Prime ideals in Galois extensions This is with reference to Proposition 1 in Robert Ash's notes.
I don't think the Dedekind assumption is necessary. 

Explicitly, if $A$ is an integral domain with fraction field $K$ and $L/K$ is Galois, let $B$ be the integral closure of $A$ in $L$. Then, I claim that any element of the Galois group $G$ of $L/K$ acts on the set of prime ideals lying over a prime $P\subseteq A$.

Let $Q$ be a prime ideal in $B$ lying over $P$. Then, if $\sigma \in G$ and $\sigma(x) \sigma(y) \in \sigma(Q)$ then, $\sigma(xy)\in  \sigma(Q)$. Applying, $\sigma ^{-1}$, we get $xy\in Q$ and hence either $x$ or $y$ is in $Q$ and thus, either $\sigma(x)$ or $\sigma(y)$ is in  $\sigma(Q)$. 
Finally, $\sigma$ is an automorphism and hence preserves intersections and also fixes $A$ and hence $P$, so $\sigma(Q) \cap A=  \sigma(Q) \cap \sigma(A) = \sigma(Q\cap A)= \sigma(P)=P$.
Are there any holes in this argument?
I am doubting since I have seen three sets of number theory notes which use the Dedekind assumption.
 A: There are no holes, that is good as far as I can see.
Here's another way to see why this is true. $\text{Aut}_K(L)$ acts by $\mathcal{O}_K$-algebra automorphisms on $\mathcal{O}_L$, where $\sigma \in \text{Aut}_K(L)$ sends $a \in \mathcal{O}_L$ to $\sigma(a)$, which is integral over $\mathcal{O}_K$ by the same relation, and therefore contained in $\mathcal{O}_L$. Write 
$$\phi : \text{Aut}_K(L) \rightarrow \text{Ring}(\mathcal{O}_L, \mathcal{O}_L)$$
for the corresponding structure map. 
$$\psi : \text{Aut}_K(L)^{\text{op}} \rightarrow \text{Top}(\text{Spec}(\mathcal{O}_L), \text{Spec}(\mathcal{O}_L))$$
sending $\sigma \in \text{Aut}_K(L)$ to $\text{Spec}(\phi(\sigma))$, where $\text{Spec}: \text{Ring} \rightarrow \text{Top}$ is the (contravariant) spectral space functor. Let $i : \mathcal{O}_K \rightarrow \mathcal{O}_L$ be the inclusion map.
Now for why $\psi(\sigma)$ sends primes over $\mathfrak{p}$ to primes over $\mathfrak{p}$ for each prime $\mathfrak{p} \in \text{Spec}(\mathcal{O}_K)$.  $\phi(\sigma) \circ i = i$ for each $\sigma \in \text{Aut}_K(L)$. Applying $\text{Spec}$, $\text{Spec}(i) \circ \psi(\sigma) = \text{Spec}(i)$. Therefore, for each prime $\mathfrak{q}$ lying over $\mathfrak{p}$, $\psi(\sigma)(\mathfrak{q})$ lies over $\text{Spec}(i) \circ \psi(\sigma)(\mathfrak{q}) = \text{Spec}(i)(\mathfrak{q}) = \mathfrak{p}$. So $\text{Aut}_K(L)$ induces an action on the primes lying above $\mathfrak{p}$. 
