.
• Thus there is exactly one intermediate field of degree 2 (a quadratic extension).

• Let \(L = {\mathbb{Q}}(\zeta, \sqrt[4] 2)\).

• Note \({\mathbb{Q}}(\zeta) = {\mathbb{Q}}(i, \sqrt 2)\)

  • \({\mathbb{Q}}(i, \sqrt{2})\subseteq {\mathbb{Q}}(\zeta)\)
    • \(\zeta_8^2 = i\), and \(\zeta_8 = \sqrt{2}^{-1}+ i\sqrt{2}^{-1}\) so \(\zeta_8 + \zeta_8 ^{-1}= 2/\sqrt{2} = \sqrt{2}\).
  • \({\mathbb{Q}}(\zeta) \subseteq {\mathbb{Q}}(i, \sqrt{2})\):
    • \(\zeta = e^{2\pi i / 8} = \sin(\pi/4) + i\cos(\pi/4) = {\sqrt 2 \over 2}\qty{1+i}\).

• Thus \(L = {\mathbb{Q}}(i, \sqrt{2})(\sqrt[4]{2}) = {\mathbb{Q}}(i, \sqrt 2, \sqrt[4] 2) = {\mathbb{Q}}(i, \sqrt[4]{2})\).

  • Uses the fact that \({\mathbb{Q}}(\sqrt 2) \subseteq {\mathbb{Q}}(\sqrt[4] 2)\) since \(\sqrt[4]{2}^2 = \sqrt{2}\)

• Conclude \begin{align*} [L: {\mathbb{Q}}] = [L: {\mathbb{Q}}(\sqrt[4] 2)] ~[{\mathbb{Q}}(\sqrt[4] 2): {\mathbb{Q}}] = 2 \cdot 4 = 8 \end{align*} using the fact that the minimal polynomial of \(i\) over any subfield of \({\mathbb{R}}\) is always \(x^2 + 1\), so \(\min_{{\mathbb{Q}}(\sqrt[4] 2)}(i) = x^2 + 1\) which is degree 2.

It suffices to show that \begin{align*} r\in R, ~t_1, t_2\in \operatorname{Tor}(M) \implies rt_1 + t_2 \in \operatorname{Tor}(M) .\end{align*}

We have \begin{align*} t_1 \in \operatorname{Tor}(M) &\implies \exists s_1 \neq 0 \text{ such that } s_1 t_1 = 0 \\ t_2 \in \operatorname{Tor}(M) &\implies \exists s_2 \neq 0 \text{ such that } s_2 t_2 = 0 .\end{align*}

Since \(R\) is an integral domain, \(s_1 s_2 \neq 0\). Then \begin{align*} s_1 s_2(rt_1 + t_2) &= s_1 s_2 r t_1 + s_1 s_2t_2 \\ &= s_2 r (s_1 t_1) + s_1 (s_2 t_2) \quad\text{since $R$ is commutative} \\ &= s_2 r(0) + s_1(0) \\ &= 0 .\end{align*}

Let \(R = {\mathbb{Z}}/6{\mathbb{Z}}\) as a \({\mathbb{Z}}/6{\mathbb{Z}}{\hbox{-}}\)module, which is not an integral domain as a ring.

Then \([3]_6\curvearrowright[2]_6 = [0]_6\) and \([2]_6\curvearrowright[3]_6 = [0]_6\), but \([2]_6 + [3]_6 = [5]_6\), where 5 is coprime to 6, and thus \([n]_6\curvearrowright[5]_6 = [0] \implies [n]_6 = [0]_6\). So \([5]_6\) is not a torsion element.

So the set of torsion elements are not closed under addition, and thus not a submodule.

Suppose \(R\) has zero divisors \(a,b \neq 0\) where \(ab = 0\). Then for any \(m\in M\), we have \(b\curvearrowright m \coloneqq bm \in M\) as well, but then \begin{align*} a\curvearrowright bm = (ab)\curvearrowright m = 0\curvearrowright m = 0_M ,\end{align*} so \(m\) is a torsion element for any \(m\).

