Different ideal and Kähler differentials

Question

Let $A$ be a Dedekind domain with fraction field $K$, $L/K$ be a finite separable extension and $B$ be the integral closure of $A$ in $L$, which is also a Dedekind domain. Assume all residue extensions are separable, e.g. if $A=\mathbb{Z}$.

Let $\delta_{L/K}$ be the different ideal, which is the inverse fractional ideal of the trace-dual of $B$. We can compute the different as the annihilator of the module of Kähler differentials of $B$ over $A$. So the zero locus of the different ideal is the set of points on which the Kähler differentials vanish.

But we also know that those primes are the primes that are ramified over some prime in $A$. And if we draw $\operatorname{Spec}(A)$ as a horizontal line (call it $x$-axis) and project down the regular curve $\operatorname{Spec}(B)$, those primes are the points with vertical on which the curve has vertical slope. "The $x$-coordinate of the tangent vector vanishes", and Dedekin's central theorem regarding different and ramification tells us moreover that it vanishes with the right order.

Is everything I have said so far correct? If so, is there any relation between the Kähler differentials and the $x$-coordinate of the "tangent vector" to the curve $B$? And if this is the case, how does one connect the definition of the different with the trace-dual and the inverse fractional ideal to this geometric interpretation?

Points where I am stuck: I don't know how to interpret the trace pairing in a geometric way, so I don't know how to interpret trace-dual ideals in a geometric way.

Extra (secondary) question: $B$ may not be generated by a single element over $A$, but $\Omega_{B/A}$ can always be generated by a single element. What is the intuition behind this? I get that if $B=A[\alpha]$ then $\Omega_{B/A}$ is generated by $d\alpha$, but if this is not the case, who is the generator and why can we expect to have a single generator?

Answer 1

If $f:X\rightarrow Y$ is a flat morphism of schemes of finite type over a field $k$, it follows $f$ is etale iff $\Omega^1_{X/Y}=0$. If a similar statement is valid for for the relative situation you get something similar in your situation.

Definition. We say the morphism $f$ is unramified at $x$ iff $\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$ and $\kappa(y) \subseteq \kappa(x)$ is separable. We may define $f$ to be "etale" iff it is flat and unramified at all points.

If $A:=\mathcal{O}_K, B:=\mathcal{O}_L$ with $K \subseteq L$ a finite extension of number fields it follows any residue field extension $\mathcal{O}_K/\mathfrak{p} \subseteq \mathcal{O}_L/\mathfrak{q}$ is separable. If $Y:=Spec(A), X:=Spec(B)$ and $f:X\rightarrow Y$ is the canonical map, it follows the support $Z$ of $\Omega^1_{X/Y}$ is a closed subscheme defined by the different $\delta_{L/K}$. The map $f$ is finite and hence $Y-f(Z):=U$ is an open subscheme of $Y$. The induced map $f_U: V:=f^{-1}(U) \rightarrow U$ is flat and $\Omega^1_{V/U}=0$ by definition. Hence if a relative version of Exercise III.10.3 in Hartshorne is true, it follows $f_U$ is etale.

If $\mathfrak{p}\in U$ is a prime where the fiber $f^{-1}(\mathfrak{p})$ is unramified, we get an isomorphism

E1. $f^{-1}(\mathfrak{p})\cong Spec(B/\mathfrak{p}B)$

where $B/\mathfrak{p}B\cong B/\mathfrak{p}_1\cdots \mathfrak{p}_d$

where $\mathfrak{p}_i \neq \mathfrak{p}_j$ are distinct maximal ideals in $B$. It follows

E2 $B/\mathfrak{p}B\cong B/\mathfrak{p}_1\oplus \cdots \oplus B/\mathfrak{p}_d:=R_1\oplus \cdots \oplus R_d$

Let $B_1:=B/\mathfrak{p}B$ and $A_1:=A/\mathfrak{p}$.

We get

$\Omega^1_{B_1/A_1}\cong \Omega^1_{R_1/A_1}\oplus \cdots \oplus \Omega^1_{B_d/A_1}=0$ since the field extensions $A_1 \subseteq B_i$ are separable.

Question: "Is everything I have said so far correct? If so, is there any relation between the Kähler differentials and the x-coordinate of the "tangent vector" to the curve B? And if this is the case, how does one connect the definition of the different with the trace-dual and the inverse fractional ideal to this geometric interpretation?"

The support $Z \subseteq X$ of the module of Kahler differentials is canonically determined by the module of Kahler differentials and the image $f(Z)\subseteq Y$ is canonically a closed subscheme of $Y$ determined by an ideal $I$ in $A$. The discriminant $D(L/K)$ in $A$ is an ideal giving a characterization of primes in $A$ that ramify in $B$. Hence the two ideals $D(L/K)$ and $I$ have the same support. Maybe they are the same ideal. A problem is that if $\mathfrak{p}\in V(D(L/K))$ it follows the fiber $\pi^{-1}(\mathfrak{p})$ is "ramified", but not all closed points $\mathfrak{q}\in \pi^{-1}(\mathfrak{p})$ need to be ramified.

Ref. your other question on the "geometry" of the map $f$:

Geometric interpretation of ramification of prime ideals.

Answer 2

This isn't a perfect answer, but my intuition is consider complex curves, and a ramification $z \to z^n$ with galois group $G=\mathbb{Z}/n$. It acts via multiplying with roots of unity.

Then we see that any meromorphic function with pole of order less than $n$ has that if we sum its conjugates, all the poles will vanish.

The reason this intuition isn't perfect is that $Q_p$ i.e even the local case, has many interesting totally ramified galois extensions, while I don't think we have this in the complex case.