0
$\begingroup$

I'm working on the following problem, and would really appreciate some clarification.

Problem: Let $K$ be a number field with $[K:\mathbb{Q}] = n$ with ring of integers $\mathcal{O}_k$. Let $M$ be a $\mathbb{Z}$-submodule in $K$ with $\dim_n(M \otimes_\mathbb{Z} \mathbb{Q}) = n$. Why is $M$ free? Let $\alpha_1, \dots, \alpha_n$ be a $\mathbb{Z}$-basis of $M$. Show that the discriminant $d(M) := d(\alpha_1, \dots, \alpha_n) \in \mathbb{Z}$ and that it is independent of the choice of the basis.

It is clear to me that a $\mathbb{Z}$-basis of $M$ (if $M$ is free) is also a $\mathbb{Q}$-basis of $K$, thus we can talk about the discriminant. It is also clear to me that $d(M)$ must be independent of choice of basis. However I have two questions:

i.) Why does $M$ have to be free? It is torsion-free, so it would suffice to show that is is finitely generated (or to show that it is a finitely generated $\mathcal{O}_k$-module). But why is this the case?

ii.) Why does $d(M)$ have to be an element of $\mathbb{Z}$? Suppose we have the basis $\alpha_1, \dots, \alpha_n$ of $M$, then $\alpha_i = \displaystyle \frac{b_i}{c_i}$ for some $b_i \in \mathcal{O}_k, c_i \in \mathbb{Z}$. Then $$d(\alpha_1, \dots, \alpha_n) = \underbrace{d(b_1, \dots, b_n)}_{\in \mathbb{Z}} \cdot \det(A)^2$$, where $A$ is change of basis matrix, i.e. $\det(A) = \displaystyle\frac{1}{c_1 \cdot \cdots \cdot c_n}$. But $\displaystyle\frac{d(b_1, \dots, b_n)}{c_1^2 \cdot \cdots \cdot c_n^2}$ does not necessarily have to lie in $\mathbb{Z}$.

Thank you very much for any help and hints!

$\endgroup$
  • 1
    $\begingroup$ Further comment: the given criterion does not guarantee that $ M $ is finitely generated. Indeed, take $ M $ to be $ K $ itself! $\endgroup$ – Starfall Oct 31 '16 at 11:17
  • $\begingroup$ @Starfall Of course! How did I not see that. Maybe this is why I had been so confused. Thx. $\endgroup$ – johnnycrab Oct 31 '16 at 12:57
1
$\begingroup$

For your first question, show that the span of a $\mathbf Q $ basis of $ K $ lying in $ \mathcal O_K $ has finite index in the ring of integers. Deduce that the ring of integers is a submodule of a free $ \mathbf Z $-module, and thus is itself free. Then, show that $ aM $ lies in the ring of integers for some integer $ a $, conclude.

For your second question, prove that the discriminant is the determinant of the trace form matrix w.r.t. some basis. This matrix has integer entries, and thus its determinant is also an integer.

$\endgroup$
  • $\begingroup$ Thanks. For the second part: why does the matrix have integer entries, if $M$ is an arbitrary module, and not $\mathcal{O}_K$ itself? I thought that the trace is in $\mathbb{Z}$ is only neceassarily true for elements in the ring of integers. $\endgroup$ – johnnycrab Oct 31 '16 at 10:41
  • $\begingroup$ If it is an arbitrary module, the discriminant need not be integral. $\endgroup$ – Starfall Oct 31 '16 at 10:46
  • $\begingroup$ Hm, but then I don't understand why $d(M)$ is in $\mathbb{Z}$. Would you mind explaining it a bit further? $\endgroup$ – johnnycrab Oct 31 '16 at 10:55
  • $\begingroup$ The claim in the question is simply not true. Consider the integer span of $ 1/2 $ and $ \sqrt{3}/2 $ in $ \mathbf Q(\sqrt {3}) $, for instance. The discriminant of this basis is $ 3/4 $. $\endgroup$ – Starfall Oct 31 '16 at 11:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.