Simple Dehomogenization in a Polynomial Ring
I’ve endeavored to be as general with this example as possible so that it may generalize to the proof of Proposition (Proj is a Scheme). I also wrote this before I wrote those notes, and so the notes might be a little more well-organized. It might be good pedagogy to struggle through this example first because I included some more motivational remarks here, then read those notes, and then come back to this example with a better idea of the structure of the thing.
Consider again. Note that an element of degree at least one, as in Proposition (Proj is a Scheme). We can see that is , and the zero-graded peice of is the set of ratios of homogeneous elements of total degree zero. Consider the map induced by the quotient map; this map is clearly surjective and we wish to show it is injective as well. If , with and , then , because . Then and we are done.
We further wish to show that . Note a point is a relevent homogeneous prime ideal which does not contain ; we wish to define a map given by the rule . Since , this does not give the unit ideal; furthermore (by an appropriate isomorphism theorem) is an integral domain, and thus is a prime ideal of , and so the map is well defined. We wish to show this map is a homeomorphism. First, we demonstrate that it has an inverse given by taking a prime ideal to the ideal generated by the “homogenezation” of elements in , where the homogenezation of an element is given by writing as a sum of homogeneous terms and multiplying each by times the difference between the maximum of the degrees of the terms of and the degree of that term. That is:
This is a well defined map; moreover, following it by the quotient map recovers the original polynomial, and given a homogeneous polynomial, quotienting by and then homogenizing yields the same polynomial. We now show that is a homeomorphism; i.e. and are continuous. First we show for ; it suffices to show that for all ideals of there is a homogeneous ideal such that . But then is just the ideal generated by the homogenezation of every element of , as contains if and only if the ideal generated by the homogenezation of contains the ideal generated by the homogenezation of and does not contain (the if is a little difficult to see; suppose that is an ideal who’s homogenezation contains the ideal generated by the homogenezation of every element of , but does not contain . Then applying gives the intended result, because and are inverses not just on the set of prime ideals but also between and the homogeneous prime ideals of which do not contain .) A similar argument shows is continuous.
Finally, we extend above to an isomorphism on structure sheaves. This is easy; the map defined above extends to the degree zero peice of any localization of , and is the zero degree peice of a localization of for all open . This holds because is the set of all locally constant functions of total degree zero which have denominator not contained in for all , which is the degree zero peice of the localization of by a multiplicatively closed set of functions of degree such that .