About the ramification locus of a morphism with zero dimensional fibers

Question

This question arises from my somewhat frustrating attempts to understand what etale means (in the world of algebraic varieties for now) and marry the more advanced algebraic geometry references and the ones from differential geometry. The particular example I am studying, is the Lyasko-Looijenga morphism.

Say that $f:X\rightarrow Y$ is a finite map of affine varieties with zero-dimensional fibers. Finiteness of the map means that the coordinate ring $A(X)=S$ is a finitely generated $A(Y)=R$ module. Now say the point $y\in Y$ corresponds to the maximal ideal $P\subset R$.

I understand how the ideal $PS$ in $S$ cuts the variety $f^{-1}(y)$ but might not be radical, so we define $S/PS$ to be the algebraic fiber but we know it is not exactly the coordinate ring of $f^{-1}(y)$. Evenmore, we know that $S/PS$ is an $R/P$-vector space of dimension $k$ at most as big as the rank of $S$ as an $R$-module (and equal if in particular $S$ is a free $R$-module).

This in turn implies that the fiber $f^{-1}(y)$ will contain at most $k$-many points (the coordinate ring of a finite variety is a vector space of dimension as big as its cardinality). I am trying to understand when this number is achieved:

The approach in Shafarevich is to consider the extension $f^{*}k(Y)\subset k(X)$ and define $\operatorname{deg}f$ to be its degree; when the cardinality of the fiber equals that degree: $$\#f^{-1}(y)=\operatorname{deg}f$$ we say that $f$ is unramified at $y$. The differential geometry approach is just to check the Jacobian at $y$; $f$ is unramified at $y$ precisely when $$J_f\ (y)\neq 0.$$

Question 1: I am asking for some help to understad the equivalence here. How does the Jacobian of $f$ affect the cardinality of $f^{-1}(y)$ in relation to the degre of the extension $f^{*}k(Y)\subset k(X)$?

Question 2: What about the ring $S/PS$? It encodes all the information we need for $f^{-1}(y)$. How can we understand here which points are fatter than others? What is the situation for $\operatorname{Spec}S/PS$?

Question 3: Is it apparent that the degree of the extension $f^{*}k(Y)\subset k(X)$ is equal to the rank of $S$ as an $R$-module, when $S$ is free over $R$? What happens when $S$ is not free?

Remark: The reason that I am not just satisfied with the Jacobian definition (the fiber has full cardinality when $J_f\ (y)\neq 0$ ) is because it is easier to compute the generic size of the fiber via $S/PS$. In particular, when -as in my case- $S$ can be shown to be a free $R$-module, the rank can be calculated via the Hilbert series of the two rings.

On the other hand, it is much easier to actually find the ramification locus using the Jacobian definition. Therein lies my current conundrum...

Answer 1

I will only try amd answer question 1 above. For the sake of convenience, I will assume that $A,B$ are integral domains, finite type over an algebraically closed field $k$. Furthermore, we are given an inclusion of rings $f : A \hookrightarrow B$ such that $B$ is module finite over $A$.

Definition 1 (Standard definition): Let $\mathfrak{m}$ be a maximal ideal of $A$. We say that $f$ is unramified at $\mathfrak{m}$ if $$\Omega^1_{B \otimes_A k(\mathfrak{m})/k(\mathfrak{m})} = 0,$$ where $k(\mathfrak{m}) = A/\mathfrak{m}$ .

<p><strong>Definition 2 (Jacobian definition)</strong>:  Write $B = A[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$. Let $\mathfrak{m}$ be a maximal ideal of $A$. We say that $f$ is <strong><em>unramified</em></strong> at $\mathfrak{m}$ if $\operatorname{Jac}_f(\mathfrak{m})$ has full rank, where
$$\operatorname{Jac}_f = \left(\begin{array}{ccc} \partial f_1/\partial x_1 &amp; \ldots &amp; \partial f_r/\partial x_1 \\

\vdots & & \vdots \ \partial f_1/\partial x_n & \ldots & \partial f_r/\partial x_n \end{array} \right)$$ and $\operatorname{Jac}_f(\mathfrak{m}) = \operatorname{Jac}_f \otimes_A A/\mathfrak{m}$.