• Suppose toward a contradiction \(\operatorname{Tor}(M)\) has rank \(n \geq 1\).
• Then \(\operatorname{Tor}(M)\) has a linearly independent generating set \(B = \left\{{\mathbf{r}_1, \cdots, \mathbf{r}_n}\right\}\), so in particular \begin{align*} \sum_{i=1}^n s_i \mathbf{r}_i = 0 \implies s_i = 0_R \,\forall i .\end{align*}
• Let \(\mathbf{r}\) be any of of these generating elements.
• Since \(\mathbf{r}\in \operatorname{Tor}(M)\), there exists an \(s\in R\setminus 0_R\) such that \(s\mathbf{r} = 0_M\).
• Then \(s\mathbf{r} = 0\) with \(s\neq 0\), so \(\left\{{\mathbf{r}}\right\} \subseteq B\) is not a linearly independent set, a contradiction.

• Let \(n = \operatorname{rank}M\), and let \(\mathcal B = \left\{{\mathbf{r}_i}\right\}_{i=1}^n \subseteq R\) be a generating set.
• Let \(\tilde M \coloneqq M/\operatorname{Tor}(M)\) and \(\pi: M \to M'\) be the canonical quotient map.

Note that the proof follows immediately.

• Suppose that \begin{align*} \sum_{i=1}^n s_i (\mathbf{r}_i + \operatorname{Tor}(M)) = \mathbf{0}_{\tilde M} .\end{align*}

• Then using the definition of coset addition/multiplication, we can write this as \begin{align*} \sum_{i=1}^n \qty { s_i \mathbf{r}_i + \operatorname{Tor}(M)} = \qty{ \sum_{i=1}^n s_i \mathbf{r}_i} + \operatorname{Tor}(M) = 0_{\tilde M} .\end{align*}

• Since \(\tilde{\mathbf{x}} = 0 \in \tilde M \iff \tilde{\mathbf{x}} = \mathbf{x} + \operatorname{Tor}(M)\) where \(\mathbf{x} \in \operatorname{Tor}(M)\), this forces \(\sum s_i \mathbf{r}_i \in \operatorname{Tor}(M)\).

• Then there exists a scalar \(\alpha\in R^{\bullet}\) such that \(\alpha \sum s_i \mathbf{r}_i = 0_M\).

• Since \(R\) is an integral domain and \(\alpha \neq 0\), we must have \(\sum s_i \mathbf{r}_i = 0_M\).

• Since \(\left\{{\mathbf{r}_i}\right\}\) was linearly independent in \(M\), we must have \(s_i = 0_R\) for all \(i\).

• Write \(\pi(\mathcal B) = \left\{{\mathbf{r}_i + \operatorname{Tor}(M)}\right\}_{i=1}^n\) as a set of cosets.

• Letting \(\mathbf{x} \in M'\) be arbitrary, we can write \(\mathbf{x} = \mathbf{m} + \operatorname{Tor}(M)\) for some \(\mathbf{m} \in M\) where \(\pi(\mathbf{m}) = \mathbf{x}\) by surjectivity of \(\pi\).

• Since \(\mathcal B\) is a basis for \(M\), we have \(\mathbf{m} = \sum_{i=1}^n s_i \mathbf{r}_i\), and so \begin{align*} \mathbf{x} &= \pi(\mathbf{m}) \\ &\coloneqq\pi\qty{ \sum_{i=1}^n s_i \mathbf{r}_i} \\ &= \sum_{i=1}^n s_i \pi(\mathbf{r}_i) \quad\text{since $\pi$ is an $R{\hbox{-}}$module morphism}\\ &\coloneqq\sum_{i=1}^n s_i \mathbf{(}\mathbf{r}_i + \operatorname{Tor}(M)) ,\end{align*} which expresses \(\mathbf{x}\) as a linear combination of elements in \(\mathcal B'\).

