Let $R$ be a ring (we will below more specify it) and $A$ a $R$-algebra. I'm looking for an transparent explanation on connection between the relative discriminant and etaleness for certain $R$-algebras. In particular I want to understand why the discriminant ideal "measures" in appropriate way if a $R$ -algebra $A$ is etale.
As I mentioned we have to specify more precisely which ring morphisms $R \to A$ I intend to talk. I have two scenarios in mind but one of the aims of this question question is how far this can be generalized.
Scenario 1 from algebraic number theory: Let $L/K$ be an an finite extension of number fields with $O_K, O_L$ their rings of integers. Let ${σ_1, ..., σ_n}$ be the set of embeddings of $L$ into $\mathbb{C}$ which are the identity on $K$. If $b_1, ..., b_n$ is any basis of $L$ over $K$, let $d(b_1, ..., b_n)$ be the square of the determinant of the $n$ by $n$ matrix whose $(i,j)$-entry is $σ_i(b_j)$.
Then, the relative discriminant of $L/K$ is the ideal $\Delta_{L/K}$ generated by the $d(b_1, ..., b_n)$ as ${b_1, ..., b_n}$ varies over all integral bases of $L/K$. (i.e. bases with the property that $b_i \in O_K$ for all $i$.) Alternatively, the relative discriminant of $L/K$ is the norm of the different of $L/K$.When $K = \mathbb{Q}$, the relative discriminant $\Delta{L}/\mathbb{Q}$ is the principal ideal of $\mathbb{Q}$ generated by the absolute discriminant $\Delta K$.
Now consider the ring map $\phi:O_K \to O_L$. The claim is that if the discriminant ideal $\Delta_{L/K}$ ist unit ideal (ie contain $_K$), then the map $\phi:O_K \to O_L$ is etale.
A consequence is that if not generally unit ideal $\Delta_{L/K}$ and we take a nonzero $d \in \Delta_{L/K}$ and localize he map $\phi:O_K \to O_L$ with respect to $d$ then the $O_K[d^{-1}]$-algebra $O_L[d^{-1}]$ becomes etale.
Scenario 2: that was a part of my old question a couple weeks ago: $R$ was assumed to be a Dedekind ring with fraction field $F=Frac(R)$ and $F \subset E$ a finite Galois extension and $A$ integral closure of $R$ in $E$. The thirt part asked for the existence of a $f \in R$ such that after localizing $R[f] \subset A[f]$ becomes etale.
According reuns answer the choice of $f$ works as follows: by primitive element theorem $E=F(a)≅F[x]/(g(x))$ with $g \in R[x]_{monic}$ and the candidate for $f$ was $d^{-1}$ with $d:= Dics(g)$ (ie the discriminant of the polynomial $g$.
I know that in this case one can show that this polynomial discriminant $d$ generates the discriminant $\Delta_{E/F}$ defined in the way as in in scenario 1 described.
unfortunately I neither understood the reason why this choice does it's job nor the striking relation between the discriminant ideal and etaleness in detail.
The definition of etaleness I know is this one: A ring map $R \to A$ is called etale if for every prime $\mathfrak{p}$ of $R$ and everyprime $\mathfrak{q}$ of $A$ lying over $\mathfrak{p}$ the localization map of local rings $R_{\mathfrak{p}} \to R_{\mathfrak{q}}$ is etale. That means that the field extension $R/\mathfrak{p} \to A/R_{\mathfrak{q}}$ is finite separable and $\mathfrak{q}= \mathfrak{p}A_{\mathfrak{q}}$.
Could anybody explain how this relation between etaleness and "invertibility" of relative discriminant ideal comes from. I still haven't study the proof of this equivalence (etale iff discrim ideal unit ideal; I would be pleasant for a reference for a transparent proof or sketch outlining the idea) so i hope that the proof itself maybe provide the reason where the invertibility of discriminant comes into game.
Other question ist how far this effect of discriminant ideal can be generalized. Can if generally be true that for evry ring map $R \to A$ as far a meaningful "discriminant ideal" $\Delta_{A/R}$ can be defined (e.g. axiomatic way) then localizing at any not trivial element & non zero divisor element of the discriminant ideal would imduce an etale map?