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I am trying to understand the definition of étale morphism in Mumford Chapter III Section 5, which I find confusing. I would appreciate any clarifications.

A morphism $f: X \to Y$ of finite type is étale, if for all $x \in X$, there are open neighbourhoods $U \subset X$ of $x$ and $V \subset Y$ of $f(x)$ such that $f(U) \subseteq V$ and such that $f$ restricted to $U$ looks like: $$ \begin{array} &U & \xrightarrow{\text{open immersion}} &\operatorname{Spec}R[X_1, .., X_n]/(f_1, ..., f_n) \\ \downarrow\rlap{\scriptstyle\text{res} \, f} & & \quad\downarrow{} \\ V & \xrightarrow{\phantom{open immersion}} & \operatorname{Spec} R \end{array} $$ where $\det (\partial f_i/ \partial x_j) (x) \neq 0$.

  1. What is the map $V \to \operatorname{Spec} R$? In particular, does this have to be an open immersion as well?

  2. How do I make sense of $\det (\partial f_i/ \partial x_j) (x)$?

Thank you.

Johnny T.
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    In your second question, do you mean to ask how to make sense of the nonvanishing condition? Because the determinant itself is the ordinary Jacobian determinant. – Tabes Bridges Jul 14 '20 at 18:11

1 Answers1

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The intention is that $V=\operatorname{Spec} R$, as you can see by consulting other sources (i.e., Stacks). (Even if it's only an open immersion, you can take an open affine neighborhood of the image of $x$ contained in the affine open and then refine the statement.) The goal here is to define an etale morphism as one that affine-locally looks like a map of schemes induced by an etale morphism of rings.

The condition that $\det(\partial f_i/\partial x_j)\neq 0$ is the condition that the morphism $\operatorname{Spec} R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)\to\operatorname{Spec} R$ is smooth of relative dimension zero at $x$.

There are famously many different ways to formulate the definition etale morphism, and if you're really interested in exploring the concept, you should endeavor to do a little work with a number of these charaterizations. For instance, a lemma a little later on in the first link from Stacks gives 10 equivalent characterizations. This question provides a good number of examples to work with, and the essential intuition is that an etale morphism should look like a map from a covering space in topology - making this precise is interesting for us in algebraic geometry, since our spaces have more structure that we have to think about.

KReiser
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  • Thank you, I will just consider $V = \operatorname{Spec}R$. Is the part $U \to \operatorname{Spec} R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)$ being an open immersion necessary? or is it a similar story as for the $V$? (I was trying to see if I can do this by considering a distinguished open subset of $\operatorname{Spec} R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)$) – Johnny T. Jul 16 '20 at 09:13
  • Yes, we do need the top horizontal map to be an open immersion. Again, keep in mind the goal with this definition: we want to make our map look locally like the map on the right. – KReiser Jul 16 '20 at 09:30
  • Sorry, I think my question was not clear. I meant can we take as a definition that for each $x$ there exist $ x \in U = \operatorname{Spec} R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)$ and $V = \operatorname{Spec} R$ such that $f(U) \subset V$ etc? – Johnny T. Jul 16 '20 at 09:38
  • (instead of identifying $U$ with an open subset of $\operatorname{Spec} R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)$) – Johnny T. Jul 16 '20 at 09:47
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    Yes, this is fine, and it's item (7) in the link to the lemma about the ten conditions in my post. – KReiser Jul 16 '20 at 09:53
  • I was assuming this without any further thoughts: $\operatorname{Spec} R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)\to\operatorname{Spec} R$ is the morphism induced by the natural map $R \to R[X_1,\cdots,X_n]/(f_1,\cdots,f_n)$. It doesn't seem to say anywhere explicitly that this is the case... How do we actually know this? – Johnny T. Jul 24 '20 at 21:40
  • What else would it be? This is frequently referred to as the "natural map" or the "canonical map" and the only reasonable choice is the map which sends $r\mapsto r$. Perhaps the source you're looking at didn't write this as explicitly as they should have, but this is what they meant. – KReiser Jul 24 '20 at 21:49