1. Reduced Schemes
Let X be a scheme. We say that X is a reduced scheme if every local ring OX,x is reduced.1
Lemma 1.1. Let F be a sheaf of sets on a topological space X. For every open set U ⊂ X, the map
F (U ) → Y
x∈U
Fx
is injective.
Proof. For each x ∈ U, let ρx: F (U ) → Fx be the map sending a section s of F over U to its stalk at x. Hence the map F (U ) →Q
x∈UFx is given by s 7→ {ρx(s)}x∈U.
Suppose s, s0 ∈ F (U ) so that ρx(s) = ρx(s0) for all x ∈ U. For each x ∈ U, we can find Vx so that s|Vx = s0|Vx. Then {Vx: x ∈ U } forms an open covering of U so that s|Vx = s0|Vx. Since F is a sheaf, s = s0.
Proposition 1.1. A scheme X is reduced if and only if OX(U ) is a reduced ring for all open subset U of X.
Proof. Suppose that X is reduced. Let s be a section of OX(U ), where U is an open subset of X so that sn= 0 for some n > 0. For each u ∈ U, ρu: OX(U ) → OX,ube the map sending a section s over U to its stalk at u. Then ρu(sn) = 0 in OX,u for all u ∈ U. By Lemma 1.1, we know the map OX(U ) → Q
x∈XOX,x sending s → {ρx(s)} is injective. Hence s = 0.
This shows that OX(U ) is reduced.
Conversely, suppose that OX(U ) is a reduced ring for all open subset U of X. Let x ∈ U.
Pick an element in OX,x with a representative (V, f ). Suppose f is nonzero. Then f is not nilpotent in OX(V ). Hence (V, f ) is not equivalent to a nilpotent element in OX,x. Hence OX,x is reduced.
Proposition 1.2. An affine scheme Spec A is reduced if and only if A is reduced.
Proof. Let A be a reduced ring. To show that Spec A is reduced, we only have to show that Ax is reduced for all x ∈ Spec A. Let a/b be an element in Ax. Suppose that (a/b)n= 0 in Ax. Then there exists s ∈ A \ x so that san = 0, then (sa)n = 0. This implies that sa is nilpotent in A. Hence sa = 0 by A being reduced. By definition a/b = (sa)/(sb) = 0 in Ax. This shows that Ax is reduced.
Suppose that X = Spec A is reduced. By Proposition 1.1, for every open set U, OX(U ) is reduced. Let U = X. We obtain that Γ(X, OX) = A is reduced. Proposition 1.3. Let X = Spec A be a scheme. T.F.A.E:
(1) X is reduced.
(2) There exists an affine open covering X =S
iUi such that Γ(Ui, OX) is reduced.
(3) For every affine open U ⊂ X, OX(U ) is reduced.
(4) For every open U ⊂ X, OX(U ) is reduced.
Proof. (1) is equivalent to (2):
Let X be a reduced scheme. Since X is a scheme, X can be covered by affine open sets.
Choose {Ui} so that Ui are affine and {Ui} forms an open covering of X. We know OX(Ui)
1A ring A is reduced if fn= 0 for some n > 0, then f = 0. In other words, A has no nonzero nilpotent elements.
1
2
is reduced for each i by Proposition 1.1. Conversely, if X is covered by {Ui} with Ui affine open. Let Ui= Spec Ai for some ring Ai. If x ∈ Ui, OX,x = (Ai)x. Since Γ(Ui, OX) = Ai is reduced, so is (Ai)x. This shows that OX,x is reduced for all x ∈ X. Hence X is reduced.
(3) is equivalent to (1):
Suppose that X is reduced. Then OX(U ) is reduced for affine open by Proposition 1.1.
Conversely, suppose OX(U ) is reduced for every affine open. Choose an affine open covering {Ui} of X. Then Γ(Ui, OX) is reduced by assumption. By the equivalence of (2) and (1), we find X is reduced.
That (4) is equivalent to (1) is proved in Proposition 1.1.