Proof. Example Sheet 3. □

Proof.

• (i) Let $𝔭O_L=P_1^{e_1}\cdots P_r^{e_r}$, $0\ne P_i\subseteq O_L$ distinct prime ideals, $e_i>0$. Define

  $$R:=O_L∕𝔭O_L\stackrel{\text{CRT}}{=}\prod _{i=1}^r{O}_L∕P_i^{e_i}.$$

  If $𝔭$ ramifies, then $O_L∕𝔭O_L$ has nilpotents. Hence

  $$\mathrm{\Delta }({\overline{x}}_1,\dots ,{\overline{x}}_n)=0\forall {\overline{x}}_i\in O_L∕𝔭O_L.$$

  Then using the fact that

  
  commutes, we get that

  $$\mathrm{\Delta }(x_1,\dots ,x_n)\equiv 0mod𝔭\forall x_i\in O_L∕𝔭O_L.$$

• (ii) $𝔭$ unramified implies $R=O_L∕𝔭O_L$ is a product of finite extensions of $k$. By Lemma 12.1, we get that the Trace form is non-degenerate, hence for ${\overline{x}}_1,\dots ,{\overline{x}}_n$ a basis of $O_L∕𝔭O_L$ as a $k$ vector space, we have $\mathrm{\Delta }({\overline{x}}_1,\dots ,{\overline{x}}_n)\ne 0$. So thee exist $x_1,\dots ,x_n\in O_L$ such that

  $$\mathrm{\Delta }(x_1,\dots ,x_n)⁄\equiv 0mod𝔭.\square$$

Proof. Let $x_1,\dots ,x_n\in O_L$ a $K$-basis for $L∕K$. Set

$$d:=\mathrm{\Delta }(x_1,\dots ,x_n)=\det ({\mathrm{Tr}}_{L∕K}(x_i{x}_j)),$$

which is non-zero since separable.

For $x\in D_{L∕K}^{-1}$ write $x={\sum }_{j=1}^r{\lambda }_j{x}_j$ with ${\lambda }_j\in K$. We show ${\lambda }_j\in \frac{1}{d}{O}_K$. We have

$${\mathrm{Tr}}_{L∕K}(x{x}_i)=\sum _{j=1}^n{\lambda }_j{\mathrm{Tr}}_{L∕K}(x_i{x}_j)\in O_K.$$

Set $A_{ij}={\mathrm{Tr}}_{L∕K}(x_i{x}_j)$. Multiplying by $\mathrm{Adj}(A)\in M_n(O_K)$, we get

$$d\left(\begin{array}{c}\hfill {\lambda }_1\hfill \\ \hfill ⋮\hfill \\ \hfill {\lambda }_n\hfill \end{array}\right)=\mathrm{Adj}(A)\left(\begin{array}{c}\hfill {\mathrm{Tr}}_{L∕K}(x{x}_1)\hfill \\ \hfill ⋮\hfill \\ \hfill {\mathrm{Tr}}_{L∕K}(x{x}_n)\hfill \end{array}\right)$$

Since ${\lambda }_i\in \frac{1}{d}{O}_K$, we have $x\in \frac{1}{d}{O}_L$. Thus $D_{L∕K}^{-1}\le \frac{1}{d{O}_K}$, so $D_{L∕K}^{-1}$ is a fractional ideal. □

Proof. First assume $O_K$, $O_L$ are PIDs. Let $x_1,\dots ,x_n$ be an $O_K$-basis for $O_L$ and $y_1,\dots ,y_n$ be the dual basis with respect to trace form. Then $y_1,\dots ,y_n$ is a basis for $D_{L∕K}^{-1}$. Let ${\sigma }_1,\dots ,{\sigma }_n:L\to \overline{K}$ be the distinct embeddings. Have

$$\sum _{i=1}^n{\sigma }_i(x_j){\sigma }_i(y_k)=\mathrm{Tr}(x_j{y}_k)={\delta }_{jk}.$$

But

$$\mathrm{\Delta }(x_1,\dots ,x_n)=\det {({\sigma }_i(x_j))}^2.$$

Thus

$$\mathrm{\Delta }(x_1,\dots ,x_n)\mathrm{\Delta }(y_1,\dots ,y_n)=1.$$

Write $D_{L∕K}^{-1}=\beta O_L$ since $\beta \in L$. Then

$$\begin{array}{llllll}\hfill d_{L∕K}^{-1}& =(\mathrm{\Delta }{(x_1,\dots ,x_n)}^{-1})& \hfill & & \hfill & \\ \hfill & =(\mathrm{\Delta }(y_1,\dots ,y_n))& \hfill & & \hfill & \\ \hfill & =(\mathrm{\Delta }(\beta x_1,\dots ,\beta x_n))& \hfill & \text{change of basis matrix is invertible in }{O}_K& \hfill & & \hfill & \\ \hfill & =N_{L∕K}({\beta }^2)\mathrm{\Delta }(x_1,\dots ,x_n)& \hfill & \text{change of basis matrix is }[\mathrm{mult}(\beta )]& \hfill & & \hfill & \end{array}$$

