S.-T. Yau College Student Mathematics Contests 2022

(a) Let $p(x)=a_n{x}^n+\cdots +a_{1}x+a_0\in R[x]$$p(x)=a_n{x}^n+\cdots +a_{1}x+a_0\in R[x]$p(x)=a_(n)x^(n)+cdots+a_(1)x+a_(0)in R[x]p(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0} \in R[x]$p(x)=a_n{x}^n+\cdots +a_{1}x+a_0\in R[x]$ be a polynomial over an integral domain $R$$R$RR$R$. Let $K$$K$KK$K$ denote the fraction field of $R$$R$RR$R$. Suppose $a/b\in K$$a/b\in K$a//b in Ka / b \in K$a/b\in K$ is a root of $p(x)$$p(x)$p(x)p(x)$p(x)$, where $a,b\in R$$a,b\in R$a,b in Ra, b \in R$a,b\in R$ and are relatively prime. Then, show that $a\mid a_0$$a\mid a_0$a∣a_(0)a \mid a_{0}$a\mid a_0$ and $b\mid a_n$$b\mid a_n$b∣a_(n)b \mid a_{n}$b\mid a_n$.

(b) Prove that $\mathbf{Q}(\sqrt{2},\sqrt{3})=\mathbf{Q}(\sqrt{2}+\sqrt{3})$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\mathbf{Q}(\sqrt{2}+\sqrt{3})$Q(sqrt2,sqrt3)=Q(sqrt2+sqrt3)\mathbf{Q}(\sqrt{2}, \sqrt{3})=\mathbf{Q}(\sqrt{2}+\sqrt{3})$\mathbf{Q}(\sqrt{2},\sqrt{3})=\mathbf{Q}(\sqrt{2}+\sqrt{3})$.

(a) Easy.

(b) It suffices to show $\mathbf{Q}(\sqrt{2},\sqrt{3})\subset \mathbf{Q}(\sqrt{2}+\sqrt{3})$$\mathbf{Q}(\sqrt{2},\sqrt{3})\subset \mathbf{Q}(\sqrt{2}+\sqrt{3})$Q(sqrt2,sqrt3)subQ(sqrt2+sqrt3)\mathbf{Q}(\sqrt{2}, \sqrt{3}) \subset \mathbf{Q}(\sqrt{2}+\sqrt{3})$\mathbf{Q}(\sqrt{2},\sqrt{3})\subset \mathbf{Q}(\sqrt{2}+\sqrt{3})$.

Let $\alpha =\sqrt{2}+\sqrt{3}$$\alpha =\sqrt{2}+\sqrt{3}$alpha=sqrt2+sqrt3\alpha=\sqrt{2}+\sqrt{3}$\alpha =\sqrt{2}+\sqrt{3}$. It is a root of

By (a), $p(x)$$p(x)$p(x)p(x)$p(x)$ has no rational roots. We claim that $p(x)$$p(x)$p(x)p(x)$p(x)$ is irreducible over $\mathbf{Z}$$\mathbf{Z}$Z\mathbf{Z}$\mathbf{Z}$. Otherwise, $p(x)=(a+bx+c{x}^2)(d+$$p(x)=\left(a+bx+c{x}^2\right)(d+$p(x)=(a+bx+cx^(2))(d+p(x)=\left(a+b x+c x^{2}\right)(d+$p(x)=(a+bx+c{x}^2)(d+$ $ex+f{x}^2$$ex+f{x}^2$ex+fx^(2)e x+f x^{2}$ex+f{x}^2$ ). A direct computation will yield such a decomposition is impossible. Hence, $p(x)$$p(x)$p(x)p(x)$p(x)$ is a minimal polynomial $\alpha$$\alpha$alpha\alpha$\alpha$ over $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$. As $\mathbf{Q}(\alpha )$$\mathbf{Q}(\alpha )$Q(alpha)\mathbf{Q}(\alpha)$\mathbf{Q}(\alpha )$ is a vector subspace of $\mathbf{Q}(\sqrt{2},\sqrt{3})$$\mathbf{Q}(\sqrt{2},\sqrt{3})$Q(sqrt2,sqrt3)\mathbf{Q}(\sqrt{2}, \sqrt{3})$\mathbf{Q}(\sqrt{2},\sqrt{3})$, we obtain

This implies the statement.