Notation: Let \(0_R\) denote \(0\in R\) regarded as a ring element, and \(\mathbf{0} \in R\) denoted \(0_R\) regarded as a module element (where \(R\) is regarded as an \(R{\hbox{-}}\)module over itself)

An \(R{\hbox{-}}\)module \(M\) is free if any of the following conditions hold:

• \(M\) admits an \(R{\hbox{-}}\)linearly independent spanning set \(\left\{{\mathbf{b}_\alpha}\right\}\), so \begin{align*}m\in M \implies m = \sum_\alpha r_\alpha \mathbf{b}_\alpha\end{align*} and \begin{align*}\sum_\alpha r_\alpha \mathbf{b}_\alpha = 0_M \implies r_\alpha = 0_R\end{align*} for all \(\alpha\).
• \(M\) admits a decomposition \(M \cong \bigoplus_{\alpha} R\) as a direct sum of \(R{\hbox{-}}\)submodules.
• There is a nonempty set \(X\) an monomorphism \(X\hookrightarrow M\) of sets such that for every \(R{\hbox{-}}\)module \(N\), every set map \(X\to N\) lifts to a unique \(R{\hbox{-}}\)module morphism \(M\to N\), so the following diagram commutes:

Equivalently, \begin{align*} \mathop{\mathrm{Hom}}_{\mathsf{Set}}(X, \mathop{\mathrm{Forget}}(N)) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) .\end{align*}

• Define the annihilator: \begin{align*} \operatorname{Ann}(m) \coloneqq\left\{{r\in R {~\mathrel{\Big\vert}~}r\cdot m = 0_M}\right\} {~\trianglelefteq~}R .\end{align*}
  • Note that \(mR \cong R/\operatorname{Ann}(m)\).
• Define the torsion submodule: \begin{align*} M_t \coloneqq\left\{{m\in M {~\mathrel{\Big\vert}~}\operatorname{Ann}(m) \neq 0}\right\} \leq M \end{align*}
• \(M\) is torsionfree iff \(M_t = 0\) is the trivial submodule.

• Let the following be an SES where \(F\) is a free \(R{\hbox{-}}\)module: \begin{align*} 0 \to N \to M \xrightarrow{\pi} F \to 0 .\end{align*}

• Since \(F\) is free, there is a generating set \(X = \left\{{x_\alpha}\right\}\) and a map \(\iota:X\hookrightarrow F\) satisfying the 3rd property from (a).

  • If we construct any map \(f: X\to M\), the universal property modules will give a lift \(\tilde f: F\to M\)

• Identify \(X\) with \(\iota(X) \subseteq F\).

• For every \(x\in X\), the preimage \(\pi^{-1}(x)\) is nonempty by surjectivity. So arbitrarily pick any preimage.

• \(\left\{{\iota(x_\alpha)}\right\} \subseteq F\) and \(\pi\) is surjective, so choose fibers \(\left\{{y_\alpha}\right\} \subseteq M\) such that \(\pi(y_\alpha) = \iota(x_\alpha)\) and define \begin{align*} f: X&\to M \\ x_\alpha &\mapsto y_\alpha .\end{align*}

• The universal property yields \(h: F\to M\):

• It remains to check that it’s a section.
  • Write \(f= \sum r_i x_i\), then since both maps are \(R{\hbox{-}}\)module morphism, by \(R{\hbox{-}}\)linearity we can write \begin{align*} (\pi \circ h)(f) &= (\pi \circ h)\qty{ \sum r_i x_i } \\ &= \sum r_i (\pi \circ h)(x_i) ,\end{align*} but since \(h(x_i) \in \pi^{-1}(x_i)\), we have \((\pi \circ h)(x_i) = x_i\). So this recovers \(f\).

• Free implies projective

  • Universal property of projective objects: for every epimorphism \(\pi:M\twoheadrightarrow N\) and every \(f:P\to N\) there exists a unique lift \(\tilde f: P\to M\):

  

  • Construct \(\phi\) in the following diagram using the same method as above (surjectivity to pick elements in preimage):

Link to Diagram

• Now take the identity map, then commutativity is equivalent to being a section.

