A scheme simple over Spec(A)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:46:46Z http://mathoverflow.net/feeds/question/94113 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94113/a-scheme-simple-over-speca A scheme simple over Spec(A)? Hugo Chapdelaine 2012-04-15T13:37:40Z 2012-05-15T15:28:48Z <p>What does it mean to say that a scheme $X$ is <strong>simple</strong> over $Spec(A)$ ?</p> <p>I stumbled on this terminology in a paper of S. Lubkin entitled "On a conjecture of Andre Weil".</p> http://mathoverflow.net/questions/94113/a-scheme-simple-over-speca/94121#94121 Answer by Nick Bagley for A scheme simple over Spec(A)? Nick Bagley 2012-04-15T15:06:08Z 2012-04-16T18:30:31Z <p>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.</p> http://mathoverflow.net/questions/94113/a-scheme-simple-over-speca/97016#97016 Answer by Hugo Chapdelaine for A scheme simple over Spec(A)? Hugo Chapdelaine 2012-05-15T15:28:48Z 2012-05-15T15:28:48Z <p>I have copied A. Stasinsky's comment who quoted a passage in the introduction of SGA1:</p> <p>"/.../ et de faire un ajustage terminologique, le mot <strong>morphisme simple</strong> ayant notamment \'et\'e remplac\'e entre-temps par celui de <strong>morphisme lisse</strong>, qui ne pr\^ete pas aux m\^emes confusions."</p>