Problem 2. Let $R$$R$RR$R$ be an integral domain with the fraction field $K$$K$KK$K$. An $R$$R$RR$R$-module $P$$P$PP$P$ is projective if there is an $R$$R$RR$R$-module $Q$$Q$QQ$Q$ such that $P\oplus Q\cong F$$P\oplus Q\cong F$P o+Q~=FP \oplus Q \cong F$P\oplus Q\cong F$ for some free $R$$R$RR$R$-module $F$$F$FF$F$. A fractional ideal $A$$A$AA$A$ is an $R$$R$RR$R$-submodule of $K$$K$KK$K$ such that $A=d^{-1}I$$A=d^{-1}I$A=d^(-1)IA=d^{-1} I$A=d^{-1}I$ for some ideal $I$$I$II$I$ of $R$$R$RR$R$ and a nonzero element $d\in R$$d\in R$d in Rd \in R$d\in R$. A fractional ideal $A$$A$AA$A$ is called invertible if $AB=R$$AB=R$AB=RA B=R$AB=R$ for some fractional ideal $B$$B$BB$B$.

Show that an invertible fractional ideal $A$$A$AA$A$ is a projective $R$$R$RR$R$-module.

Solution: Assume that $A$$A$AA$A$ is an invertible fractional ideal. Let $A^{-1}$$A^{-1}$A^(-1)A^{-1}$A^{-1}$ its inverse. Then

for some $a_1,\dots ,a_n\in A$$a_1,\dots ,a_n\in A$a_(1),dots,a_(n)in Aa_{1}, \ldots, a_{n} \in A$a_1,\dots ,a_n\in A$ and $a_1^{\mathrm{\prime }},\dots ,a_n^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$$a_1^{\mathrm{\prime }},\dots ,a_n^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$a_(1)^('),dots,a_(n)^(')inA^(')a_{1}^{\prime}, \ldots, a_{n}^{\prime} \in A^{\prime}$a_1^{\mathrm{\prime }},\dots ,a_n^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ since $A{A}^{-1}=R$$A{A}^{-1}=R$AA^(-1)=RA A^{-1}=R$A{A}^{-1}=R$. Let $S$$S$SS$S$ be a free $R$$R$RR$R$-module of rank $n$$n$nn$n$, say, generated by $y_1,\dots ,y_n$$y_1,\dots ,y_n$y_(1),dots,y_(n)y_{1}, \ldots, y_{n}$y_1,\dots ,y_n$. Define $\phi :S\to A$$\phi :S\to A$varphi:S rarr A\varphi: S \rightarrow A$\phi :S\to A$ by $\phi \left(y_i\right)=a_i$$\phi \left(y_i\right)=a_i$varphi(y_(i))=a_(i)\varphi\left(y_{i}\right)=a_{i}$\phi \left(y_i\right)=a_i$ and $\psi :A\to S$$\psi :A\to S$psi:A rarr S\psi: A \rightarrow S$\psi :A\to S$ by $\phi (c)=c(a_1^{\mathrm{\prime }}{y}_1+\cdots a_n^{\mathrm{\prime }}{y}_n)$$\phi (c)=c\left(a_1^{\mathrm{\prime }}{y}_1+\cdots a_n^{\mathrm{\prime }}{y}_n\right)$varphi(c)=c(a_(1)^(')y_(1)+cdotsa_(n)^(')y_(n))\varphi(c)=c\left(a_{1}^{\prime} y_{1}+\cdots a_{n}^{\prime} y_{n}\right)$\phi (c)=c(a_1^{\mathrm{\prime }}{y}_1+\cdots a_n^{\mathrm{\prime }}{y}_n)$. This makes sense because $c{a}_i^{\mathrm{\prime }}\in R$$c{a}_i^{\mathrm{\prime }}\in R$ca_(i)^(')in Rc a_{i}^{\prime} \in R$c{a}_i^{\mathrm{\prime }}\in R$. Obviously, $\phi \psi =i{d}_A$$\phi \psi =i{d}_A$varphi psi=id_(A)\varphi \psi=i d_{A}$\phi \psi =i{d}_A$, hence $A$$A$AA$A$ is a direct summand of $S$$S$SS$S$. In other words, $A$$A$AA$A$ is a projective module.

Problem 3. Give a direct proof that the Lie algebra $\mathfrak{s}(4,\mathbf{C})$$\mathfrak{s}(4,\mathbf{C})$s(4,C)\mathfrak{s}(4, \mathbf{C})$\mathfrak{s}(4,\mathbf{C})$ is isomorphic to the Lie algebra $\mathrm{\Im }(6,\mathbf{C})$$\mathrm{\Im }(6,\mathbf{C})$ℑ(6,C)\mathfrak{\Im}(6, \mathbf{C})$\mathrm{\Im }(6,\mathbf{C})$. (You should construct a Lie algebra homomorphism and prove that it is an ismorphism; you should not use Dynkin diagrams or the classification theory of simple Lie algebras.)

