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Jun 18, 2010 at 9:22 comment added kaddar Excuse me. Yes, of course, $dim(X)=dim (S)$.
Jun 14, 2010 at 21:20 comment added Qing Liu In the first statement you may want to add dim(X)=dim(S), as closed immersions are finite but rarely open.
Jun 11, 2010 at 20:28 comment added kaddar Read " source" instead "target" below!
Jun 11, 2010 at 20:07 comment added kaddar The (strong) normalization is finite but no open. The weak normalization is open.....
Jun 11, 2010 at 19:53 history answered kaddar CC BY-SA 2.5