Exercise $20$ (Chap. $3$) of Marcus' Number Fields asks you to show the following assertion:
Let $K$ and $L$ be two number fields with $K\subset L$. Let $R$ be the ring of algebraic integers of $K$ and $S$ the ring of algebraic integers of $L$. Fix a prime ideal $P\subset R$. We know that $S/PS$ is an $n$-dimensional vector space over $R/P$. Call a set of elements of $S$ independent mod $P$ iff the corresponding elements in $S/PS$ are linearly independent over $R/P$.
If $PS=\prod_{i=1}^rQ_i^{e_i}$ is the factorization of $PS$ into prime ideals of $S$, for every $i$ fix a set $B_i=\left \{\beta_{ik}\right \}_{1\leq k\leq f_i}$ (where $f_i=f\left (Q_i\mid P\right )$) corresponding to a basis for $S/Q_i$ over $R/P$.
For each $i=1,\cdots ,r$ and for each $j=1,\cdots ,e_i$ fix an element $\alpha_{ij}\in \left (Q_i^{j-1}\setminus Q_i^j\right )\cap \bigcap_{h\neq i}Q_h^{e_h}$ (which is possible by the Chinese Remainder Theorem).
The set $\left \{\alpha_{ij}\beta_{ik}:1\leq i\leq r,1\leq j\leq e_i,1\leq k\leq f_i\right \}$ contains $\sum_{i=1}^re_if_i=n$ elements. Show that they are independent mod $P$.
The book gives you the following hint:
Take any nontrivial linear dependence and consider it mod $Q_i$ for each $i$, then consider it mod $Q_i^2$, etc.)
Following the hint, we consider a linear dependence $\sum_{i=1}^r\sum_{j=1}^{e_i}\sum_{k=1}^{f_i}r_{ijk}\alpha_{ij}\beta_{ik}\in PS$ where the $r_{ijk}$ are in $R$. We want to show that in fact they are in $P$.
We fix $i$. We have:
- $PS\subset Q_i$.
- $\alpha_{ij}\in Q_i$ for $j\geq 2$
- $\alpha_{ih}\in Q_i$ for $h\neq i$.
Therefore $\sum_{k=1}^{f_i}r_{i1k}\alpha_{i1}\beta_{ik}\equiv 0\pmod {Q_i}$ and since $\alpha_{i1}\not\in Q_i$, then $Q_i$ being prime allows us to conclude that $\sum_{k=1}^{f_i}r_{i1k}\beta_{ik}\equiv 0\pmod {Q_i}$. By the definition of the $\beta_{ik}$ we have $r_{i1k}\in P$ for every $i$ and $k$ (because $i$ was arbitrary).
I do not know how to continue. The previous argument suggests to keep fixing $i$ and proving the asserion by induction on $j$. The base step was the previous one. So, suppose that $r_{ijk}\in P$ for every $1\leq j\leq m-1< e_i$. We have:
- $PS\subset Q_i^m$
- $\alpha_{ij}\in Q_i^m$ for every $j\geq m+1$
- $\alpha_{ih}\in Q_i^m$ for $i\neq h$
Therefore we have $\sum_{j=1}^{m}\sum_{k=1}^{f_i}r_{ijk}\alpha_{ij}\beta_{ik}\equiv 0\pmod {Q_i^m}$. Now, by induction, we can drop the $j<m$ since $P\subset PS\subset Q_i^m$, and now $\sum_{k=1}^{f_i}r_{imk}\alpha_{im}\beta_{ik}\equiv 0\pmod {Q_i^m}$
And at this step I got stuck. We know that $\alpha_{im}\not\in Q_i^m$, but $Q_i^m$ is not a prime ideal, therefore we cannot drop $\alpha_{im}$ as we did previously on the base step. Even if we could, the $\beta_{ik}$ are not defined in terms of an independence mod $Q_i^m$ but on an independence mod $Q_i$. What should I do?
I think I have arrived to a possible solution (correct me if I am wrong).
Clearly it suffices to show that $r:=\sum_{k=1}^{f_i}r_{imk}\beta_{ik}\equiv 0\pmod {Q_i}$. We know that $\alpha_{im}r\in Q_i^{m}$ and by the definition of $\alpha_{im}$, it belongs to $Q_i^{m-1}$ but not to $Q_i^m$. Therefore the prime decomposition of the ideal $\left (\alpha_{im}\right )$ can be written as $Q_i^{m-1}A$ with $A$ an ideal not divisible by $Q_i$. Since $Q_i^m\mid \left (\alpha_{im}r\right )=Q_i^{m-1}A\left (r\right )$, then $Q_i\mid \left (r\right )$, therefore $r\in Q_i$ and we are done.