Solution: We have the representation $\mathfrak{s}\mathfrak{l}(4,\mathbf{C})\to \mathfrak{g}\mathfrak{l}(W)$$\mathfrak{s}\mathfrak{l}(4,\mathbf{C})\to \mathfrak{g}\mathfrak{l}(W)$sl(4,C)rarrgl(W)\mathfrak{s l}(4, \mathbf{C}) \rightarrow \mathfrak{g l}(W)$\mathfrak{s}\mathfrak{l}(4,\mathbf{C})\to \mathfrak{g}\mathfrak{l}(W)$ where $W={\wedge }^2{\mathbf{C}}^4$$W={\wedge }^2{\mathbf{C}}^4$W=^^^(2)C^(4)W=\wedge^{2} \mathbf{C}^{4}$W={\wedge }^2{\mathbf{C}}^4$. Let $e_1,e_2,e_3,e_4$$e_1,e_2,e_3,e_4$e_(1),e_(2),e_(3),e_(4)e_{1}, e_{2}, e_{3}, e_{4}$e_1,e_2,e_3,e_4$ be the standard basis of ${\mathbf{C}}^4$${\mathbf{C}}^4$C^(4)\mathbf{C}^{4}${\mathbf{C}}^4$. Then $e_1\wedge e_2\wedge e_3\wedge e_4$$e_1\wedge e_2\wedge e_3\wedge e_4$e_(1)^^e_(2)^^e_(3)^^e_(4)e_{1} \wedge e_{2} \wedge e_{3} \wedge e_{4}$e_1\wedge e_2\wedge e_3\wedge e_4$ gives an identification ${\wedge }^4{\mathbf{C}}^4\cong \mathbf{C}$${\wedge }^4{\mathbf{C}}^4\cong \mathbf{C}$^^^(4)C^(4)~=C\wedge^{4} \mathbf{C}^{4} \cong \mathbf{C}${\wedge }^4{\mathbf{C}}^4\cong \mathbf{C}$. We define a complex symmetric bilinear form $S:W\times W\to \mathbf{C}$$S:W\times W\to \mathbf{C}$S:W xx W rarrCS: W \times W \rightarrow \mathbf{C}$S:W\times W\to \mathbf{C}$ by

for $\theta ,\tau \in W$$\theta ,\tau \in W$theta,tau in W\theta, \tau \in W$\theta ,\tau \in W$. By writing out an explicit orthogonal basis, you can show that $S$$S$SS$S$ is non-degenerate. Then, one checks that for any $A\in \mathfrak{S}\mathfrak{l}(4,\mathbf{C})$$A\in \mathfrak{S}\mathfrak{l}(4,\mathbf{C})$A inSl(4,C)A \in \mathfrak{S l}(4, \mathbf{C})$A\in \mathfrak{S}\mathfrak{l}(4,\mathbf{C})$, we have

Note here that $\mathfrak{B}\mathfrak{l}(4,\mathbf{C})$$\mathfrak{B}\mathfrak{l}(4,\mathbf{C})$Bl(4,C)\mathfrak{B l}(4, \mathbf{C})$\mathfrak{B}\mathfrak{l}(4,\mathbf{C})$ acts trivially on ${\wedge }^4{\mathbf{C}}^4$${\wedge }^4{\mathbf{C}}^4$^^^(4)C^(4)\wedge^{4} \mathbf{C}^{4}${\wedge }^4{\mathbf{C}}^4$.

Hence, we obtain a Lie algebra homomorphism $\mathfrak{g}\mathfrak{l}(4,\mathbf{C})\to \mathfrak{g}\mathfrak{d}(6,\mathbf{C})$$\mathfrak{g}\mathfrak{l}(4,\mathbf{C})\to \mathfrak{g}\mathfrak{d}(6,\mathbf{C})$gl(4,C)rarrgd(6,C)\mathfrak{g l}(4, \mathbf{C}) \rightarrow \mathfrak{g} \mathfrak{d}(6, \mathbf{C})$\mathfrak{g}\mathfrak{l}(4,\mathbf{C})\to \mathfrak{g}\mathfrak{d}(6,\mathbf{C})$.This must be an isomorphism since both Lie algebras have the same dimension and $\mathfrak{l}(4,\mathbf{C})$$\mathfrak{l}(4,\mathbf{C})$l(4,C)\mathfrak{l}(4, \mathbf{C})$\mathfrak{l}(4,\mathbf{C})$ is simple.

Problem 4. Let $A={\mathcal{O}}_K$$A={\mathcal{O}}_K$A=O_(K)A=\mathcal{O}_{K}$A={\mathcal{O}}_K$ be the ring of integers of a number field $K$$K$KK$K$. Given a nonzero ideal $\mathfrak{a}\subset A$$\mathfrak{a}\subset A$asub A\mathfrak{a} \subset A$\mathfrak{a}\subset A$ and an arbitrary nonzero element $a\in \mathfrak{a}$$a\in \mathfrak{a}$a inaa \in \mathfrak{a}$a\in \mathfrak{a}$, show that there exists $b\in \mathfrak{a}$$b\in \mathfrak{a}$b inab \in \mathfrak{a}$b\in \mathfrak{a}$ such that $a$$a$aa$a$ and $b$$b$bb$b$ generate $\mathfrak{a}$$\mathfrak{a}$a\mathfrak{a}$\mathfrak{a}$ (in particular, every ideal is 2-generated.

