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.
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 $B$.