About Inner product space

Question

So this $\delta$ function is degree function over Euclidean domain A which is simply defined as

Now my question is why det L is well defined in first picture and how to get corollary 3.9 and remark that integral lattice is iff $\delta$(det$L$) =0 ?

Answer 1

The way we define $\det(L)$ is the following.

Let $v_1,\ldots,v_n$ be a basis for $L$ as a free $A$-module. Then let $v_1',\ldots,v_n'$ be a different basis for $L$. Let $P$ be the basis change matrix from $\{v_i'\}$ to $\{v_i\}$, so that if $v\in L$ and $v$ has coordinates $x=(x_1,\ldots,x_n)$ with respect to the $v_i'$, then $v$ has coordinates $Px$ with respect to the $v_i$.

Then $P=(p_{ij})$, where $v_j' = \sum_{i=1}^n p_{ij}v_i$.

Thus $$B(v_i',v_j') = B\left(\sum_{k=1}^n p_{ki}v_k,\sum_{\ell=1}^n p_{\ell j}v_\ell\right)=\sum_{k,\ell}p_{ki}p_{\ell j}B(v_k,v_\ell).$$ Thus as matrices, if we let $C$ be the matrix $C_{k\ell}=B(v_k,v_\ell)$, and $C'$ be the matrix $C'_{ij}=B(v_i',v_j')$, we have $C' = P^T CP$. Thus $\det(C')=\det(C)\det(P)^2$. Since $P$ is an invertible matrix with entries in $A$, $\det(P)\in U(A)$, and $\det(P)^2\in U(A)^2$.

Thus $\det(L)$ is a well-defined class in $K^\times/U(A)^2$.

You mentioned that you figured out Corollary 3.9, so I'll leave this here.