Solution: Since ${\mathcal{O}}_K$${\mathcal{O}}_K$O_(K)\mathcal{O}_{K}${\mathcal{O}}_K$ is a Dedekind domain, every nonzero ideal of it has a unique factorization as a product of nonzero prime ideals. Write $aA={\mathfrak{p}}_1^{n_1}\cdots {\mathfrak{p}}_r^{n_r}({\mathfrak{p}}_i$$aA={\mathfrak{p}}_1^{n_1}\cdots {\mathfrak{p}}_r^{n_r}\left({\mathfrak{p}}_i\right.$aA=p_(1)^(n_(1))cdotsp_(r)^(n_(r))(p_(i):}a A=\mathfrak{p}_{1}^{n_{1}} \cdots \mathfrak{p}_{r}^{n_{r}}\left(\mathfrak{p}_{i}\right.$aA={\mathfrak{p}}_1^{n_1}\cdots {\mathfrak{p}}_r^{n_r}({\mathfrak{p}}_i$ 's are distinct nonzero prime ideals of $A,n_i\in \mathbf{N}$$A,n_i\in \mathbf{N}$A,n_(i)inNA, n_{i} \in \mathbf{N}$A,n_i\in \mathbf{N}$ ). Since $a\in \mathfrak{a}$$a\in \mathfrak{a}$a inaa \in \mathfrak{a}$a\in \mathfrak{a}$, we have $\mathfrak{a}$$\mathfrak{a}$a\mathfrak{a}$\mathfrak{a}$ divides $aA$$aA$aAa A$aA$, so that $\mathfrak{a}={\mathfrak{p}}_1^{m_1}\cdots {\mathfrak{p}}_r^{m_r}$$\mathfrak{a}={\mathfrak{p}}_1^{m_1}\cdots {\mathfrak{p}}_r^{m_r}$a=p_(1)^(m_(1))cdotsp_(r)^(m_(r))\mathfrak{a}=\mathfrak{p}_{1}^{m_{1}} \cdots \mathfrak{p}_{r}^{m_{r}}$\mathfrak{a}={\mathfrak{p}}_1^{m_1}\cdots {\mathfrak{p}}_r^{m_r}$ with $0\le m_i\le n_i$$0\le m_i\le n_i$0 <= m_(i) <= n_(i)0 \leq m_{i} \leq n_{i}$0\le m_i\le n_i$. For each $i$$i$ii$i$, pick $x_i\in {\mathfrak{p}}_i^{m_i}\mathrm{\setminus }{\mathfrak{p}}_i^{m_i+1}$$x_i\in {\mathfrak{p}}_i^{m_i}\mathrm{\setminus }{\mathfrak{p}}_i^{m_i+1}$x_(i)inp_(i)^(m_(i))\\p_(i)^(m_(i)+1)x_{i} \in \mathfrak{p}_{i}^{m_{i}} \backslash \mathfrak{p}_{i}^{m_{i}+1}$x_i\in {\mathfrak{p}}_i^{m_i}\mathrm{\setminus }{\mathfrak{p}}_i^{m_i+1}$. Since ${\mathfrak{p}}_i^{m_i+1}$${\mathfrak{p}}_i^{m_i+1}$p_(i)^(m_(i)+1)\mathfrak{p}_{i}^{m_{i}+1}${\mathfrak{p}}_i^{m_i+1}$ and ${\mathfrak{p}}_j^{m_j+1}$${\mathfrak{p}}_j^{m_j+1}$p_(j)^(m_(j)+1)\mathfrak{p}_{j}^{m_{j}+1}${\mathfrak{p}}_j^{m_j+1}$ are coprime when $i\ne j$$i\ne j$i!=ji \neq j$i\ne j$, by the Chinese Remainder Theorem, the system of congruences $b\equiv x_i(mod{\mathfrak{p}}_i^{m_i+1}),1\le i\le r$$b\equiv x_i\left(mod{\mathfrak{p}}_i^{m_i+1}\right),1\le i\le r$b-=x_(i)(modp_(i)^(m_(i)+1)),1 <= i <= rb \equiv x_{i}\left(\bmod \mathfrak{p}_{i}^{m_{i}+1}\right), 1 \leq i \leq r$b\equiv x_i(mod{\mathfrak{p}}_i^{m_i+1}),1\le i\le r$ has a solution $b\in A$$b\in A$b in Ab \in A$b\in A$.

