Why isn't every finite locally free morphism etale?

I have been reading the notes here. There, a finite locally free morphism of schemes is defined as as a morphism of schemes $f: X \rightarrow Y$ which is finite and for which the sheaf $f_{*} \mathcal{O}_{X}$ is locally free as an $\mathcal{O}_{Y}$-module. A finite etale morphism is then defined as a morphism of schemes $f: X \rightarrow Y$ which is finite locally free, and for which the fiber over any point $q \in Y$ is an etale $\kappa(q)$-algebra.

I am struggling to understand what the part about the fibers being etale algebras adds at all. It seems like every finite locally free morphism would trivially satisfy this.

Take $f: X \rightarrow Y$ to be a finite locally free morphism of schemes. For any $q \in Y$, choose an affine $\operatorname{spec}A$ around $q$ small enough so that $f_{*} \mathcal{O}_{X}$ is free, say of degree $d$. Then since $f$ is finite, it is in particular affine, so the morphism looks locally like $\operatorname{spec}(A^{\oplus d}) \rightarrow \operatorname{spec}A$. Then the fiber over $q$ is just $\operatorname{spec}(\kappa(q)^{\oplus d})$. This seems to trivially be an etale $\kappa(q)$-algebra, since tensoring with some algebraic closure $\Omega$ would give $d$ copies of $\Omega$.

Where is the flaw in my reasoning? Or is it just the case that every finite locally free morphism is etale?

Further to this, the same notes say that for a finite etale morphism, the degree of the morphism at a point $q \in Y$ is the same as the cardinality of the fiber over $q$. But again the above reasoning seems to show this for any finite locally free morphism. Where does etale come into it at all?

Solution 1:

Here is an example of a finite locally free morphism which is not etale: take spec of the natural inclusion $\Bbb F_2(t^2)\subset \Bbb F_2(t)$. This fails to be etale because it's a non-separable field extension.

The flaw in your reasoning is that your identification of $f_*\mathcal{O}_X$ as $A^d$ locally is only as a module - it needs to be as a ring to say anything interesting. (You should also note that even in the case of $\Bbb R\subset\Bbb C$, your idea is wrong - this is a point mapping to a point, while your reasoning would have two points mapping to one point.)

Solution 2:

It is my impression that this is a student site, where PhD students in algebraic geometry and other fields can ask questions on details. It is difficult to find books giving "all" details, and this forum is a good place to discuss this during a "pandemic".

Question:"Take $f:X \rightarrow Y$ to be a finite locally free morphism of schemes. For any $q \in Y$, choose an affine $Spec(A)$ around $q$ small enough so that $f_∗\mathcal{O}_X$ is free, say of degree $d$. Then since $f$ is finite, it is in particular affine, so the morphism looks locally like $Spec(A^{\oplus d})→Spec(A)$. Then the fiber over $q$ is just $Spec(κ(q)^{\oplus d})$. This seems to trivially be an etale $κ(q)$-algebra, since tensoring with some algebraic closure $Ω$ would give $d$ copies of $Ω$. Where is the flaw in my reasoning? Or is it just the case that every finite locally free morphism is etale?"

Answer: If $B:=A[t]/(P(t))$ where $P(t):=t^d+a_1t^{d-1}+\cdots + a_d \in A[t]$, let $S:=Spec(B), T:=Spec(A)$ and $\pi:S\rightarrow T$ be the canonical map. Since there is an isomorphism

$B\cong A\{1,\overline{t},\cdots , \overline{t}^{d-1}\}$ as $A$-modules, it follows $B$ is a free $A$-module of rank $d$ hence the map $\pi$ is "finite locally free" as map of schemes. If $x\in T$ is any point with residue field $\kappa(x)$ we may pass to the fiber at $x$:

$\pi^{-1}(x):=Spec(\kappa(x)[t]/\overline{P(t)})$

where $P_x(t):=\overline{P(t)}\subseteq \kappa(x)[t]$ is the polynomial we get when we "reduce the coefficients modulo $x$". It follows the polynomial $P_x(t)$ may be written as s product

$P_x(t)=\prod_{i=1}^n P_{i,x}(t)^{l_i}$ where $P_{i,x}(t)$ is an irreducible polynomial for all $i$. If $l_i\geq 2$ for some $i$ or if $P_{i,x}(t)$ has repeated roots in the algebraic closure $\overline{\kappa(x)}$ it follows the map $\pi$ is not etale. The fiber ring at $x$ is the direct sum

F1. $\kappa(x)[t]/(P_{1,x}(t)^{l_1}) \oplus \cdots \oplus \kappa(x)[t]/(P_{n,x}(t)^{l_n})$

Your claim: "Then the fiber over $q$ is just $Spec(\kappa(q)^{\oplus d})$.

