# Integral Closure, Galois extension,and Dedekind Domain

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.

• 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. – Madarb Dec 9 '18 at 7:38
• 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. – Kento Dec 9 '18 at 7:56
• 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 :) – Madarb Dec 9 '18 at 8:00
• 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. – Kento Dec 9 '18 at 8:07
• @reuns I think you don't necessarily have $\sigma(B) = B$. A possible counterexample is given below. – pisco Dec 9 '18 at 12:53

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 $$B$$.
• 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 $\sigma(B) \ne B$. – reuns Dec 9 '18 at 23:22
• And replacing $B$ by $\bigcap_{\sigma \in G} \sigma(B)$ makes the claim true – reuns Dec 10 '18 at 18:19