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Jan 27, 2012 at 9:41 comment added S. Carnahan You just need to define the local ring at the glued spot. If you choose to specify that the intersection is transverse, then you will get a separated irreducible scheme.
Jan 26, 2012 at 17:00 comment added HNuer What do you mean by glue together points? Will this still be separated and irreducible?
Jan 26, 2012 at 0:21 comment added M P Ok, then glue together two distinct points in $\mathbb{A}^2$.
Jan 26, 2012 at 0:10 comment added J.C. Ottem MP: I think he wants an affine example.
Jan 26, 2012 at 0:04 comment added M P Blow up two or more distinct points on a line in $\mathbb{A}^2$.
Jan 25, 2012 at 23:58 history asked HNuer CC BY-SA 3.0