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Jun 9, 2013 at 12:22 vote accept Daniel Barter
Jun 5, 2013 at 1:26 comment added François Brunault See the answers to mathoverflow.net/questions/41624/…
Jun 5, 2013 at 1:04 history edited Daniel Barter CC BY-SA 3.0
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Jun 5, 2013 at 0:57 comment added Allen Knutson A ring $R$ is $\mathbb Z$-graded iff it carries an action of the multiplicative group (proof: define $R_i$ to be the subspace where $z\cdot$ acts by $z^i$). Then, that group will also act on ${\rm Spec}\ R$. Then for the first question, you need to look past the obvious inclusion $R_0 \to R$ to the less obvious quotient $R \to R_0$. This only works because $R$ is $\mathbb N$-graded, not just $\mathbb Z$-graded.
Jun 5, 2013 at 0:51 history edited Daniel Barter CC BY-SA 3.0
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Jun 4, 2013 at 21:59 answer added David Mahone timeline score: 1
Jun 4, 2013 at 21:46 answer added pinaki timeline score: 4
Jun 4, 2013 at 21:37 comment added Daniel Barter Hmm this sounds interesting as it is much closer to the "classical" definition of projective space. Two thinks I am a little unsure about though: How is $ {\rm spec} R_0 $ closed subset of $ {\rm spec } R $? How do you mod out by scaling?
Jun 4, 2013 at 21:28 comment added Allen Knutson This is more about $\rm Proj$ than about its structure sheaf, but I much prefer the definition ${\rm Proj}\ R = ({\rm Spec}\ R \setminus {\rm Spec} R_0)/$scaling to the gluing construction. (And in fact, I essentially never glue schemes together along open subschemes.)
Jun 4, 2013 at 21:00 history asked Daniel Barter CC BY-SA 3.0