There is a SES

• Let \(m+M_t \in M/M_t\) be arbitrary. Suppose this is a torsion element, the claim is that it must be the trivial coset. This will follow if \(m\in M_t\)
• Since this is torsion, there exists \(r\in R\) such that \begin{align*} M_t = r(m + M_t) \coloneqq(rm) + M_t \implies rm\in M_t .\end{align*}
• Then \(rm\) is torsion in \(M\), so there exists some \(s\in R\) such \(s(rm) = 0_M\).
• Then \((sr)m = 0_M\) which forces \(m\in M_t\)

By the correspondence theorem, submodules of \(M/N\) biject with submodules \(A\) of \(M\) containing \(N\).

So

• \(M\) is maximal:

• \(\iff\) no such (proper, nontrivial) submodule \(A\) exists

• \(\iff\) there are no (proper, nontrivial) submodules of \(M/N\)

• \(\iff M/N\) is simple.

Identify \({\mathbb{Z}}{\hbox{-}}\)modules with abelian groups, then by (a), \(N\) is maximal \(\iff\) \(M/N\) is simple \(\iff\) \(M/N\) has no nontrivial proper subgroups.\

By Cauchy’s theorem, if \({\left\lvert {M/N} \right\rvert} = ab\) is a composite number, then \(a\divides ab \implies\) there is an element (and thus a subgroup) of order \(a\). In this case, \(M/N\) contains a nontrivial proper cyclic subgroup, so \(M/N\) is not simple. So \({\left\lvert {M/N} \right\rvert}\) can not be composite, and therefore must be prime.

• Let \(G = \left\{{x \in {\mathbb{C}}{~\mathrel{\Big\vert}~}x^n=1 \text{ for some }n\in {\mathbb{N}}}\right\}\), and suppose \(H < G\) is a proper submodule.

• Since \(H\neq G\), there is some \(p\) and some \(k\) such that \(\zeta_{p^k}\not\in H\).

  • Otherwise, if \(H\) contains every \(\zeta_{p^k}\) it contains every \(\zeta_n\)

Then there must be a prime \(p\) such that the \(\zeta_{p^k} \not \in H\) for all \(k\) greater than some constant \(m\) – otherwise, we can use the fact that if \(\zeta_{p^k} \in H\) then \(\zeta_{p^\ell} \in H\) for all \(\ell \leq k\), and if \(\zeta_{p^k} \in H\) for all \(p\) and all \(k\) then \(H = G\).

But this means there are infinitely many elements in \(G\setminus H\), and so \(\infty = [G: H] = {\left\lvert {G/H} \right\rvert}\) is not a prime. Thus by (b), \(H\) can not be maximal, a contradiction.

• Let \(A \in \operatorname{GL}(m, {\mathbb{F}})\).

• Since \(A^n\) is diagonalizable, \(\min_{A^n}(x) \in {\mathbb{F}}[x]\) is separable and thus factors as a product of \(m\) distinct linear factors: \begin{align*} \min_{A^n}(x) = \prod_{i=1}^m (x-\lambda_i), \quad \min_{A^n}(A^n) = 0 \end{align*}

  where \(\left\{{\lambda_i}\right\}_{i=1}^m \subset {\mathbb{F}}\) are the distinct eigenvalues of \(A^n\).

• Moreover \(A\in \operatorname{GL}(m,{\mathbb{F}}) \implies A^n \in \operatorname{GL}(m,{\mathbb{F}})\): \(A\) is invertible \(\iff \operatorname{det}(A) = d \in {\mathbb{F}}^{\times}\), and so \(\operatorname{det}(A^n) = \operatorname{det}(A)^n = d^n \in {\mathbb{F}}^{\times}\) using the fact that the determinant is a ring morphism \(\operatorname{det}: \operatorname{Mat}(m\times m) \to{\mathbb{F}}\) and \({\mathbb{F}}^{\times}\) is closed under multiplication.

• So \(A^n\) is invertible, and thus has trivial kernel, and thus zero is not an eigenvalue, so \(\lambda_i \neq 0\) for any \(i\).

• Since the \(\lambda_i\) are distinct and nonzero, this implies \(x^k\) is not a factor of \(\mu_{A^n}(x)\) for any \(k\geq 0\). Thus the \(m\) terms in the product correspond to precisely \(m\) distinct linear factors.

