Example for Etale Morphism

algebraic-geometryschemes

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)?

Best Answer

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.