aa r X i v : . [ m a t h . AG ] A p r A Class of Noncommutative Spectra
Keqin LiuDepartment of MathematicsThe University of British ColumbiaVancouver, BCCanada, V6T 1Z2March, 2011
Abstract
We construct a class of noncommutative spectra and give the basicproperties of the class of noncommutative spectra.
How to select a class of noncommutative rings such that we can rewritealgebraic geometry in the context of the class of noncommutative rings? Thepurpose of this paper is to present our answer to this interesting problem. Themain result of this paper is to construct a class of noncommutative spectra whichare called trispectra. Trispectra come from a class of nonncommutative ringswhich are called Hu-Liu trirings by us. In section 1, we introduce Hu-Liu triringsand give the basic properties of Hu-Liu trirings. The most important example ofHu-Liu trirings is triquaternions which is defined in section 1. Triquaternions,which are regarded as a kind of new numbers by us, can be used to replacecomplex numbers to develop the counterpart of complex algebraic geometry. Insection 2, we introduce prime triideals and prove some basic facts about primetriideals. In section 3, we use prime triideals to characterize the trinilradical.In the last section of this paper, we make trispectra into a topological space byintroducing extended Zariski topology.Throughout this paper, the word “ring” means an associative ring with anidentity. A ring R is also denoted by ( R, + , · ) to indicate that + is the additionand · is the multiplication in the ring R . The word “triring” always means“Hu-Liu triring”. Let A and B be two subsets of a ring ( R, + , · ). We shall use A + B and AB to denote the following subsets of RA + B := { a + b | a ∈ A, b ∈ B } , AB := { ab | a ∈ A, b ∈ B } . Definition 1.1
A ring R with a multiplication · is called a Hu-Liu triring ifthe following three properties hold. (i)
There exist two commutative subrings R and R of the ring ( R , + , · ) , calledthe even part and odd part of R respectively, such that R = R ⊕ R (as Abelian groups) and R R ⊆ R , R R + R R ⊆ R , R R = 0; (1) (ii) There exists a binary operation ♯ on the odd part R such that ( R , + , ♯ ) is a commutative ring and the two associative products · and ♯ satisfy the triassociative law : x ( α ♯ β ) = ( xα ) ♯ β, (2)( α ♯ β ) x = α ♯ ( βx ) , (3) where x ∈ R and α , β ∈ R ; (iii) For each x ∈ R , we have R x = x R . (4)A Hu-Liu triring R = R ⊕ R is sometimes denoted by( R = R ⊕ R , + , · , ♯ ), where the associative product ♯ on the odd part R iscalled the local product , and the identity 1 ♯ of the ring ( R , + , ♯ ) is calledthe local identity of the triring R .Clearly, a commutative ring is a Hu-Liu triring with zero odd part. The firstand the most important example of Hu-Liu trirings which is not a commutativering is triquaternions whose definition is given in the following example. Example
Let Q = R ⊕ R i ⊕ R j ⊕ R k be a 4-dimensional real vector space,where R is the field of real numbers. Then ( Q = Q ⊕ Q , + , · , ♯ ) is a Hu-liutriring, where Q = R ⊕ R i is the even part, Q = R j ⊕ R k is the odd part,the ring multiplication · and the local product ♯ are defined by the followingmultiplication tables: · i j k i j ki i − k − jj j − k k k j ♯ j kj j kk k − j Q is called the triquaternions . Remark
For convenience, a Hu-Liu triring will be simply called a triring inthe remaining part of this paper.
