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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.

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$I$ 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.

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  • $\begingroup$ 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. $\endgroup$
    – scd
    Dec 18 '20 at 17:41
  • $\begingroup$ 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 $\Bbb{R}^n/a\Bbb{R}\cong \Bbb{R}^{n-1}$. $\endgroup$
    – reuns
    Dec 18 '20 at 18:07
  • $\begingroup$ Could you please also give some background about why $K\otimes_\mathbb{Q}\mathbb{R}=\mathbb{R}[x]/f(x)$?. $\endgroup$
    – scd
    Dec 19 '20 at 7:43
  • $\begingroup$ $\mathbb{Q}[x]/(f(x))\otimes_\mathbb{Q}\mathbb{R}=\mathbb{R}[x]/(f(x))$ this is essentially the definition, we take a $\Bbb{Q}$-basis of $\mathbb{Q}[x]/(f(x))$ and we enlarge the coefficients to $\Bbb{R}$ $\endgroup$
    – reuns
    Dec 19 '20 at 8:11
  • $\begingroup$ Now I see, thanks a lot! $\endgroup$
    – scd
    Dec 19 '20 at 10:26

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