Answer: No, formula F1 gives a description of the fiber ring in this case, and it is not a direct sum of $d$ copies of $\kappa(x)$ as you claim.

Rings of integers in number fields: At this link

Tensoring is thought as both restricting and extending?

you will find a discussion of the calculation of the fiber of the canonical map

$\pi: C:=Spec(\mathcal{O}_L)\rightarrow S:=Spec(\mathcal{O}_K)$

for a finite extension of number fields $K\subseteq L$.

Example. If $K=\mathbb{Q}$ is the rational numbers, it follows $\mathcal{O}_L$ is a free $\mathbb{Z}$-module of finite rank (this result is proved in an introductory book on algebraic number theory). Hence the map $\pi$ is finite and locally free. The fiber of $\pi$ at a non-zero prime $(p)$ in $S$ is "ramified" in general. By definition it follows

$\pi^{-1}(p):=\mathbb{Z}/(p)\mathbb{Z} \otimes_{\mathbb{Z}} \mathcal{O}_L \cong $

F1. $\mathcal{O}_L/(p)\mathcal{O}_L \cong \mathcal{O}_L/\mathfrak{p_1}^{l_1}\cdots \mathfrak{p_n}^{l-n} \cong \mathcal{O}_L/\mathfrak{p_1}^{l_1}\oplus \cdots \oplus \mathcal{O}_L/\mathfrak{p_n}^{l_n}:=R_1\oplus \cdots \oplus R_n$.

The decomposition in F1 is proved in any book on algebraic number theory. The rings $R_i$ are Artinian local rings for all $i=1,..,n$ and the schematic fiber $\pi^{-1}((p))$ is the disjoint union

F2. $\pi^{-1}((p))\cong Spec(R_1)\cup \cdots \cup Spec(R_n)$

The integers $l_i\geq 2$ in general and the field extension $\mathbb{Z}/(p)\mathbb{Z} \subseteq \mathcal{O}_L/\mathfrak{p}_i$ is a non trivial finite extension in general.

Hence the fiber is not a finite direct sum of copies of $\mathbb{Z}/(p)\mathbb{Z}$.

We could define the map $\pi$ to be "etale at a point $x\in C$" iff the module of relative Kahler differentials $\Omega(\pi)_{x}$ is zero at $x$.

Note: There is a problem in positive characteristic. If $char(\kappa(x))=p>0$ and $f(x_1,..,x_n)\in \kappa(x)[x_1^p,..,x_n^p]$ it follows the differential

$df:=\sum_i \frac{\partial f}{\partial x_i}dx_i=0$. let $A:=\kappa(x)[x_1,..,x_n]/(f)$. It follows

$\Omega^1_{A/\kappa(x)}\cong A\{dx_1,..,dx_n\}$ is a free $A$-module of rank $n$. Intuitively $Spec(A)$ is a hypersurface in $\mathbb{A}^n_{\kappa(x)}$ and should have dimension $n-1$. Hence to use the module of Kahler differentials to define "etale" in characteristic $p$ can lead to problems.

In Hartshorne, Exercise II.3.5 you prove that any finite morphism $f:X\rightarrow Y$ of schemes is closed, hence the image $f(Z)$ of any closed subscheme $Z\subseteq X$ is closed in $Y$. If the relative cotangent sheaf $\Omega(f):=\Omega^1_{X/Y}$ is zero on an open subscheme $U\subseteq X$, it follows the complement $Z(f):=X-U$ is closed and hence $D(f):=f(Z(f))\subseteq Y$ is a closed subscheme of $Y$. Hence in the case when $f$ is finite and locally free we may in many cases use the relative cotangent sheaf $\Omega(f)$ to construct an open subscheme $V:=Y-D(f)$ with the property that the induced morphism $f_V: \pi^{-1}(V)\rightarrow V$ is etale. Hence intuitively your locally trivial finite morphism $f$ should be etale over an open subscheme $V\subseteq Y$. In Hartshorne III.10.7 there is a "generic smoothness" theorem valid for a morphism $f$ of algebraic varieties over an algebraically closed field of characteristic zero. In this situation "etale" means "smooth of relative dimension $0$".

