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".
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asked Apr 15, 2012 at 13:37
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7$\begingroup$ "Simple" means "smooth" in this context. $\endgroup$– nafCommented Apr 15, 2012 at 14:22
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6$\begingroup$ 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." $\endgroup$– A StasinskiCommented Apr 15, 2012 at 15:11
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$\begingroup$ Why not post this as an answer? $\endgroup$– Martin BrandenburgCommented Apr 15, 2012 at 19:43
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$\begingroup$ 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. $\endgroup$– user5117Commented May 14, 2012 at 20:39
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2 Answers
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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."
answered May 15, 2012 at 15:28
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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.
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1$\begingroup$ As explained in the above comments, "simple" meant something else in this context. This other usage is, however, largely obsolete. $\endgroup$ Commented Apr 15, 2012 at 15:20
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4$\begingroup$ I meant "simple=smooth" is obsolete. $\endgroup$ Commented Apr 15, 2012 at 15:21