Exercise $3.21$ of Marcus’ “Number Fields” (an excellent book) has been giving me trouble on and off for a while now. I think I’ve finally got it but I’m suspicious of my solution since it nowhere requires “large $N$“. I would appreciate feedback and any suggestions for alternative solutions to the exercise (that does not mean alternative proofs of the theorem, that means alternative proofs that follow the template Marcus suggests). There is also a further result about when this is the exact power of $p$ dividing the discriminant (when all ramification indices are coprime with $p$); does anyone know if the strategy of this exercise can be used to prove that as well?
The setup: we have a number field $K$ with number ring $R$, a prime integer $p$ and prime factorisation $pR=Q_1^{e_1}Q_2^{e_2}\cdots Q_r^{e_r}$ with inertial degrees $f_1,\cdots,f_r$. Let $n=[K:\Bbb Q]$. There exist $\alpha_{i,j}\in(Q_i^{j-1}\setminus Q_i^j)\prod_{h\neq i}Q_h^N$ and $\beta_{i,k}\in R$ for $1\le i \le r$, $1\le j\le e_i$, $1\le k\le f_i$ such that the set of products $\{\alpha_{i,j}\beta_{i,k}\}$ is a set of $n$ independent elements mod $p$. Here, $N$ is any integer larger than every single $e_i$ but can be chosen as large as please: Marcus seems to suggest we need a larger choice of $N$ than merely $\max e_\bullet$.
We need to show the discriminant of the $\alpha\beta$ is divisible by $p^{\sum_i(e_i-1)f_i}$. My solution:
Let $L$ be the normal closure of $K/\Bbb Q$ and $Z$ any prime of $L$ over $p$. If $j>1$ then any product $\gamma=\alpha_{i,j}\beta_{i,k}\cdot\alpha_{i’,j’}\beta_{i’,k’}$ will be an element of every prime of $K$ over $p$, so any $\sigma\in\mathsf{Gal}(L;\Bbb Q)$ will satisfy: $\sigma^{-1}(Z)$ is a prime of $L$ over $p$ so $\gamma\in\sigma^{-1}(Z)\cap R\subset\sigma^{-1}(Z)$ so $\sigma(\gamma)\in Z$.
It follows from additive closure of $Z$ that $\mathrm{Tr}^K_{\Bbb Q}(\gamma)\in Z\cap\Bbb Z=p\Bbb Z$. It then follows all the $\sum_i(e_i-1)f_i$ rows of the matrix of traces of such $\gamma$ with $j>1$ are divisible by $p$, so the determinant is divisible by $p^{\sum_i(e_i-1)f_i}$.
Is this legitimate? Can it be improved to show the stronger theorem, that this is the exact power of $p$ dividing the discriminant if all $e_i$ are indivisible by $p$? Many thanks.