Your question: "Or is it just the case that every finite locally free morphism is etale?"

Answer: In many cases there should be an open subscheme $V\subseteq Y$ where the induced morphism $f_V:f^{-1}(V) \rightarrow V$ is etale. You must look for a relative version of HH. Corr III.10.7, the "Stacks Project" is a good place to start.

Given any morphism $\pi:X\rightarrow S$ of schemes, one uses the ideal sheaf of the diagonal to define the notion differentially smooth. The morphism $\pi$ is "differentially smooth" iff it is flat, $\Omega^1_{X/S}$ is locally free and the canonical map

$Sym^*_{\mathcal{O}_X}(\Omega_{X/S})\rightarrow \oplus_{i\geq 0} \mathcal{I}^i/\mathcal{I}^{i+1}$

is an isomorphism of $\mathcal{O}_X$-modules. Here $\mathcal{I} \subseteq \mathcal{O}_{X\times_S X}$ is the ideal of the diagonal. It could be this is a useful notion of smoothness in characteristic $p>0$. This notion is mentioned in [EGA, IV4 no. 32]. You find a copy at numdam.org:

http://www.numdam.org/item/PMIHES_1967__32__5_0/

Example. Note that for any monic polynomial $P(t)\in A[t]$ with $B:=A[t]/(P(t))$ it follows for any prime ideal $\mathfrak{p}\subseteq A$ there is an isomorphism $C:=\kappa(\mathfrak{p})\otimes_A B$ \cong $\kappa(\mathfrak{p})[t]/(P_p(t))$, where $P_p(t)$ is the polynomial we get when we reduce the coefficients of $P(t)$ "modulo $\mathfrak{p}$". It follows from Matsumuras book that

$C\otimes_B \Omega^1_{B/A} \cong \Omega^1_{C/\kappa(\mathfrak{p})}\cong C\{dt\} /C\{P_p'(t)dt\}$.

The module $\Omega(\mathfrak{p}):=C\otimes \Omega^1_{B/A}$ is the restriction of $\Omega^1_{B/A}$ to the fiber ring $C$, and if $P_p'(t)=0$ in $\kappa(\mathfrak{p})[t]$ it follows $\Omega(\mathfrak{p})$ does not give information on the polynomial $P_p(t)$. Hence in positive characteristic we must use another definition to define "smooth" and "etale".

Question: "I am struggling to understand what the part about the fibers being etale algebras adds at all. It seems like every finite locally free morphism would trivially satisfy this."

Answer: for any finite extension $K \subseteq L$ of number fields, it follows the morphism $\pi: S:=Spec(\mathcal{O}_L) \rightarrow T:=Spec(\mathcal{O}_K)$ is always integral, locally free (and ramified when $K=\mathbb{Q}$). Hence it is not etale in general. There is an open subscheme $U \subseteq T$ with the property that $\pi_U:\pi^{-1}(U) \rightarrow U$ is etale. The fiber $\pi^{-1}((p))$ in formula F1 above is unramified iff $l_i=1$ for all $i$. The module of Kahler differentials $\Omega:=\Omega^1_{\mathcal{O}_L/\mathcal{O}_K}$ is zero on the open subscheme $\pi^{-1}(U)$. Note moreover that the field extension

$\mathcal{O}_K/\mathfrak{p} \subseteq \mathcal{O}_L/\mathfrak{q}$

is always separable since it is a finite extension of finite fields (here the ideals $\mathfrak{p}, \mathfrak{q}$ are maximal ideals). The ramified prime ideals in $\mathcal{O}_K$ are related to the discriminant of the extension $L/K$ (see in "Algebraic number theory", page 49, by Neukirch). In Neukirch he defines the discriminant using a basis for $L$ over $K$, but it can also be defined using the module of Kahler differentials $\Omega$.

Question: "I noticed that sometimes people are leaving two separate answers for the same question.Sometimes when I answer and leave up to three different methods as the answer, should I be splitting those up into three different answers? Is this being done to pad upvotes, or is it frowned upon or should it be avoided?"

Answer: "It is a matter of understandability. In most cases, I try to combine multiple approaches into one answer. However, if the approaches are long and involved, I will split them into separate answers to reduce confusion and improve readability. I strive to give one answer per question, but on occasion, that is not the best approach."

I agree to this principle, but some users have computers that are slow and where it is difficult to write longer posts (posts with more than 5000 characters) - I believe this forum should allow such users to write multiple posts if this is needed.