4
$\begingroup$

Let $A$: Dedekind domain, $K$: $\operatorname{Frac}(A)$, $B$: Dedekind domain with $A \subset B$, $L$: $\operatorname{Frac}(B)$

Let $L/K$: galois extension with galois group: $G$.

$B^G=\{b \in B \mid \sigma(b)=b \text{ for all } \sigma \in G\}=A$ $\implies B$ is the integral closure of $A$ in $L$

Is this true? I already prove the converse, but not sure if this holds. Thank you in advance.

$\endgroup$
  • $\begingroup$ I think it's not true. Try to take B=A. Then the fixed ring is A, but unless the extension is trivial, this is not the integral closure. $\endgroup$ – Madarb Dec 9 '18 at 7:38
  • $\begingroup$ when A=B, then L=K. Since A is dedekind, A is integrally closed and B(=A) is the integral closure of A in L holds, i think. $\endgroup$ – Kento Dec 9 '18 at 7:56
  • $\begingroup$ I don't think I understood what you say. The claim you want to prove, as I understand it, is that if B is a subring of L, for which the fixed ring under the action of G is A, then B is the integral closure. It is not the true if you take B=A. If I got you wrong, please explain what you wanted to prove :) $\endgroup$ – Madarb Dec 9 '18 at 8:00
  • 1
    $\begingroup$ you understand it right. but i do not get that it does not hold when B=A. when B=A,the extension L/K is trivial, which leads to the conclusion. i think. please tell me if i get something wrong. $\endgroup$ – Kento Dec 9 '18 at 8:07
  • 1
    $\begingroup$ @reuns I think you don't necessarily have $\sigma(B) = B$. A possible counterexample is given below. $\endgroup$ – pisco Dec 9 '18 at 12:53
4
$\begingroup$

This is not true.

Let $L/K$ be a Galois extension of number field, let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$ that splits into more than one primes in $\mathcal{O}_L$: $$\mathfrak{p}\mathcal{O}_L = \mathfrak{P}_1 \cdots \mathfrak{P}_r$$

Let $A = (\mathcal{O}_K)_{\mathfrak{p}}$, the localization at $\mathfrak{p}$ and $B = (\mathcal{O}_L)_{\mathfrak{P}_1}$. Both are Dedekind domains. It is easily seen that $B^G = A$, but the integral closure of $A$ in $L$ is the localization of $\mathcal{O}_L$ at all $\mathfrak{P}_1, \cdots, \mathfrak{P}_r$, which is a proper subset of $B$.

$\endgroup$
  • 1
    $\begingroup$ So $K = \mathbf{Q}, L = \mathbf{Q}(i),p=5= (2+i)(2-i), A=\mathbf{Z}_{(5)} = \{\frac{u}{v}, (u,v) \in \mathbf{Z}^2, 5 \nmid v \}$, $B = \mathbf{Z}[i]_{(2+i)} = \{\frac{u}{v}, (u,v) \in \mathbf{Z}[i]^2, v \not \in (2+i)\mathbf{Z}[i] \}$, $\sigma(c+id) = c-id, G= \{\sigma^2,\sigma\}$ then $\frac1{5^n} \not \in B$ so $B^G = \mathbf{Z}_{(5)}$ but $\frac{2+i}{5}=\frac{1}{2-i} \in B$ is not integral over $\mathbf{Z}_{(5)}$. $A,B$ are Dedekind domains since they have only one prime ideal. The problem is that $\sigma(B) \ne B$. $\endgroup$ – reuns Dec 9 '18 at 23:22
  • $\begingroup$ And replacing $B$ by $\bigcap_{\sigma \in G} \sigma(B)$ makes the claim true $\endgroup$ – reuns Dec 10 '18 at 18:19

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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