Proof. $L∕K$ separable tells us that $L=K(\alpha )$ for some $\alpha \in L$. Consider the matrix $A$ for $(\bullet ,\bullet )$ in the $K$-basis for $L$ given by $1,\alpha ,\dots {\alpha }^{n-1}$.

Then $A_{ij}={\mathrm{Tr}}_{L∕K}({\alpha }^{i+j})={[B{B}^{\top }]}_{ij}$ where

So

(Vandermonde determinant), which is non-zero since ${\sigma }_i(\alpha )\ne {\sigma }_j(\alpha )$ for $i\ne j$ by separability. □

Proof. $O_L$ a subring of $L$, hence $O_L$ is an integral domain.

Need to show:

• (i) $O_L$ is Noetherian.
• (ii) $O_L$ is integrally closed in $L$.
• (iii) Every $\ne 0$ prime ideal $P$ in $O_L$ is maximal.

Proofs:

• (i) Let $e_1,\dots ,e_n\in L$ be a $K$-basis for $L$. Upon scaling by $K$, we may assume $e_i\in O_L$ for all $i$.

  Let $f_i\in L$ be the dual basis with respect to the trace form $(\bullet ,\bullet )$. Let $x\in O_L$, and write $x={\sum }_{i=1}^n{\lambda }_i{f}_i,{\lambda }_i\in K$. Then ${\lambda }_i={\mathrm{Tr}}_{L∕K}(x{e}_i)\in O_K$.

  (For any $z\in O_L$, ${\mathrm{Tr}}_{L∕K}(z)$ is a sum of elements in $\overline{K}$ which are integral over $O_K$. Hence ${\mathrm{Tr}}_{L∕K}(z)\in K$ is integral over $O_K$, hence ${\mathrm{Tr}}_{L∕K}(z)\in O_K$.)

  Thus $O_L\subseteq O_K{f}_1+\cdots +O_K{f}_n\subseteq L$. Since $O_K$ is Noetherian, $O_L$ is finitely generated as an $O_K$-module, hence $O_L$ is Noetherian.

• (ii) Example Sheet 2.
• (iii) Let $P$ be a non-zero prime ideal of $O_L$, and $p:=P\cap O_K$ be a prime ideal of $O_K$. Let $0\ne x\in P$. Then $x$ satisfies an equation

  $$x^n+a_{n-1}{x}^{n-1}+\cdots +a_0=0,a_i\in O_K,$$

  with $a_0\ne 0$. Then $a_0\in P\cap O_K$ is a non-zero element of $p$, hence $p$ is non-zero, hence $p$ is maximal.

  We have $O_K∕p\hookrightarrow O_L∕P$, and $O_L∕P$ is a finite dimensional vector space over $O_K∕p$. Since $O_L∕P$ is an integral domain and finite, it is a field. □

Proof. $x{O}_{K{,}_{(p)}}={(p{O}_{K,(p)})}^{v_p(x)}$ by definition of $v_p(x)$.

Lemma follows from property of localisation

for all prime ideals $p$. □

Proof. By Lemma 10.4 for any $0\ne x\in O_K$ and $i=1,\dots ,r$ we have $v_{P_i}(x)=e_i{v}_p(x)$. Hence, up to equivalence, $|\bullet {|}_{P_i}$ extends $|\bullet {|}_p$.

Now suppose $|\bullet |$ is an absolute value on $L$ extending $|\bullet {|}_p$. Then $|\bullet |$ is bounded on $\mathbb{Z}$, hence is non-archimedean. Let $R=\{x\in L||x|\le 1\}\le L$ be the valuation ring for $L$ with respect to $|\bullet |$. Then $O_K\subseteq R$, and since $R$ is integrally closed in $L$ (Lemma 6.8), we have $O_L\subseteq R$. Set

(where $m_R$ is the maximal ideal of $R$).

Hence $P$ a prime ideal in $O_L$. It is non-zero since $p\subseteq P$. Then $O_{L,(p)}\subseteq R$, since $s\in O_L\setminus P⟹|s|=1$.

