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Jul 3, 2013 at 13:22 comment added Martin Brandenburg I understand the idea of this definition, but somehow it is cheating. It only says what $I$ should do, but doesn't present $I$ as an object of algebraic geometry in the usual sense. Also, as already noted in the comment, there are even set-theoretical obstructions for this point of view.
Jun 16, 2013 at 15:37 comment added S. Carnahan Since non-empty schemes have lots of geometric points, there is some set-theoretic subtlety in defining $I$. One can always restrict inputs to low levels of $V$ to get a more digestable but slightly less universal object.
Jun 15, 2013 at 23:31 history answered S. Carnahan CC BY-SA 3.0