• We can now construct a polynomial that annihilates \(A\), namely \begin{align*} q_A(x) \coloneqq\min_{A^n}(x^n) = \prod_{i=1}^m (x^n-\lambda_i) \in {\mathbb{F}}[x], \end{align*}

  where we can note that \(q_A(A) = \min_{A^n}(A^n) = 0\), and so \(\min_A(x) \divides q_A(x)\) by minimality.

• This reduces to showing that no pair \(x^n-\lambda_i, x^n-\lambda_j\) share a root. and that \(x^n-\lambda_i\) does not have multiple roots.

• For the first claim, we can factor \begin{align*} x^n - \lambda_i = \prod_{k=1}^n (x - \lambda_i^{1\over n} e^{2\pi i k \over n}) \coloneqq\prod_{k=1}^n (x-\lambda^{1\over n} \zeta_n^k) ,\end{align*} where we now use the fact that \(i\neq j \implies \lambda_i^{1\over n} \neq \lambda_j^{1\over n}\). Thus no term in the above product appears as a factor in \(x^n - \lambda_j\) for \(j\neq i\).

• For the second claim, we can check that \({\frac{\partial }{\partial x}\,}\qty{x^n - \lambda_i} = nx^{n-1}\neq 0\in {\mathbb{F}}\), and \(\gcd(x^n-\lambda_i, nx^{n-1}) = 1\) since the latter term has only the roots \(x=0\) with multiplicity \(n-1\), whereas \(\lambda_i\neq 0 \implies\) zero is not a root of \(x^n-\lambda_i\).

But now since \(q_A(x)\) has exactly distinct linear factors in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu[x]\) and \(\min_A(x) \divides q_A(x)\), \(\min_A(x) \in {\mathbb{F}}[x]\) can only have distinct linear factors, and \(A\) is thus diagonalizable over \({\mathbb{F}}\).

Letting \(\mathbf{v}\) be fixed, since \(\left\{{A^j \mathbf{v}}\right\}\) spans \(V\) we have A \begin{align*} B\mathbf{v} = \sum_{j=0}^{n-1}c_j A^j \mathbf{v} .\end{align*}

So let \(p(x) = \sum_{j=0}^{n-1}c_jx^j\). Then consider how \(B\) acts on any basis vector \(A^k \mathbf{v}\).

We have \begin{align*} BA^k \mathbf{v} &= A^k B\mathbf{v} \\ &= A^k p(A) \mathbf{v} \\ &= p(A) A^k \mathbf{v} ,\end{align*}

so \(B = p(A)\) as operators since their actions agree on every basis vector in \(V\).

• If \(\left\{{A^j \mathbf{v}_k {~\mathrel{\Big\vert}~}0\leq j \leq n-1}\right\}\) is linearly independent, this means that \(A\) does satisfy any polynomial of degree \(d < n\).

• So \(\deg m_A(x) = n\), and since \(m_A(x)\) divides \(\chi_A(x)\) and both are monic degree polynomials of degree \(n\), they must be equal.

• Let \(A\curvearrowright k[x]\) by \(A \curvearrowright p(x) \coloneqq p(A)\). This induces an invariant factor decomposition \(V =\cong \bigoplus k[x]/(f_i)\).

• Since the product of the invariant factors is the characteristic polynomial, the largest invariant factor is the minimal polynomial, and these two are equal, there can only be one invariant factor and thus the invariant factor decomposition is \begin{align*} V\cong \frac{k[x]}{(\chi_A(x))} \end{align*} as an isomorphism of \(k[x]{\hbox{-}}\)modules.

• So \(V\) is a cyclic \(k[x]\) module, which means that \(V = k[x]\curvearrowright\mathbf{v}\) for some \(\mathbf{v}\in V\) such that \(\operatorname{Ann}(\mathbf{v}) = \chi_A(x)\), i.e. there is some element \(\mathbf{v}\in V\) whose orbit is all of \(V\).

