# Minkowski Theory: The embedding of fractional ideal is a full rank lattice.

I encountered this question when studying the proof towards Minkowski bound. Let $$K/\mathbb{Q}$$ be a number field of degree $$n$$. Let $$r$$ and $$s$$ be the number of real and complex embedding of $$K$$ respectively, then $$n=r+2s$$. Let $$\sigma:K\to \mathbb{R}^r\times\mathbb{C}^s\cong \mathbb{R}^n$$ be the usual embedding. The literature suggests that for any fractional ideal $$I$$ of $$\mathcal{O}_K$$, $$\sigma(I)$$ is a full rank lattice, i.e. a lattice of rank $$n$$. I have difficulty proving this fact.

I knew that $$\mathcal{O}_K$$ is a free $$\mathbb{Z}$$-module of rank $$n$$. Suppose $$\{v_1.\dots,v_n\}$$ is a $$\mathbb{Z}$$-basis of $$\mathcal{O}_K$$ and $$I$$ a non-zero ideal, then for a non-zero $$a\in I$$, $$\{av_i\}\subset I$$ is $$\mathbb{Z}$$-linearly independent. Furthermore, $$\sigma(I)$$ is discrete.

$$I$$ contains $$a O_K$$ and $$a O_K$$ has finite index in $$I$$ so you already have the answer : discrete and full rank subgroup.

Otherwise

$$K= \Bbb{Q}[x]/(f)$$, factorize $$f=\prod_j f_j \in \Bbb{R}[x]$$ so that $$r_1=\# \{ f_j,\deg(f_j)=1\},r_2=\# \{ f_j,\deg(f_j)=2\}$$.

Then $$K\otimes_\Bbb{Q}\Bbb{R} = \Bbb{R}[x]/(f)\cong \prod_j \Bbb{R}[x]/(f_j) = \Bbb{R}^{r_1}\Bbb{C}^{r_2}$$ The isomorphism is given by the real/complex embeddings.

On the other hand $$O_K= O_K\otimes_\Bbb{Z}\Bbb{Z}$$ is a lattice in $$K\otimes_\Bbb{R}\Bbb{R}$$.

For a fractional ideal $$I$$ then $$m I$$ is a non-zero ideal for some integer, it has finite index $$N(mI)$$ in $$O_K$$ so $$mI$$ and $$I$$ are lattices too.

• Thank you very much for the answer. I have a questions: Are you using a result like "finite index subgroup of free Z-module is free" in the first and last paragraph in your answer? I don't yet know such a result and want to check if I misunderstood.
– scd
Dec 18 '20 at 17:41
• You can say that lattice = discrete full rank subgroup of $\Bbb{R}^n$. They are free (generated by $n$ elements), the proof is by induction, take $l\ne 0\in L$, then $L\cap l\Bbb{R} = a \Bbb{Z}$ and $L/a\Bbb{Z}$ is a discrete full rank subgroup of $\Bbb{R}^n/a\Bbb{R}\cong \Bbb{R}^{n-1}$. Dec 18 '20 at 18:07
• Could you please also give some background about why $K\otimes_\mathbb{Q}\mathbb{R}=\mathbb{R}[x]/f(x)$?.
– scd
Dec 19 '20 at 7:43
• $\mathbb{Q}[x]/(f(x))\otimes_\mathbb{Q}\mathbb{R}=\mathbb{R}[x]/(f(x))$ this is essentially the definition, we take a $\Bbb{Q}$-basis of $\mathbb{Q}[x]/(f(x))$ and we enlarge the coefficients to $\Bbb{R}$ Dec 19 '20 at 8:11
• Now I see, thanks a lot!
– scd
Dec 19 '20 at 10:26