By the congruence relation above, we see that $b\in {\mathfrak{p}}_i^{m_i}\mathrm{\setminus }{\mathfrak{p}}_i^{m_i+1}$$b\in {\mathfrak{p}}_i^{m_i}\mathrm{\setminus }{\mathfrak{p}}_i^{m_i+1}$b inp_(i)^(m_(i))\\p_(i)^(m_(i)+1)b \in \mathfrak{p}_{i}^{m_{i}} \backslash \mathfrak{p}_{i}^{m_{i}+1}$b\in {\mathfrak{p}}_i^{m_i}\mathrm{\setminus }{\mathfrak{p}}_i^{m_i+1}$ as well. So for each $1\le i\le r$$1\le i\le r$1 <= i <= r1 \leq i \leq r$1\le i\le r$, the order of ${\mathfrak{p}}_i$${\mathfrak{p}}_i$p_(i)\mathfrak{p}_{i}${\mathfrak{p}}_i$ in the prime ideal factorization of $bA$$bA$bAb A$bA$ is exactly $m_i$$m_i$m_(i)m_{i}$m_i$. In other words

where ${\mathfrak{q}}_j$${\mathfrak{q}}_j$q_(j)\mathfrak{q}_{j}${\mathfrak{q}}_j$ 's are nonzero prime ideals different from those ${\mathfrak{p}}_i$${\mathfrak{p}}_i$p_(i)\mathfrak{p}_{i}${\mathfrak{p}}_i$ 's (if they exist). Therefore

as desired.

Problem 5. Let $p$$p$pp$p$ be a prime number and ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ be a primitive $p$$p$pp$p$-th root of unity. Let $K=\mathbf{Q}\left({\zeta }_p\right)$$K=\mathbf{Q}\left({\zeta }_p\right)$K=Q(zeta_(p))K=\mathbf{Q}\left(\zeta_{p}\right)$K=\mathbf{Q}\left({\zeta }_p\right)$.

(a) Show that ${\mathrm{\Phi }}_p=\sum _{i=0}^{p-1}{X}^i$${\mathrm{\Phi }}_p=\sum _{i=0}^{p-1} X^i$Phi_(p)=sum_(i=0)^(p-1)X^(i)\Phi_{p}=\sum_{i=0}^{p-1} X^{i}${\mathrm{\Phi }}_p=\sum _{i=0}^{p-1}{X}^i$ is the minimal polynomial of ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ over $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$.

(b) Compute the trace ${\mathrm{Tr}}_{K/\mathbf{Q}}(1-{\zeta }_p)$${\mathrm{Tr}}_{K/\mathbf{Q}}\left(1-{\zeta }_p\right)$Tr_(K//Q)(1-zeta_(p))\operatorname{Tr}_{K / \mathbf{Q}}\left(1-\zeta_{p}\right)${\mathrm{Tr}}_{K/\mathbf{Q}}(1-{\zeta }_p)$ and the norm ${\mathcal{N}}_{K/\mathbf{Q}}(1-{\zeta }_p)$${\mathcal{N}}_{K/\mathbf{Q}}\left(1-{\zeta }_p\right)$N_(K//Q)(1-zeta_(p))\mathcal{N}_{K / \mathbf{Q}}\left(1-\zeta_{p}\right)${\mathcal{N}}_{K/\mathbf{Q}}(1-{\zeta }_p)$.

(c) Show that $(1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$$\left(1-{\zeta }_p\right){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$(1-zeta_(p))O_(K)nnZ=pZ\left(1-\zeta_{p}\right) \mathcal{O}_{K} \cap \mathbf{Z}=p \mathbf{Z}$(1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$ and deduce that for all $y\in {\mathcal{O}}_K$$y\in {\mathcal{O}}_K$y inO_(K)y \in \mathcal{O}_{K}$y\in {\mathcal{O}}_K$, we have

(d) Determine explicitly the ring of integers of $K$$K$KK$K$.

