Suppose that X is irreducible

1. Irreducible Scheme

A topological space X is irreducible if X is nonempty and can not be a union of two proper closed subsets. A nonempty closed subset Z of X is irreducible if Z equipped with the induced topology from X is an irreducible topological space. A point ξ of an irreducible closed subset Z of X is called a generic point if Z = {ξ}.

Proposition 1.1. A topological space X is irreducible if and only if the intersection of any two nonempty open subsets of X is nonempty.

Proof. Suppose that X is irreducible. Assume that there exist two nonempty open subsets U1, U2 so that U1 ∩ U₂ = ∅. Denote Zi = X \ Ui for i = 1, 2. Then Z1 and Z2 are proper closed subsets of X. Since U₁∩ U₂ = ∅, Z₁∪ Z₂ = X. This leads to a contradiction that X is irreducible. Hence any two nonempty open subsets of X is nonempty.

Conversely, suppose that the intersection of any two nonempty open subsets of X is nonempty. Assume that X = Z₁∪ Z₂ is a union of closed subsets of X. Denote U_i = X \ Z_i for i = 1, 2. Then U1∩ U₂= ∅. This shows that either U1 = ∅ or U2= ∅. This is equivalent to say that Z₁ = X or Z₂ = X. Hence X is irreducible.

 Definition 1.1. A scheme is irreducible if its underlying topological space is irreducible.

Lemma 1.1. Let X = Spec A be the spectrum of a ring A and D(f ) be the distinguished open set associated with f ∈ A. Then D(f ) = ∅ if and only if f is nilpotent.

Proof. Let f be nilpotent. Then fⁿ = 0 for some n > 0. In other words, fⁿ ∈ p for all p ∈ X. Since p is prime, f ∈ p. We see that f ∈ p for all p ∈ X. Hence p ∈ V (f ) for all p∈ X. We obtain X = V (f ). Hence D(f ) = ∅.

Conversely, suppose D(f ) = ∅. Then X = V (f ). This shows that f ∈ p for all p. Hence f ∈T

p∈Xp= Nil(A). (The nilradical of A.) Hence f is nilpotent.

 Proposition 1.2. Let X = Spec A be the spectrum of a ring A. Then X is irreducible if and only if its nil radical Nil(A) is a prime ideal.

Proof. Let us assume that ξ = Nil(A) is a prime ideal. Then the Nil(A) is the intersection of all prime ideals of A. Hence p ∈ V (ξ) for all p ∈ X. Hence X = V (ξ). Notice that V (ξ) is a closed subset of X containing ξ. Then V (ξ) contains the closure of ξ. Moreover, if V (I) is a closed subset of X containing ξ, then ξ ⊃ I. Thus V (ξ) ⊂ V (I). This implies that V (ξ) is in fact the closure of ξ. In other words, ξ is the generic point. Hence all nonempty open subsets of X contain ξ. Hence the intersection of any two nonempty open subsets of X contains ξ.

Suppose that X is irreducible. Let f g ∈ Nil(A). To show that Nil(A) is a prime, we need to show that either f ∈ Nil(A) or g ∈ Nil(A). Then D(f g) = D(f ) ∩ D(g). If both D(f ) and D(g) are nonempty, then f and g are not nilpotent. Since X is irreducible and D(f ) and D(g) are nonempty open subsets of X, D(f ) ∩ D(g) = D(f g) is nonempty. This implies that f g is not nilpotent, i.e. f g 6∈ Nil(A). This leads to the contradiction to the assumption that f g ∈ Nil(A). Thus either D(f ) = ∅ or D(g) = ∅. Hence either f or g is nilpotent, i.e.

f ∈ Nil(A) or g ∈ Nil(A). 

Proposition 1.3. Let X = Spec A. For p ∈ Spec A, V (p) is an irreducible closed subset.

Conversely, every irreducible closed subset of X is of the form V (p) for some p ∈ X.

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Proof. Let Z = V (I) be a nonempty closed subset of X where I is a radical ideal of X.

Suppose I is not prime. Then we can choose f, g ∈ A so that f g ∈ I but f, g 6∈ I. In this case, we consider (f ) + I and (g) + I. Then V ((f ) + I) and V ((g) + I) are proper closed subsets of V (I). Moreover,

V ((f ) + I) ∪ V ((g) + I) = V ((f g) + I) = V (I) = Z.

This implies that Z is not irreducible.

If Z is not irreducible, Z is a union of two proper closed subsets V (I₁) and V (I₂). Assume that I₁ and I₂ are radical ideals of A. Then I₁ ⊃ I and I₂ ⊃ I. Choose f_i ∈ I_i\ I. Then (fi) + I ⊃ I for i = 1, 2. Then V ((fi) + I) ⊃ V (Ii). We see that

V ((f₁f₂) + I) = V ((f₁) + I) ∪ V ((f₂) + I) = V (I).

Thenp(f1f2) + I = I. Since (f1f2) ⊂ (f1f2) + I ⊂p(f1f2) + I, we see that f1f2 ∈ I. This

shows that I is not a prime. 

