Algebraic Geometry, University of Zurich, FS 2014 SELECTED SOLUTIONS TO HOMEWORK ASSIGNMENT VI MARTIN GALLAUER ALVES DE SOUZA C- (1) By definition, Xii = Xi and uii = id. It remains to prove ujk uij = uik on the intersection Xij ∩Xik . On the coordinate rings this corresponds to the equalities # # # −1 −1 −1 −1 u# ik (tkl ) := til tik = til tij tik tij =: uij (tjl tjk ) =: uij ujk (tkl ) for each l. (2) The Xi are affine opens of X, their intersections Xij are also affine, and the canonical morphism OX (Xi ) ⊗ OX (Xj ) → OX (Xij ), k[tik , k 6= i] ⊗ k[tjk , k 6= j] −→ k[tik , k 6= i; t−1 ij ] f ⊗ g 7−→ f · u# ij (g), is surjective. Indeed, the choice f = tik , g = 1 yields the variables tik , i 6= k; and f = 1, g = tji yields t−1 ij . # (3) Denote by ϕi the isomorphism tn t0 k[tij , j 6= i] −→ k[ , . . . , ] ti ti tj tij 7−→ . ti Now let p ⊂ k[t0 , . . . , tn ] be a homogeneous prime ideal not containing ti , i. e. a point of D+ (ti ) ⊂ Proj(k[t0 , . . . , tn ]). Then also pti ∩ k[ tt0i , . . . , ttni ] is prime and defines a point in Spec(k[ tt0i , . . . , ttni ]). We know (Lecture notes 2.76) that this yields an identification of affine schemes t0 tn ∼ ∼ D+ (ti ) −→ Spec(k[ , . . . , ]) −→ Spec(k[tij , j 6= i]) = Xi . αi ti ti ϕi Fix j different from i. The first isomorphism αi sends D+ (ti ) ∩ D+ (tj ) to D(tj /ti ), hence ϕi αi sends it to D(tij ) = Xij . It now suffices to prove that on this intersection D+ (ti ) ∩ D+ (tj ), we have the equality uij ϕi αi = ϕj αj . (In fact, we will then have shown more generally that Pn is isomorphic as a scheme to Proj(k[t0 , . . . , tn ]), since they are obtained by glueing the same data.) On the coordinate rings this equality corresponds to the equality for any l: −1 tj # # # tl # −1 αi# ϕ# u (t ) = α ϕ (t t ) = α ( )= jl il i ij i i ij i ti ti −1 ti ti tl # tl αj ( ) = αj# ( ) = αj# ϕ# j (tjl ). tj tj tj tj