(a) Consider $g(X)={\mathrm{\Phi }}_p(X+1)=(X+1{)}^{p-1}+\cdots +X+1$$g(X)={\mathrm{\Phi }}_p(X+1)=(X+1{)}^{p-1}+\cdots +X+1$g(X)=Phi_(p)(X+1)=(X+1)^(p-1)+cdots+X+1g(X)=\Phi_{p}(X+1)=(X+1)^{p-1}+\cdots+X+1$g(X)={\mathrm{\Phi }}_p(X+1)=(X+1{)}^{p-1}+\cdots +X+1$. Then $g(X)=\frac{1}{X}[(X+1{)}^p-1]=\sum _{i=0}^{p-1}(\genfrac{}{}{0ex}{}{p}{i+1})X^i$$g(X)=\frac{1}{X}\left[(X+1{)}^p-1\right]=\sum _{i=0}^{p-1} (\genfrac{}{}{0ex}{}{p}{i+1})X^i$g(X)=(1)/(X)[(X+1)^(p)-1]=sum_(i=0)^(p-1)((p)/(i+1))X^(i)g(X)=\frac{1}{X}\left[(X+1)^{p}-1\right]=\sum_{i=0}^{p-1}\binom{p}{i+1} X^{i}$g(X)=\frac{1}{X}[(X+1{)}^p-1]=\sum _{i=0}^{p-1}(\genfrac{}{}{0ex}{}{p}{i+1})X^i$. Notice that for all $0\le i\le p-2$$0\le i\le p-2$0 <= i <= p-20 \leq i \leq p-2$0\le i\le p-2$, we have $p|(\genfrac{}{}{0ex}{}{p}{i+1})$$p\left|(\genfrac{}{}{0ex}{}{p}{i+1})\right.$p|((p)/(i+1)):}p \left\lvert\,\binom{ p}{i+1}\right.$p|(\genfrac{}{}{0ex}{}{p}{i+1})$, but $p^2+(\genfrac{}{}{0ex}{}{p}{1})=p$$p^2+(\genfrac{}{}{0ex}{}{p}{1})=p$p^(2)+((p)/(1))=pp^{2}+\binom{p}{1}=p$p^2+(\genfrac{}{}{0ex}{}{p}{1})=p$. By the Eisenstein's criterion, $g(X)$$g(X)$g(X)g(X)$g(X)$ is irreducible over $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$, so is ${\mathrm{\Phi }}_p(X)=g(X-1)$${\mathrm{\Phi }}_p(X)=g(X-1)$Phi_(p)(X)=g(X-1)\Phi_{p}(X)=g(X-1)${\mathrm{\Phi }}_p(X)=g(X-1)$. Since ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ is a root of $X^p-1=(X-1){\mathrm{\Phi }}_p(X)$$X^p-1=(X-1){\mathrm{\Phi }}_p(X)$X^(p)-1=(X-1)Phi_(p)(X)X^{p}-1=(X-1) \Phi_{p}(X)$X^p-1=(X-1){\mathrm{\Phi }}_p(X)$ but not of $(X-1)$$(X-1)$(X-1)(X-1)$(X-1)$, we have ${\mathrm{\Phi }}_p\left({\zeta }_p\right)=0$${\mathrm{\Phi }}_p\left({\zeta }_p\right)=0$Phi_(p)(zeta_(p))=0\Phi_{p}\left(\zeta_{p}\right)=0${\mathrm{\Phi }}_p\left({\zeta }_p\right)=0$. Therefore, ${\mathrm{\Phi }}_p(X)$${\mathrm{\Phi }}_p(X)$Phi_(p)(X)\Phi_{p}(X)${\mathrm{\Phi }}_p(X)$, being monic, is the minimal polynomial of ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ over $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$.

(b) For each $1\le i\le p-1$$1\le i\le p-1$1 <= i <= p-11 \leq i \leq p-1$1\le i\le p-1$, we have ${\zeta }_p^i\in K$${\zeta }_p^i\in K$zeta_(p)^(i)in K\zeta_{p}^{i} \in K${\zeta }_p^i\in K$ is also a root of $X^p-1=(X-1){\mathrm{\Phi }}_p(X)$$X^p-1=(X-1){\mathrm{\Phi }}_p(X)$X^(p)-1=(X-1)Phi_(p)(X)X^{p}-1=(X-1) \Phi_{p}(X)$X^p-1=(X-1){\mathrm{\Phi }}_p(X)$ but not of $(X-1)$$(X-1)$(X-1)(X-1)$(X-1)$, we have ${\mathrm{\Phi }}_p\left({\zeta }_p^i\right)=0$${\mathrm{\Phi }}_p\left({\zeta }_p^i\right)=0$Phi_(p)(zeta_(p)^(i))=0\Phi_{p}\left(\zeta_{p}^{i}\right)=0${\mathrm{\Phi }}_p\left({\zeta }_p^i\right)=0$. Thus $K/\mathbf{Q}$$K/\mathbf{Q}$K//QK / \mathbf{Q}$K/\mathbf{Q}$ is a Galois extension whose Galois group is $\mathrm{Gal}(K/\mathbf{Q})=\{{\sigma }_i;{\sigma }_i\left({\zeta }_p\right)={\zeta }_p^i,1\le i\le p-1\}$$\mathrm{Gal}(K/\mathbf{Q})=\left\{{\sigma }_i;{\sigma }_i\left({\zeta }_p\right)={\zeta }_p^i,1\le i\le p-1\right\}$Gal(K//Q)={sigma_(i);sigma_(i)(zeta_(p))=zeta_(p)^(i),1 <= i <= p-1}\operatorname{Gal}(K / \mathbf{Q})=\left\{\sigma_{i} ; \sigma_{i}\left(\zeta_{p}\right)=\zeta_{p}^{i}, 1 \leq i \leq p-1\right\}$\mathrm{Gal}(K/\mathbf{Q})=\{{\sigma }_i;{\sigma }_i\left({\zeta }_p\right)={\zeta }_p^i,1\le i\le p-1\}$. So