Thus

$$d_{L∕K}^{-1}=N_{L∕K}{(D_{L∕K}^{-1})}^2{d}_{L∕K}$$

so

$$N_{L∕K}(D_{L∕K})=d_{L∕K}.$$

In general, localise at $S=O_K\setminus 𝔭$ and use $S^{-1}{D}_{L∕K}=D_{S^{-1}}{O}_L∕S^{-1}{O}_K$. Then $S^{-1}{d}_{L∕K}=d_{S^{-1}{O}_L∕S^{-1}{O}_K}$. Details omitted. □

Proof. Let $\alpha ={\alpha }_1,\dots ,{\alpha }_n$ be the roots of $g$. Write

$$\frac{g(X)}{X-\alpha }={\beta }_{n-1}{X}^{n-1}+\cdots +{\beta }_{1}X+{\beta }_0$$

with ${\beta }_i\in O_L$ and ${\beta }_{n-1}=1$. We claim

$$\sum _{i=1}^n\frac{g(X)}{X-{\alpha }_i}\frac{{\alpha }_i^n}{g^{\prime }({\alpha }_i)}=X^r$$

for $0\le r\le n-1$.

Indeed the difference is a palynomial of degree $<n$, which vanishes for $X={\alpha }_1,\dots ,{\alpha }_n$. Equate coefficients of $X^s$, which gives

$${\mathrm{Tr}}_{L∕K}\left(\frac{{\alpha }^r{\beta }_s}{g^{\prime }(\alpha )}\right)={\delta }_{rs}.$$

Since $1,\alpha ,\dots ,{\alpha }^{n-1}$ is an $O_K$ basis for $O_L$, $D_{L∕K}^{-1}$ has an $O_K$ basis

$$\frac{{\beta }_0}{g^{\prime }(\alpha )},\frac{{\beta }_1}{g^{\prime }(\alpha )},\dots ,\underset{\frac{1}{g^{\prime }(\alpha )}}{\underbrace{\frac{{\beta }_{n-1}}{g^{\prime }(\alpha )}}}.$$

Note all of these are $O_L$ multiples of the last term, since the ${\beta }_i$ are in $O_L$. So $D_{L∕K}^{-1}=\frac{1}{(g^{\prime }(\alpha ))}$, hence $D_{L∕K}=(g^{\prime }(\alpha ))$. □

Proof. Let $x\in L$, $𝔭\subseteq O_K$. Then

$$\mathrm{Tr}L∕K(x)=\sum _{P|𝔭}{\mathrm{Tr}}_{L_P∕K_{𝔭}}(x)\text{(∗)}$$

(of Corollary 10.10).

Let $r(P)=v_P(D_{L∕K})$, $s(P)=v_P(D_{L_P}∕K_{𝔭})$.

• $\subseteq$ (i.e. $r(P)\ge s(P)$). Let $x\in L$ with $v_P(x)\ge -s(P)$ for all $P$. Then ${\mathrm{Tr}}_{L_P∕K_{𝔭}}(xy)\in O_{K_{𝔭}}$, for all $y\in L$ and for all $P$. Using ($\ast$) we get

  $${\mathrm{Tr}}_{L∕K}(xy)\in O_{K_{𝔭}}\forall y\in O_L,\forall P.$$

  Thus

  $${\mathrm{Tr}}_{L∕K}(xy)\in O_K\forall y\in O_L$$

  so $D_{L∕K}\subseteq {\prod }_P{D}_{L_P}∕K_{𝔭}$.

• $\supseteq$ (i.e. $r(P)\le s(P)$). Fix $P$ and let $x\in P^{-r(P)}\setminus P^{-r(P)+1}$. Then $v_P(x)=-r(P)$, $v_{P^{\prime }}(x)\ge 0$ for all $P^{\prime }\ne P$. By ($\ast$) , we have

  $${\mathrm{Tr}}_{L_P∕K_{𝔭}}(xy)={\mathrm{Tr}}_{L∕K}(xy)-\sum _{\begin{array}{c}{P}^{\prime }|𝔭\\ P^{\prime }\ne P\end{array}}{\mathrm{Tr}}_{L_P∕K_{𝔭}}(xy)\forall y\in O_L$$

  hence

  $${\mathrm{Tr}}_{L_P∕K_{𝔭}}(xy)\in O_{K_{𝔭}}\forall y\in O_{L_P}.$$

  Hence , i.e. $-v_P(x)=r(P)\le s(P)$. So $D_{L∕K}\supseteq {\prod }_P{D}_{L_P}∕K_{𝔭}$. □

Proof. Apply $N_{L∕K}$ to $D_{L∕K}={\prod }_{P|𝔭}{D}_{L_P}∕K_{𝔭}$. □