But $O_{L,(p)}$ is a discrete valuation ring, hence a maximal subring of $L$, so $O_{L,(p)}=R$. Hence $|\bullet |$ is equivalent to $|\bullet {|}_p$. Since $|\bullet |$ extends $|\bullet {|}_p$, $P\cap O_K=p$ so $P_1^{e_1}\cdots P_r^{e_r}\subseteq P$, so $P=P_i$ for some $i$. □

Proof. Case 1: $|\bullet |$ non-archimedean. Then $|\bullet |{|}_{\mathbb{Q}}$ is equivalent to $|\bullet {|}_p$ for some prime $p$ by Ostrowski’s Theorem. Theorem 10.5 gives that $|\bullet |$ is equivalent to $|\bullet {|}_p$ for some $𝔭\subseteq O_K$ a prime ideal with $𝔭|p$.

Case 2: $|\bullet |$ archimedean. See Example Sheet 2. □

Proof. Let $M=L{K}_{𝔭}=\mathrm{Im}({\pi }_P)\subseteq L_P$.

Write $L=K(\alpha )$ then $M=K_{𝔭}(\alpha )$. Hence $M$ is a finite extension of $K_{𝔭}$ and $[M:K_{𝔭}]\le [L:K]$. Moreover $M$ is complete (Theorem 6.1) and since $L\subseteq M\subseteq L_P$, we have $M=L_P$. □

Proof. Example Sheet 2. □

Proof. Write $L=K(\alpha )$ and let $f(X)\in K[X]$ be the minimal polynomial of $\alpha$. Then we have

where $f_i(X)\in K_{𝔭}[X]$ are distinct irreducible (separable). Since $L\cong K[X]∕f(X)$,

Set $L_i=K_{𝔭}[X]∕f_i(X)$ a finite extension of $K_{𝔭}$. Then $L_i$ contains both $K_{𝔭}$ and $L$ (use $K[X]∕f(x)\to K_{𝔭}[X]∕f_i(X)$ injective since morphism of fields). Moreover $L$ is dense inside $L_i$ (approximate coefficients of $K_{𝔭}[X]∕f_i(X)$ with an element of $K[X]∕f_i(X)$).

The theorem follows from the following three claims:

• (1) $L_i\cong L_P$ for some prime $P$ of $O_L$ dividing $𝔭$.
• (2) Each $P$ appears at most once.
• (3) Each $P$ appears at least once.

Proof of claims:

• (1) Since $[L_i:K_{𝔭}]<\infty$, there is a unique absolute value on $L_i$ extending $|\bullet {|}_{𝔭}$. Theorem 10.5 gives us that $|\bullet |{|}_L$ is equivalent to $|\bullet {|}_P$ for some $P|𝔭$. Since $L$ is dense in $L$ and $L_i$ is complete, we have $L_i\cong L_P$.
• (2) Suppose $\phi :L_i\to L_j$ is an isomorphism preserving $L$ and $K_{𝔭}$; then

  $$\phi :K_{𝔭}[X]∕f_i(X)\to K_{𝔭}[X]∕f_i(X)$$

  takes $x$ to $x$ and hence $f_i=f_i$.

• (3) By Lemma 10.7, the natural map ${\pi }_P:L{\otimes }_K{K}_{𝔭}\to L_P$ is surjective for any prime $P|𝔭$.

  Since $L_P$ is a field, ${\pi }_P$ factors through $L_i$ for some $i$, and hence $L_i\cong L_P$ by surjectivity of ${\pi }_P$. □

Proof. Let $B_1,\dots ,B_r$ be bases for $L_{P_1},\dots ,L_{P_r}$ as $K_{𝔭}$-vector spaces. Then $B={\bigcup }_i{B}_i$ is a basis for $L{\otimes }_K{K}_{𝔭}$ over $K_{𝔭}$. Let ${[\mathrm{mult}(x)]}_B$ (respectively ${[\mathrm{mult}(x)]}_{B_i}$) denote the matrix for $\mathrm{mult}(x):L{\otimes }_K{K}_{𝔭}\to L{\otimes }_K{K}_{𝔭}$ (respectively $L_{P_i}\to L_{P_i}$) with respect to the basis $B$ (respectively $B_i$). Then

hence