and

(c) Notice that ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ is a root of a monic ${\mathrm{\Phi }}_p\in \mathbf{Z}[X]$${\mathrm{\Phi }}_p\in \mathbf{Z}[X]$Phi_(p)inZ[X]\Phi_{p} \in \mathbf{Z}[X]${\mathrm{\Phi }}_p\in \mathbf{Z}[X]$, so ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ is integral over $\mathbf{Z}$$\mathbf{Z}$Z\mathbf{Z}$\mathbf{Z}$. For each $m\in \mathbf{Z}$$m\in \mathbf{Z}$m inZm \in \mathbf{Z}$m\in \mathbf{Z}$, let $z=m\prod _{i=2}^{p-1}(1-{\zeta }_p^i)\in {\mathcal{O}}_K$$z=m\prod _{i=2}^{p-1} \left(1-{\zeta }_p^i\right)\in {\mathcal{O}}_K$z=mprod_(i=2)^(p-1)(1-zeta_(p)^(i))inO_(K)z=m \prod_{i=2}^{p-1}\left(1-\zeta_{p}^{i}\right) \in \mathcal{O}_{K}$z=m\prod _{i=2}^{p-1}(1-{\zeta }_p^i)\in {\mathcal{O}}_K$. Then $z(1-{\zeta }_p)=m\prod _{i=1}^{p-1}(1-{\zeta }_p^i)=mp$$z\left(1-{\zeta }_p\right)=m\prod _{i=1}^{p-1} \left(1-{\zeta }_p^i\right)=mp$z(1-zeta_(p))=mprod_(i=1)^(p-1)(1-zeta_(p)^(i))=mpz\left(1-\zeta_{p}\right)=m \prod_{i=1}^{p-1}\left(1-\zeta_{p}^{i}\right)=m p$z(1-{\zeta }_p)=m\prod _{i=1}^{p-1}(1-{\zeta }_p^i)=mp$. Thus $p\mathbf{Z}\subset (1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}$$p\mathbf{Z}\subset \left(1-{\zeta }_p\right){\mathcal{O}}_K\cap \mathbf{Z}$pZsub(1-zeta_(p))O_(K)nnZp \mathbf{Z} \subset\left(1-\zeta_{p}\right) \mathcal{O}_{K} \cap \mathbf{Z}$p\mathbf{Z}\subset (1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}$. On the other hand, suppose $z\in {\mathcal{O}}_K$$z\in {\mathcal{O}}_K$z inO_(K)z \in \mathcal{O}_{K}$z\in {\mathcal{O}}_K$ satisfies
$m=z(1-{\zeta }_p)\in \mathbf{Z}$$m=z\left(1-{\zeta }_p\right)\in \mathbf{Z}$m=z(1-zeta_(p))inZm=z\left(1-\zeta_{p}\right) \in \mathbf{Z}$m=z(1-{\zeta }_p)\in \mathbf{Z}$. Taking norms on both sides, we get $m^p={\mathcal{N}}_{K/\mathbf{Q}}(z)p$$m^p={\mathcal{N}}_{K/\mathbf{Q}}(z)p$m^(p)=N_(K//Q)(z)pm^{p}=\mathcal{N}_{K / \mathbf{Q}}(z) p$m^p={\mathcal{N}}_{K/\mathbf{Q}}(z)p$. Since $z$$z$zz$z$ is integral, ${\mathcal{N}}_{K/\mathbf{Q}}(z)\in \mathbf{Z}$${\mathcal{N}}_{K/\mathbf{Q}}(z)\in \mathbf{Z}$N_(K//Q)(z)inZ\mathcal{N}_{K / \mathbf{Q}}(z) \in \mathbf{Z}${\mathcal{N}}_{K/\mathbf{Q}}(z)\in \mathbf{Z}$, we have $p\mid m^p$$p\mid m^p$p∣m^(p)p \mid m^{p}$p\mid m^p$, so $p\mid m$$p\mid m$p∣mp \mid m$p\mid m$. Thus $(1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}\subset p\mathbf{Z}$$\left(1-{\zeta }_p\right){\mathcal{O}}_K\cap \mathbf{Z}\subset p\mathbf{Z}$(1-zeta_(p))O_(K)nnZsub pZ\left(1-\zeta_{p}\right) \mathcal{O}_{K} \cap \mathbf{Z} \subset p \mathbf{Z}$(1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}\subset p\mathbf{Z}$. As a consequence, $(1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$$\left(1-{\zeta }_p\right){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$(1-zeta_(p))O_(K)nnZ=pZ\left(1-\zeta_{p}\right) \mathcal{O}_{K} \cap \mathbf{Z}=p \mathbf{Z}$(1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$.