If p is a prime ideal of a ring A, then V (p) is the closure of p in Spec A.

Proposition 1.4. Let X be a scheme. Suppose X is irreducible. Then X has a unique generic point.

Proof. Suppose X = Spec A. We have seen that ξ = Nil(A) is the unique generic point.

Since X is irreducible, any affine open subset U of X is also irreducible. Suppose X (not necessarily affine) has two generic points ξ and η. Then ξ and η are also generic points of U of X. Since U is affine, ξ = η by the uniqueness of the generic point of an affine scheme. Hence we proved the uniqueness of generic point of a scheme. Now let us prove the existence.

The affine open subsets of X forms a basis for the Zariski topology of X. Since X is irreducible, the intersection of any two nonempty affine open subsets of X are nonempty.

Hence there exists a nonempty affine open set contained in the intersection of any two nonempty affine open subsets of X. Let U, V be nonempty affine open sets. Choose ξ and η so that ξ and η are generic point of U and V respectively. Choose W nonempty affine open so that W ⊂ U ∩ V. Then we know both ξ and η are generic point of W. By uniqueness of the generic point of W, ξ = η. This proves that all the affine open subsets of X share the same generic point ξ. Now let us prove that ξ is the generic point of X.

Let x be any point of X and U be any open neighborhood of x. Since X is a scheme, we can choose an affine open neighborhood V of x contained in U. Since ξ is a generic point of V, U contains ξ. Hence x lies in the closure of {ξ}. Thus X = {ξ}.

 Proposition 1.5. Let X be a scheme. Any irreducible closed subset of X has a unique generic point.

Proof. Let Z be an irreducible closed subset of X. For any affine open subset U of X, we consider the intersection U ∩ Z. Either U ∩ Z is empty or U ∩ Z is nonempty. Suppose that U ∩ Z is nonempty. Then U ∩ Z is an irreducible closed subset of U. Write U = Spec A.

Then U ∩ Z = V (I) for some ideal I. Since U ∩ Z is irreducible, I = p is a prime ideal.

Then V (p) is the closure of p in U ∩ Z. Hence p is the generic point of U ∩ Z. Let ξ be the generic point of U ∩ Z. Similar we can show that U ∩ Z share the same generic point ξ for all affine open subset U of X. Using the fact that affine open subsets forms a base for the Zariski topology of X, ξ is the generic point of Z. (The detailed proof of these statements is the same as that of the above proposition.)



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Proposition 1.6. Let X be a scheme. T.F.A.E.

(1) X is irreducible.

(2) There exists an affine open covering {Ui} so that U_i is irreducible and Ui∩ U_j 6= ∅ for all i, j.

(3) X is nonempty and every nonempty affine open U ⊂ X is irreducible.

Proof. (1) implies (2). Since X is irreducible, it has a unique generic point ξ. Since X is a scheme, we can choose an affine open covering {U_i} for X so that U_i are all nonempty.

Since Ui is nonempty, Ui contains the generic point ξ for all i. This implies that Ui∩ U_j contains ξ and hence is nonempty. Moreover any subspace of irreducible topological space is also irreducible. Hence Ui is irreducible. Similarly, (1) implies (3).

(2) implies (1). Let X = Z₁ ∪ Z₂ where Z₁ and Z₂ are closed subsets of X. For any U_i, either U_i∩ Z₁ = ∅ or U_i∩ Z₂ = ∅. Hence either U_i ⊂ Z₁ or U_i ⊂ Z₂. We may assume that there exists Ui so that Ui ⊂ Z₁. Since Ui is irreducible, Ui∩ Z₁ = Ui is an irreducible closed subset of U_i. Then U_i∩ Z₁ has a unique generic point ξ. For any U_j, we know that Ui ∩ U_j is nonempty by assumption and hence contains ξ. Then Ui∩ U_j is densed in Uj, and contained in Z₁ ∩ U_j. Notice that Z₁ ∩ U_j is a closed set containing ξ, then Z₁∩ U_j contains the closure of ξ in Uj. Since Ui∩ U_j is dense in Uj, the closure of ξ in Uj is Uj. Hence Uj ⊂ Z₁∩ U_j ⊂ U_j. We conclude that Uj ⊂ Z₁. Hence X =S

iUi⊂ Z₁ which implies that X = Z₁. Thus X is irreducible.

(3) implies (2). Since X is nonempty, we can choose an affine open covering {Ui} of X so that U_i are all nonempty. By assumption, U_i are all irreducible. Now we want to show that Ui∩ U_j 6= ∅. Suppose U_i∩ U_j is an empty set. Since Ui and Uj are affine, so is Ui` U<sub>j</sub>. Since both U<sub>i</sub> and U<sub>j</sub> are nonempty, U<sub>i</sub>` U_j is also nonempty. Since every affine open subset of X is irreducible, Ui` U_j is irreducible but this is impossible. Hence Ui∩ U_j is nonempty for any i, j.