<p><strong>Definition 3 (Shafarevich)</strong>: Let $\mathfrak{m}$ be a maximal ideal of $A$. We say that $f$ is <strong><em>unramified</em></strong> at $\mathfrak{m}$ if the number of maximal ideals in the Artinian ring $B \otimes_A A/\mathfrak{m}$ is equal to the degree of the field extension $\operatorname{Frac}(B)/\operatorname{Frac} (A)$.</p>

Let us first show that Definition 1 and 2 are equivalent. This is essentially immediate. It is a standard exercise to check that if we think of $\operatorname{Jac}_f$ as a $B$-linear map $B^{\oplus r} \to B^{\oplus n}$, then $\Omega^1_{B/A}$ is the cokernel of this linear map. Hence $\operatorname{Jac}_f(\mathfrak{m})$ has full rank if and only if $$\Omega^1_{B/A} \otimes_A A/\mathfrak{m} = \Omega^1_{B \otimes_A k(\mathfrak{m})/k(\mathfrak{m})} = 0.$$ This shows that Definitions 1 and 2 are equivalent.

Now at this stage I can only prove that Definition 1 implies 3. Define

$$ N:= \text{Number of primes in $B$ lying over $\mathfrak{m}$}$$ and $$ r := \dim_{k(\eta)} B\otimes_A k(\eta) = [\operatorname{Frac}(B): \operatorname{Frac}(A)],$$ where by $\eta$ I mean the generic point of $\operatorname{Spec} A$. By upper semicontinuity of the fiber dimension of a coherent sheaf, it follows that $\dim_{k(\mathfrak{m})} B\otimes_A k(\mathfrak{m}) \geq r$. However by assumption of $f$ being unramified at $\mathfrak{m}$ in the sense of definition $1$, we have that $B\otimes_A k(\mathfrak{m})$ is a product of $\dim_{k(\mathfrak{m})} B\otimes_A k(\mathfrak{m})$ copies of $k$. It follows that $N \geq r$. However, we also know that trivially $N \leq r$ and so we are done.

Edit: The last inequality $N \leq r$ is wrong without further hypothesis. I will have to think about my answer.

Answer 2

You have to be a bit careful here. E.g. BenLim writes in his answer that trivially $N \leq r,$ but this is not actually true in the generality under discussion here.

E.g. let $A = \{ f \in \mathbb C[t] \, | \, f(-1 ) = f(1) \} \subset B = \mathbb C[t].$

The corresponding map on Specs is the map from $\mathbb A^1$ to the nodal curve $$y^2 = x^3 + x^2$$ (Note that $A \cong \mathbb C[x,y]/(y^2 - x^3 - x^2)$ via $x = t^2 - 1,$ $y = t(t^2 - 1).$)

Then the Frac$(A) = $ Frac$(B) = \mathbb C(t)$, so the degree of the function field extension is $1$. But above the point $(x,y) = (0,0)$ of the nodal curve, there are two points of $\mathbb A^1$ (namely $t = \pm 1$).

Also, the map Spec $B \to $ Spec $A$ is unramified at every point of Spec $A$, in the sense of BenLim's Definition 1.

(Geometrically, you have a line mapping to a nodal curve, and infinitesimally, at every point in the source, this map looks like a closed immersion into the target. Hence it behaves like an immersion from a formal/infinitesimal perspective, and hence is unramified, even though it is not actually an immersion.)

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Summary: The claim that Defs. 1/2 imply Def. 3 is not true in general.

To connect BenLim's Definitions 1 and 2 to his Definition 3, you will need an additional assumption, such as flatness for the map $A \to B$ (which is what is missing in the above example).

E.g. if $A$ and $B$ are smooth over the ground field, then miracle flatness implies that $B$ is flat over $A$ (given that we have a finite morphism, so that all fibres are finite), and then Def. 3 will be equiv. to Defs. 1 and 2.

Remark: When we are looking at f.g. modules over a Noeth. ring, as here, flatness is equivalent to locally free. Rereading your question, it seems that you are happy to be in this situation (and as noted above, it is automatic if $A$ and $B$ correspond to smooth varieties).

In this case, all the definitions of unramified from BenLim's answer are equivalent. So e.g. you can use the Jacobian criterion to check for unramifiedness (equivalently, etaleness, since we are giving ourselves flatness, and etale equals unramified plus flat for morphisms of varieties, or more generally, for morphisms of Noetherian schemes), and then use the field degree to determine the actual size of the fibres.