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What does it mean to say that a scheme $X$ is simple over $Spec(A)$ ?

I stumbled on this terminology in a paper of S. Lubkin entitled "On a conjecture of Andre Weil".

flag	asked Apr 15 2012 at 13:37 Hugo Chapdelaine 3,337●3●18

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"Simple" means "smooth" in this context. – ulrich Apr 15 2012 at 14:22

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From the Introduction in SGA1: "/.../ et de faire un ajustage terminologique, le mot \og morphisme simple\fg ayant notamment \'et\'e remplac\'e entre-temps par celui de \og morphisme lisse\fg, qui ne pr\^ete pas aux m\^emes confusions." – A Stasinski Apr 15 2012 at 15:11

Ok thanks for the explanation. – Hugo Chapdelaine Apr 15 2012 at 16:56

Why not post this as an answer? – Martin Brandenburg Apr 15 2012 at 19:43

This question just floated back to the surface; it would be good if somebody wrote up A Stasinski's comment as an answer, to stop that happening repeatedly. – Artie Prendergast-Smith May 14 2012 at 20:39

Nick Bagley · Answer 1 · 2012-04-16 18:30:31Z UTC

Because a scheme is a locally ringed space X, then it is simple if its topology does not contain a nontrivial two sided ideal. As Spec(A) is referring to the spectrum, this scheme is simple if the set of all proper prime ideals of the noncommutative ring A does not contain any nontrivial two sided ideals defined by x $\cdot$ r $\in$ I, r $\cdot$ x $\in$ I, if the set of ideals (I,+) is a subgroup of an additive group (R,+). In a commutative ring, this is true for all ideals in I.

Hugo Chapdelaine · Accepted Answer · 2012-05-15 15:28:48Z UTC

I have copied A. Stasinsky's comment who quoted a passage in the introduction of SGA1:

"/.../ et de faire un ajustage terminologique, le mot morphisme simple ayant notamment \'et\'e remplac\'e entre-temps par celui de morphisme lisse, qui ne pr\^ete pas aux m\^emes confusions."

A scheme simple over Spec(A)?

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