• But then noting that monomials span \(k[x]\) as a \(k{\hbox{-}}\)module, we can write \begin{align*} V &\cong k[x] \curvearrowright\mathbf{v} \\ &\coloneqq\left\{{f(x) \curvearrowright\mathbf{v} {~\mathrel{\Big\vert}~}f \in k[x]}\right\} \\ &= \mathop{\mathrm{span}}_k \left\{{x^k \curvearrowright\mathbf{v} {~\mathrel{\Big\vert}~}k \geq 0}\right\} \\ &\coloneqq\mathop{\mathrm{span}}_k \left\{{A^k\mathbf{v} {~\mathrel{\Big\vert}~}k \geq 0}\right\} ,\end{align*} where we’ve used that \(x\) acts by \(A\) and thus \(x^k\) acts by \(A^k\).

• Moreover, we can note that if \(\ell \geq \deg \chi_A(x)\), then \(A^\ell\) is a linear combination of \(\left\{{A^j \mathrel{\Big|}0 \leq j \leq n-1}\right\}\), and so \begin{align*} V &\cong \mathop{\mathrm{span}}_k \left\{{A^\ell\mathbf{v} {~\mathrel{\Big\vert}~}\ell \geq 0}\right\} \\ &= \mathop{\mathrm{span}}_k \left\{{A^\ell \mathbf{v} {~\mathrel{\Big\vert}~}1 \leq \ell \leq n-1}\right\} .\end{align*}

• Let \(\mathbf{v} \in \Lambda\), so \(\mathbf{v} = \sum_{i=1}^n r_i \mathbf{e}_i\) where \(r_i \in {\mathbb{Z}}\) for all \(i\).

• Then if \(\mathbf{x} = \sum_{j=1}^n s_j \mathbf{e}_j \in \Lambda\) is arbitrary, we have \(s_j \in {\mathbb{Z}}\) for all \(j\) and \begin{align*} {\left\langle {\mathbf{v}},~{\mathbf{x}} \right\rangle} &= {\left\langle {\sum_{i=1}^n r_i \mathbf{e}_i},~{\sum_{j=1}^n s_j \mathbf{e}_j } \right\rangle} \\ &= \sum_{i=1}^n \sum_{j=1}^n r_i s_j {\left\langle {\mathbf{e}_i},~{\mathbf{e}_j } \right\rangle} \in {\mathbb{Z}} \end{align*} since this is a sum of products of integers (since \({\left\langle {\mathbf{e}_i},~{\mathbf{e}_j} \right\rangle} \in {\mathbb{Z}}\) for each \(i, j\) pair by assumption) so \(\mathbf{v} \in \Lambda {}^{ \vee }\) by definition.

• Suppose \(\operatorname{det}M = 0\). Then \(\ker M \neq \mathbf{0}\), so let \(\mathbf{v} \in \ker M\) be given by \(\mathbf{v} = \sum_{i=1}^n v_i \mathbf{e}_i \neq \mathbf{0}\).

• Note that \begin{align*} M\mathbf{v} = 0 &\implies \left[ \begin{array}{ccc} \mathbf{e}_1 \cdot \mathbf{e}_1 & \mathbf{e}_1 \cdot \mathbf{e}_2 & \cdots \\ \mathbf{e}_2 \cdot \mathbf{e}_1 & \mathbf{e}_2 \cdot \mathbf{e}_2 & \cdots \\ \vdots & \vdots & \ddots \end{array} \right] \left[\begin{array}{c} v_1 \\ v_2 \\ \vdots \end{array}\right] = \mathbf{0} \\ \\ &\implies \sum_{j=1}^n v_j{\left\langle {\mathbf{e}_k},~{\mathbf{e}_j} \right\rangle} = 0 {\quad \operatorname{for each fixed} \quad} k .\end{align*}

• We can now note that \({\left\langle {\mathbf{e}_k},~{\mathbf{v}} \right\rangle} = \sum_{j=1}^n v_j {\left\langle {\mathbf{e}_k},~{\mathbf{e}_j} \right\rangle} = 0\) for every \(k\) by the above observation, which forces \(\mathbf{v} = 0\) by non-degeneracy of \({\left\langle {{-}},~{{-}} \right\rangle}\), a contradiction.

