Let $K$ be a finite Galois extension of $\mathbb{Q}$ and let $A$ be a subring. We say $A$ is integrally closed in $K$ if any element $a \in K$ integral over $A$ is contained in $A$.
Does being integrally closed have an interpretation in terms of the automorphisms of $K$?
I have come across a similar observation which goes as follows:
Lemma: Let $L/K$ be a Galois extension of fields, and let $A \subset L$ be a subring. If $\sigma (A) = A$ for each $\sigma \in [L, L]_K$, then $A$ is integral over $A \cap K$.
Proof: Take $a \in A$ and let $f \in K[x]$ be the (monic) minimal polynomial for $a$ over $K$. $f(x)$ factors as $\prod_{\sigma \in [L, L]_K }(x - \sigma(a))$. Since each $\sigma(a)$ belongs to $A$, the coefficients of $f$ belong to $A$, so that they belong to $A \cap K$. Therefore $a$ is integral over $A \cap K$.
I suspect there should be some way of expressing "integrally closed" with the automorphisms, but I can't quite figure out what it should be.
Of course, this would lead to a description of $\mathcal{O}_K \subset K$ for a Galois number field $K$ in terms of the Galois group, which is the case I am interested in.