Let $K$ be a field extension of degree $n$ over $\mathbb{Q}$. We know that the ring of integers $\mathcal{O}_{K}$ is a free $\mathbb{Z}$-module of rank $n$, and so is any fractional ideal $I$ in $K$. These are examples of lattices in $K$, which are by definition the $\mathbb{Z}$-span of a basis of $K$ over $\mathbb{Q}$. I would like to fully understand the geometric picture that we get from these lattices. In some cases I think I have, but in others I am not very sure how to picture things. I couldn't find any good reference on the literature, and I also didn't find any related question on stack exchange (I only found this less general question, but it does not address the full problem I am trying to understand).


If $\alpha_{1},\ldots, \alpha_{n}\in \mathcal{O}_{K}$ is a basis of $K$ over $\mathbb{Q}$ and $\alpha_{1}^{\vee},\ldots,\alpha_{n}^{\vee}\in K$ is the dual basis with respect to the trace pairing. Let $M=\sum_{i=1}^{n} \mathbb{Z}\alpha_{i}$ and $M^{\vee}=\sum^{n}_{i=1}\mathbb{Z}\alpha_{i}^{\vee}$. Then, since any $x\in \mathcal{O}_{K}$ can be written as $$ x=\sum^{n}_{i=1}\text{Tr}_{K/ \mathbb{Q} }(x\alpha_{i})\alpha_{i}^{\vee}$$

we have that $M\subseteq M^{\vee}$. Moreover, we can use the geometric picture with the lattices to compute the cardinality of the quotient $M^{\vee} / M$. Visualize $M^{\vee}$ as a grid in $\mathbb{R}^{n}$ (up to coordinate change, we may assume $\alpha_{i}^{\vee} = e_{i}$). Then $M$ is a subgrid and the cardinality of the quotient is the number of times we can fit the fundamental mesh of the finer grid into the one of the coarser grid. Since we did that coordinate change and our finer fundamental mesh is just the unit cube, the caridnality of the quotient is just the volume of the fundamental mesh of the grid given by $M$. By an elementary argument of integral calculus, this is just the determinant of the matrix of base chagne. By the previous formula, this is precisely

$$ | \det(\text{Tr}_{K/\mathbb{Q}}(\alpha_{i}\alpha_{j})) |$$

My problem: (see EDIT below)

I want to see that the norm of an ideal $I\subseteq \mathcal{O}_{K}$ is the number of times that the fundamental mesh of the grid corresponding to $\mathcal{O}_{K}$ fits into the subgrid defined by $I$. I tried to carry on similar arguments as in the example but I couldn't figure out how. In particular, one of my main problems is the following:

Given the generator(s) of an ideal $I\subseteq \mathcal{O}_{K}$, how can I know what is its corresponding subgrid?

My attempt: (see EDIT below)

If $\mathcal{O}_{K}=\sum_{i=1}^{n}\mathbb{Z}\alpha_{i}$ and $0\neq x\in I$ has integral equation $x^{n}+a_{1}x^{n-1}+\ldots +a_{0}=0$, then $a_{0}\in I\cap \mathbb{Z}$ and we have $$ \sum^{n}_{i=1}\mathbb{Z}a_{0}\alpha_{i} \subseteq I \subseteq \mathcal{O}_{K} $$

In particular, if $I=\mathfrak{p}$ is a prime in $\mathcal{O}_{K}$ above the prime $p$ in $\mathbb{Z}$, then the grid of $I$ contains the grid with fundamental mesh $[0,p]^{n}$ (if we again see $\mathcal{O}_{K}$ as $\mathbb{Z}^{n}\subseteq \mathbb{R}^{n}$). But this won't be an equality unless $p$ is inert in $K$, right? This brought me to the following question:

If $p\mathcal{O}_{K}=\mathfrak{p}_{1}^{e_{1}}\cdots \mathfrak{p}_{g}^{e_{g}}$, is the fundamental mesh of $p\mathcal{O}_{K}$ covered like a puzzle by those of the primes in the factorization?

Any help/reference is appreciated!


Thanks to the useful comment below, I realized that my hopes were wrong: the subgrid corresponding to the ideal is not always aligned with the grid of the ring of integers. But it is still possible to do a basechange to get this geometric interpretation of the norm. If $K$ has $r_{1}$ real embeddings and $r_{2}$ pairs of complex embeddings, then we can define a map

$$ \lambda \colon K \to \mathbb{R}^{r_{1}+2r_{2}}=\mathbb{R}^{n} $$

given by $x \mapsto (\sigma_{1}(x),\ldots, \sigma_{r_{1}}(x),\text{Re}(\sigma_{r_{1}+2}(x)),\text{Im}(\sigma_{r_{1}+2}(x)), \ldots, \text{Re}(\sigma_{r_{1}+2r_{2}}(x)),\text{Im}(\sigma_{r_{1}+2r_{2}}))$.

With this definition, the following equality holds

$$\text{Vol}(\mathbb{R}^{n}/\lambda(I))=\frac{1}{2^{r_{2}}}\sqrt{|\text{disc}_{K}|}N_{K/\mathbb{Q}}(I) $$

and we get the desired geometric interpretation.

But I still couldn't figure out the answer to the second question. Is ramification also possible to picture in this lattice way with some appropiate base change? I suspect that the theorem mentioned in the comment below could be useful for this, but I couldn't find a way to apply it.

  • $\begingroup$ This is the second time I have come across an abandoned question that was asked two days ago. Go here $\longrightarrow$ math.stackexchange.com/questions/2650097/… to view a discussion that I had with someone else who asked a question not receiving any comments or answers. $\endgroup$ – Mr Pie Feb 16 '18 at 11:54
  • 2
    $\begingroup$ I'm not sure exactly how you want to see this, but may be applying the stacked bases theorem to the lattice of $\mathcal{O}_K$ and the sublattice of the ideal $I$ would help you: There is a $\Bbb{Z}$-bases $e_1,e_2,\ldots,e_n$ of $\mathcal{O}_k$ and a sequence of integers $d_1,d_2,\ldots,d_n$ such that $d_1e_1,d_2e_2,\ldots,d_ne_n$ is a basis of $I$. When you choose the gridlines according to that basis, then the boxes are aligned nicely. Different ideals require a different basis, but that's just about change of coordinates. $\endgroup$ – Jyrki Lahtonen Feb 16 '18 at 12:17
  • $\begingroup$ @JyrkiLahtonen that is a useful theorem that I didn't know, thank you. As you also point out in your comment, the key issue that I didn't expect is that you do need to make a base change sometimes in order to have the boxes nicely aligned, e.g. for $(1+i)\subseteq \mathbb{Z}[i]$ using the basis $\{ 1, 1+i \}$ one gets indeed the visualization of the norm (if I did everything correct). But I still could not figure out the second question related to ramification of primes. Is it possible to apply this theorem that you mentioned to do so? I could not find a way to do it. $\endgroup$ – Pedro Feb 19 '18 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.