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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?

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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.)

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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.

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    $\begingroup$ Since posting this answer, you have made 15 edits. This is starting to get to the point of excessive, especially as many of your edits seem like very minor cosmetic edits. Please decide what you are going to say, say it, and then leave well-enough alone. The constant revision is seen as "noise" on the site. $\endgroup$
    – Xander Henderson
    Dec 29, 2020 at 20:46
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    $\begingroup$ Yes, they may seem as minor "cosmetic edits", but I recently had a discussion on "unwritten rules" on this site. From the discussion it seems a post on this site can be deleted because of the "presentation" - this is why I edit an answer if I feel the "presentation" can be improved. Is this a problem? $\endgroup$
    – hm2020
    Dec 30, 2020 at 10:55
  • $\begingroup$ It's not true that any finite extension of number fields is ramified $\endgroup$ Jan 1, 2021 at 16:34
  • $\begingroup$ I believe Theorem III.2.18 in the Neukirch book "Algebraic number theory" proves that all finite extensions of $\mathbb{Q}$ are ramified, hence in this case it holds. $\endgroup$
    – hm2020
    Jan 1, 2021 at 17:01

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