Let $A=\mathbb{Z}\alpha_1+…+\mathbb{Z}\alpha_n$, $B=\mathbb{Z}\beta_1+…+\mathbb{Z}\beta_n$, be two lattices, s.t. $A\subseteq B$. Then $d(\alpha_1,…,\alpha_n)=c^2d(\beta_1,…,\beta_n)$, where $d$ is the discriminant and $c$ is the index of $A$ in $B$. Why this is true?
I read a proof of discriminant being square free, makes and integral basis, there I encountered this. Can anyone help?
Best Answer
I assume everything takes place in a number field $K/\Bbb Q$ with $[K:\Bbb Q]=n$. Let $\{\sigma_1, \dots, \sigma_n\}=\mathrm{Hom}(K,\overline{\Bbb Q})$ be the field homomorphisms to a fixed algebraic closure so that $\mathrm{Tr}_{K/\Bbb Q}=\sum_{i=1}^n \sigma_i$.
For a $n$ linearly independent elements $\alpha_1, \dots, \alpha_n \in K$ consider the $n \times n$ matrix $A=(\sigma_i(\alpha_j))_{i,j}$.
Then one has $A^TA=(b_{i,j})_{i,j}$ where $b_{i,j}=\sum_{k=1}^n\sigma_k(\alpha_i) \sigma_k(\alpha_j)=\mathrm{Tr}_{K/\Bbb Q}(\alpha_i\alpha_j)$
We see that $\det(A^TA)=\mathrm{Disc}(\alpha_1, \dots, \alpha_n)$
Now consider another $n$ linearly independent elements $\beta_1, \dots, \beta_n$. Let $T:K \to K$ be the linear transformation determined by $T\beta_j = \alpha_j$ for all $j$. Let $T$ be represented by the matrix $(t_{i,j}) \in \mathrm{GL}_n(\Bbb Q)$ so that we have $\sum_{j,k}t_{j,k}\beta_k=\alpha_j$
Because each $\sigma_i$ is $\Bbb Q$-algebra homomorphism we can apply it to this equation and obtain $\sum_{j,k}t_{j,k}\sigma_i(\beta_k)=\sigma_i(\alpha_j)$.
This means if we consider the matrix $B:=(\sigma_i(\beta_j))_{i,j}$, we have $TB=A$. As before with $A$, we have $\det(B^TB)=\mathrm{Disc}(\beta_1, \dots, \beta_n)$ But $TB=A$, so $\det(A^TA)=\det(T)^2\det(B^TB)$, so $\mathrm{Disc}(\alpha_1, \dots, \alpha_n) = \det(T)^2 \mathrm{Disc}(\beta_1, \dots, \beta_n)$
Now assume that $\Bbb Z\alpha_1+ \dots+\Bbb Z\alpha_n \subset \Bbb Z\beta_1 + \dots +\Bbb Z\beta_n$. Then $T$ has integer entries. We can thus find the Smith normal form of $T$, this means that there are $U,V \in \mathrm{GL}_2(\Bbb Z)$ such that $UTV=\mathrm{diag}(a_1, \dots, a_n)$ for integers $a_1, \dots, a_n$. We can think of $U$ and $V$ as automorphisms (base-changes over $\Bbb Z$) of the two lattices. We obtain from this that $\Bbb Z\beta_1 + \dots + \Bbb Z\beta_n/\Bbb Z \alpha_1 + \dots + \Bbb Z\alpha_n \cong \Bbb Z/a_1 \Bbb Z \oplus \dots \oplus \Bbb Z/a_n \Bbb Z$, so that the index of $\Bbb Z \alpha_1 + \dots + \Bbb Z\alpha_n$ in $\Bbb Z\beta_1 + \dots + \Bbb Z\beta_n$ is $\prod_{i=1}^n |a_i|=|\det(T)|$ (using that $U$ and $V$ have determinant $\pm 1$)
Plugging this into $\mathrm{Disc}(\alpha_1, \dots, \alpha_n) = \det(T)^2 \mathrm{Disc}(\beta_1, \dots, \beta_n)$ proves the result.