Now take any $y\in {\mathcal{O}}_K$$y\in {\mathcal{O}}_K$y inO_(K)y \in \mathcal{O}_{K}$y\in {\mathcal{O}}_K$. Observe that for each $1\le i\le p-1$$1\le i\le p-1$1 <= i <= p-11 \leq i \leq p-1$1\le i\le p-1$, we have

so ${\mathrm{Tr}}_{K/\mathbf{Q}}\left(y(1-{\zeta }_p)\right)=\sum _{i=1}^{p-1}{\sigma }_i\left(y(1-{\zeta }_p)\right)\in (1-{\zeta }_p){\mathcal{O}}_K$${\mathrm{Tr}}_{K/\mathbf{Q}}\left(y\left(1-{\zeta }_p\right)\right)=\sum _{i=1}^{p-1} {\sigma }_i\left(y\left(1-{\zeta }_p\right)\right)\in \left(1-{\zeta }_p\right){\mathcal{O}}_K$Tr_(K//Q)(y(1-zeta_(p)))=sum_(i=1)^(p-1)sigma_(i)(y(1-zeta_(p)))in(1-zeta_(p))O_(K)\operatorname{Tr}_{K / \mathbf{Q}}\left(y\left(1-\zeta_{p}\right)\right)=\sum_{i=1}^{p-1} \sigma_{i}\left(y\left(1-\zeta_{p}\right)\right) \in\left(1-\zeta_{p}\right) \mathcal{O}_{K}${\mathrm{Tr}}_{K/\mathbf{Q}}\left(y(1-{\zeta }_p)\right)=\sum _{i=1}^{p-1}{\sigma }_i\left(y(1-{\zeta }_p)\right)\in (1-{\zeta }_p){\mathcal{O}}_K$ as well. On the other hand, $y(1-{\zeta }_p)\in {\mathcal{O}}_K$$y\left(1-{\zeta }_p\right)\in {\mathcal{O}}_K$y(1-zeta_(p))inO_(K)y\left(1-\zeta_{p}\right) \in \mathcal{O}_{K}$y(1-{\zeta }_p)\in {\mathcal{O}}_K$, so its trace is in $\mathbf{Z}$$\mathbf{Z}$Z\mathbf{Z}$\mathbf{Z}$. Therefore ${\mathrm{Tr}}_{K/\mathbf{Q}}\left(y(1-{\zeta }_p)\right)\in (1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$${\mathrm{Tr}}_{K/\mathbf{Q}}\left(y\left(1-{\zeta }_p\right)\right)\in \left(1-{\zeta }_p\right){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$Tr_(K//Q)(y(1-zeta_(p)))in(1-zeta_(p))O_(K)nnZ=pZ\operatorname{Tr}_{K / \mathbf{Q}}\left(y\left(1-\zeta_{p}\right)\right) \in\left(1-\zeta_{p}\right) \mathcal{O}_{K} \cap \mathbf{Z}=p \mathbf{Z}${\mathrm{Tr}}_{K/\mathbf{Q}}\left(y(1-{\zeta }_p)\right)\in (1-{\zeta }_p){\mathcal{O}}_K\cap \mathbf{Z}=p\mathbf{Z}$.

(d) Let $y=b_0+b_1{\zeta }_p+\cdots +b_{p-1}{\zeta }_p^{p-1}\in {\mathcal{O}}_K$$y=b_0+b_1{\zeta }_p+\cdots +b_{p-1}{\zeta }_p^{p-1}\in {\mathcal{O}}_K$y=b_(0)+b_(1)zeta_(p)+cdots+b_(p-1)zeta_(p)^(p-1)inO_(K)y=b_{0}+b_{1} \zeta_{p}+\cdots+b_{p-1} \zeta_{p}^{p-1} \in \mathcal{O}_{K}$y=b_0+b_1{\zeta }_p+\cdots +b_{p-1}{\zeta }_p^{p-1}\in {\mathcal{O}}_K$ where $b_i\in \mathbf{Q}$$b_i\in \mathbf{Q}$b_(i)inQb_{i} \in \mathbf{Q}$b_i\in \mathbf{Q}$. Notice that for any $1\le i\le p-1,{\zeta }_p^i$$1\le i\le p-1,{\zeta }_p^i$1 <= i <= p-1,zeta_(p)^(i)1 \leq i \leq p-1, \zeta_{p}^{i}$1\le i\le p-1,{\zeta }_p^i$ is a conjugate of ${\zeta }_p$${\zeta }_p$zeta_(p)\zeta_{p}${\zeta }_p$ over $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$, so