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I'm want to understand the concept of etale morphism of schemes using following definition:

A morphism $f: X \to Y$ is etale iff it is * flat *(1), locally of finite type(2) and has the separable field condition(3):

Here (1), (2) and (3) mean:

(1) For every $x \in X$ the induced morphism $f_x ^{\#}:\mathcal{O}_{Y, f(x)} \to \mathcal{O}_X$ is flat

(2) There exist an open affine neighbourhood $U_x =Spec(R)$ of $x$ and an o. a. n. $V_{f(x)}= Spec(S)$ of $f(x)$ with $f(U_x) \subset V_{f(x)}$ such that the induced ring map $S \to R$ is of finite presentation

(3) Let $m_x \subset \mathcal{O}_{X,x}$ the unique maximal ideal of local ring $\mathcal{O}_{X,x}$ and respectively $m_{f(x)} \subset \mathcal{O}_{X,x}$ the unique maximal ideal of $\mathcal{O}_{Y,f(x)}$: Then the induced finite field extension $\mathcal{O}_{Y,f(x)}/m_{f(x)} \to \mathcal{O}_{X,x}/m_x$ is separable

I'm trying to acquire the intuition for etaleness by considering following four examples

(a) $Spec(\mathbb{C}[T, T^{-1}]) \to Spec(\mathbb{C}[T])$

(b) $Spec(\mathbb{C}[T] /(T^d - 2))\to Spec(\mathbb{C}[T])$

(c) $Spec(\mathbb{C}[T, Y]/(Y^d - T)) \to Spec(\mathbb{C}[T])$

(d) $Spec(\mathbb{C}[T, T^{-1},Y]/(Y^d - T)) \to \mathbb{C}[T]$

My attempts:

(a) Is is flat since it just a localization of $\mathbb{C}[T] $ on $T$. Or a secound argument: Open embeddings are etale.

But what about (b), (c) and (d)?

All induced ring maps $\mathbb{C}[T] \to ...$ in (b),(c), (d)$ are quotient maps, therefore of finite presentaions. Since beeing from finite presentations behaves well under base changes / localisations, condition (2) holds.

What about conditions (1) and (3)? Why it suffice to consider in all cases only the localisation on $p= (T)$?

The main problem here for me is to analyze what hapens on stalks, therefore what happens with the localisations of the ring map $\mathbb{C}[T] \to ...$ with respect an arbitrary point /prime ideal $p =(T - \lambda)$ for $\lambda \in \mathbb{C}$. I guess that one can distinguish two relevant cases 1. $\lambda=0$ and 2. $\lambda \neq 0$ arbitrary.

Could anybody explain how to argue exactly for (1) and (3)?

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  • $\begingroup$ @Ravi: Hi. I don't understand why the fact that only at $T=0$ the fiber has only one element imply that the localisation map can't be flat. Why it is flat if it has $d$ fibers? What criterion for flatness are you using? $\endgroup$ – KarlPeter Nov 24 '18 at 19:56
  • $\begingroup$ It seems that my intuition was incorrect so I've removed my comment. You might want to look at the examples on page 7 here: www2.bc.edu/brian-lehmann/papers/flatness.pdf $\endgroup$ – Ravi Nov 24 '18 at 22:29
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Example b) is not even flat, so it is not etale.

Morally, what's happening at $T=0$ in example c) is that the fiber is changing from $d$ distinct points to one point: when $d>1$, this doesn't fit with the idea of etale maps as being like covering maps. (The technical term for this is that the map ramifies at $T=0$ - as etale is equivalent to flat + unramified, this provides one proof that the map in c) is not etale.)

Before performing the computations to show that condition 3) fails for example c), I should note that the condition as written in your post at the time of this answer is not correct. It should instead read as follows:

If $f:X\to Y$ is a morphism of schemes with $y=f(x)$ for some $x\in X, y\in Y$, then $\mathfrak{m}_y\mathcal{O}_{X,x} = \mathfrak{m}_x$ and the field extension $\kappa(y)\subset\kappa(x)$ is separable.

To check flatness, we observe that $\Bbb C[T,Y]/(Y^d-T)$ is a finitely-generated free $\Bbb C[T]$ module with basis $1,Y,Y^2,\cdots,Y^{d-1}$. By computing fiber products, we see that the fiber over $T=\lambda$ is just the spectrum of the ring $R=\Bbb C[T,Y]/(Y^d-T,T-\lambda)$. But $R\cong\Bbb C[Y]/(Y^d-\lambda)$, and when $\lambda\neq0$, $Y^d-\lambda$ splits as a product of linear terms, so $R\cong \Bbb C^d$ and it is immediate to check condition 3). When $\lambda=0$, the condition on maximal ideals fails: $\mathfrak{m}_{y=0}\mathcal{O}_{X,x=0}=(Y^d)$ which is not even maximal.

For example d), the map is etale precisely because the locus of points where 1) or 3) failed was removed.

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  • $\begingroup$ Hi. Thank you for your answer. Two points are still unclear: 1. Concerning case c) that the map ramifies at $T=0$: What criterion for ramifying are you using here that allows to see that this map ramilies at $T=0$ in obvious way? This one en.wikipedia.org/wiki/Glossary_of_algebraic_geometry#unramified or the "naive" one from the theory of Riemann surfaces? Does it suffice here to say that since at $T=0$ the map has (set theoretically) another number of preimages $f^{-1}(T)$ than $d >0$ then it unramified in light of morphism of schemes? $\endgroup$ – KarlPeter Nov 25 '18 at 4:17
  • $\begingroup$ 2. By considering d) you showed that for $y=f(x)$ with $x := x_{\lambda}$ for $\lambda \neq 0$ the schematic fiber is $k(y) \times _Y X := f^{-1}(y) = Spec(R)$ for $R \cong \mathbb{C}^d$. How this concretely provides that $\mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ is flat and $k(y)=\mathcal{O}_{Y,f(x)}/m_{f(x)} \to \mathcal{O}_{X,x}/m_x= k(x)$ is separable? $\endgroup$ – KarlPeter Nov 25 '18 at 4:18
  • $\begingroup$ I know that generaly, if $Spec(R)$ is affine neighbourhood of $x$ and $Spec(S)$ is affine neighbourhood of $f(x) =y$ then the fiber $f^{-1}(y)$ is the spectrum of $S \otimes_R k(y)$, therefore a push out. Futhermore, you have shown that $f^{-1}(y)= Spec(R)=Spec(\mathbb{C}^d)$. We know that the can morphism $k(y) \to R$ is flat und flatness is stable under base change. Can the map $\mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ interpreted as the base change of this? Or did you had another argument in mind concerning the immediate verification of flatness and condition 3) for the case d)? $\endgroup$ – KarlPeter Nov 25 '18 at 4:19
  • $\begingroup$ 1) The definition you linked is correct. The idea of set-theoretic preimages is good intuition, but there are situations where it doesn't capture the correct information without refinement (consider the map of schemes corresponding to $\Bbb Q[x]\hookrightarrow Q(i)[x]$ for instance). 2) The map of residue fields is the identity, which is obviously separable. The local rings are isomorphic (they're regular local rings in varieties of the same dimension) and the map is an isomorphism. 3) The computaton performed in the $T\neq0$ case from example c) applies directly to d) with no alterations. $\endgroup$ – KReiser Nov 25 '18 at 6:13
  • $\begingroup$ Let us abbreviate $X :=Spec(\mathbb{C}[T, Y]/(Y^d - T))$ and $Y := Spec(\mathbb{C}[T])$: $\endgroup$ – KarlPeter Nov 25 '18 at 17:03

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