Mannheimer Manuskripte 177 gk-mp-9403/3 SOME CONCEPTS OF ...

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30 MARTIN SCHLICHENMAIERLemma 2. Let X = Spec(R) and {f i } i∈J a set of elements of R then the union ofthe sets X fi equals X if and only if the ideal generated by the f i equals the whole ringR.Proof. The union of the X fi covers Spec(R) iff no prime ideal of R contains all the f i .But every ideal strictly smaller than the whole ring is dominated by a maximal (andhence prime) ideal. The above can only be the case iff the ideal generated by the f i isthe whole ring. □Lemma 3. The affine scheme X = Spec(R) is a quasicompact space. This says everyopen cover of X has a finite subcover.Proof. Let X = ⋃X jj∈Jbe a cover of X. Because the basis open set X f are a basis ofthe topology, every X j can be given as union of X fi . Altogether, we get a refinement ofthe cover X = ⋃ X fi . By Lemma 2 the ideal generated by these f i is the whole ring.i∈IIn particular, 1 is a finite linear combination of the f i . Taking only these f i which occurwith a non-zero coefficient in the linear combination we get (using Lemma 2 again) thatX fik , k = 1,..,r is a finite subcover of X. Taking for every k just one element X jkcontaining X fik we obtain a finite number of sets which is a subcover from the cover westarted with. □Note that this space is not called a compact space because the Hausdorff conditionthat every distinct two points have disjoint open neighbourhoods is obviously notfulfilled.The following proposition says that the sheaf axioms (1) and (2) from App. A for thebasis open sets are fulfilled.Proposition. Let X f be coverd by {X fi } i∈I .(a) Let g,h ∈ R f = O R (X f ) with g = h as elements in R fi = O R (X fi ) for everyi ∈ I, then g = h also in R f .(b) Let g i ∈ R fi be given for all i ∈ I with g i = g j in R fi f j, then there exist ag ∈ R f with g = g i in R fi .Proof. Because X f = Spec(R f ) is again an affine scheme it is enough to show theproposition for R f = R, where R is an arbitrary ring. Let X = ⋃ X fi .(a) Let g,h ∈ R be such that they map to the same element in R fi . This can onlybe the case if in R we havef n ii · (g − h) = 0, ∀i ∈ I,i∈I
CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 31(see the construction of the ring of fractions above). Due to the quasicompactness it isenough to consider finitely many f i , i = 1,..,r. Hence, there is a N such that for everyi the element fiN annulates (g − h). There is another number M, depending on N andr, such that we have for the following ideals(f N 1 ,f N 2 ,... ,f N r ) ⊇ (f 1 ,f 2 ,... ,f r ) M .Because the X fi , i = 1,..,r are a cover of X the ideal on the right side equals (1).Hence, also the ideal on the left. Combining 1 as linear combination of the generatorwe get1 · (g − h) = (c 1 f N 1 + c 2 f N 2 · · · + c r f N r )(g − h) = 0 .This shows (a)(b) Let g i ∈ R fi , i ∈ I be given such that g i = g j in R fi f j. This says there as a N suchthat(f i f j ) N g i = (f i f j ) N g jin R. Note that every g i can be written as g∗ if k iiwith g ∗ i∈ R. Hence, if N is big enoughthe elements f N i g i are in R. Again by the quasicompactness a common N will do it forevery pair (i,j). Using the same arguments as in (a) we get1 = ∑ e i f N i , e i ∈ R .This formula corresponds to a “partition of unity”. We setg = ∑ e i f N i g i .We getf N j g = ∑ if N j e i f N i g i = ∑ ie i f N i f N j g j = f N j g j .This shows g = g j in R fj .□
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Mannheimer Manuskripte 177 gk-mp-9403/3 SOME CONCEPTS OF ...

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