???

\todo[inline]{Missing work!}

\(\implies\):

• By contrapositive, suppose there is a proper nonzero invariant subspace \(W<V\) with \(T(W) \subseteq W\), we will show the characteristic polynomial \(f \coloneqq\chi_{V, T}(x)\) is reducible.
• Since \(T(W)\subseteq W\), the restriction \(g\coloneqq\chi_{V, T}(x) \mathrel{\Big|}_W: W\to W\) is a linear operator on \(W\).

• Choose an ordered basis for \(W\), say \({\mathcal{B}}_W \coloneqq\left\{{\mathbf{w}_1, \cdots, \mathbf{w}_k}\right\}\) where \(k=\dim_F(W)\)

• Claim: this can be extended to a basis of \(V\), say \({\mathcal{B}}_V \coloneqq\left\{{\mathbf{w}_1, \cdots, \mathbf{w}_k, \mathbf{v}_1, \cdots, \mathbf{v}_j}\right\}\) where \(k+j = \dim_F(V)\).

  • Note that since \(W<V\) is proper, \(j\geq 1\).

• Restrict \(T\) to \(W\) to get \(T_W\), then let \(B = [T_W]_{{\mathcal{B}}_W}\) be the matrix representation of \(T_W\) with respect to \({\mathcal{B}}_W\).

• Now consider the matrix representation \([T]_{{\mathcal{B}}_V}\), in block form this is given by \begin{align*} [T]_{{\mathcal{B}}_V} = \begin{bmatrix} B & C \\ 0 & D \end{bmatrix} \end{align*} where we’ve used that \(W<V\) is proper to get the existence of \(C, D\) (there is at least one additional row/column since \(j\geq 1\) in the extended basis.)

  \todo[inline]{Why?}

• Now expand along the first column block to obtain \begin{align*} \chi_{T, V}(x) \coloneqq\operatorname{det}([T]_{{\mathcal{B}}_V} - xI) = \operatorname{det}(B - xI)\cdot \operatorname{det}(D - xI) \coloneqq\chi_{T, W}(x) \cdot \operatorname{det}(D-xI) .\end{align*}

• Claim: \(\operatorname{det}(D - xI) \in xF[x]\) is nontrivial

• The claim follows because this forces \(\deg(\operatorname{det}(D-xI)) \geq 1\) and so \(\chi_{T, W}(x)\) is a proper divisor of \(\chi_{T, V}(x)\).

• Thus \(f\) is reducible.

\(\impliedby\)

• Suppose \(f\) is reducible, then we will produce a proper \(T{\hbox{-}}\)invariant subspace.
• Claim: if \(f\) is reducible, there exists a nonzero, noncyclic vector \(\mathbf{v}\).
• Then \(\mathop{\mathrm{span}}_k\left\{{T^j\mathbf{v}}\right\}_{j=1}^d\) is a \(T{\hbox{-}}\)invariant subspace that is nonzero, and not the entire space since \(\mathbf{v}\) is not cyclic.

• Let \(\min_{T, F}(x)\) be the minimal polynomial of \(T\) and \(\chi_{T, F}(x)\) be its characteristic polynomial.
• By Cayley-Hamilton, \(\min_{T, F}(x)\) divides \(\chi_{T, F}\)
• Since \(\chi_{T, F}\) is irreducible, these polynomials are equal.
• Claim: \(T/F\) is diagonalizable \(\iff \min_{T, F}\) splits over \(F\) and is squarefree.
• Replace \(F\) with its algebraic closure, then \(\min_{T, F}\) splits.
• Claim: in characteristic zero, every irreducible polynomial is separable
  • Proof: it must be the case that either \(\gcd(f, f') = 1\) or \(f' \equiv 0\), where the second case only happens in characteristic \(p>0\).
  • The first case is true because \(\deg f' < \deg f\), and if \(\gcd(f, f') = p\), then \(\deg p < \deg f\) and \(p\divides f\) forces \(p=1\) since \(f\) is irreducible.
• So \(\min_{T, F}\) splits into distinct linear factors
• Thus \(T\) is diagonalizable.