Definition 1.2
Let ( R = R ⊕ R , + , · , ♯ ) be a triring, and let I be a subgroupof the additive group of R . (i) I is called a triideal of R if IR + RI ⊆ I , I = ( R ∩ I ) ⊕ ( I ∩ R ) , and I ∩ R is an ideal of the ring ( R , + , ♯ ) . (ii) I is called a subtriring of R if ∈ I , II ⊆ I , I = ( R ∩ I ) ⊕ ( I ∩ R ) and I ∩ R is a subring of the ring ( R , + , ♯ ) . Let I be a triideal of a triring ( R = R ⊕ R , + , · , ♯ ). It is clear that RI = (cid:18) RI (cid:19) ⊕ (cid:18) RI (cid:19) with (cid:18) RI (cid:19) i := R i + II for i = 0 and 1. We now define alocal product on (cid:18) RI (cid:19) by( α + I ) ♯ ( β + I ) := α ♯ β + I for α , β ∈ R . (5)Then the local product defined by (5) is well-defined, and the triassociative lawholds. Therefore, RI becomes a triring, which is called the quotient triring of R with respect to the triideal I . Definition 1.3
Let R = R ⊕ R and R = R ⊕ R be trirings. A map φ : R → R is called a triring homomorphism if φ ( x + y ) = φ ( x ) + φ ( y ) , φ ( xy ) = φ ( x ) φ ( y ) , φ (1 R ) = 1 R ,φ ( R ) ⊆ R , φ ( R ) ⊆ R ,φ ( α ♯ β ) = φ ( α ) ♯ φ ( β ) , φ (1 ♯ ) = ¯1 ♯ where x , y ∈ R , α , β ∈ R , R and R are the identities of R and R respectively,and ♯ and ¯1 ♯ are the local identities of R and R respectively. A bijective triringhomomorphism is called a triring isomorphism . We shall use R ≃ R toindicate that there exists a triring isomorphism from R to R . Let φ be a triring homomorphism from a triring R to a triring R . the kernel Kerφ and the image
Imφ of φ are defined by Kerφ := { x | a ∈ R and φ ( x ) = 0 } and Imφ := { φ ( x ) | x ∈ R } . Kerφ is a triideal of R , Imφ is a subtriring of R and φ : x + Kerφ → φ ( x ) for x ∈ R is a triring isomorphism from the quotient triring RKerφ to the subtriring
Imφ of R . If I is triideal of R , then ν : x x + I for x ∈ R is a surjective triring homomorphism from R the quotient tiring RI with kernel I . The map ν is called the natural triring homomorphism . Proposition 1.1
Let φ be a surjective homomorphism from a triring R = R ⊕ R to a triring R = R ⊕ R . (i) φ ( R i ) = R i for i = 0 and . (ii) Let S := { I | I is a triideal of R and I ⊇ Kerφ } and S := { I | I is a triideal of R } . The map
Ψ : I φ ( I ) := { φ ( x ) | x ∈ I } is a bijection from S to S , and the inverse map Ψ − : S → S is given by Ψ − : I φ − ( I ) := { x | x ∈ R and φ ( x ) ∈ I } . (iii) If I is a triideal of R containing Kerφ , then the map x + I φ ( x ) + φ ( I ) for x ∈ R is a triring isomorphism from the quotient triring RI onto the quotienttriring Rφ ( I ) . Proof
A routine check.We now prove a basic property for trirings.
Proposition 1.2
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. If x i ∈ R i for i ∈ { , } , then both Rx = R x ⊕ R x and R ♯ x are triideals of R . roof Since ( R , + , ♯ ) is a commutative ring, R ♯ x is clearly a triideal of R .Using the triassociative law and (4), we have (cid:16) R i ( R x + R x ) (cid:17) [ (cid:16) R x + R x ) R i (cid:17) ⊆ R x + R x (6)for i = 0, 1. By (6), R x ⊕ R x is an ideal of the ring ( R, + , · ). Also, R x = R ♯ (1 ♯ x ) is obviously an ideal of the commutative ring ( R , + , ♯ ).Thus, R x ⊕ R x is a triideal of R .The next position gives some operations about triideals in a triring. Proposition 1.3
Let I , J , I λ with λ ∈ Λ be triideals of a triring R . (i) The intersection I ∩ J and the sum X λ ∈ Λ I λ are triideals of R .Moreover, wehave ( I ∩ J ) i = I i ∩ J i and X λ ∈ Λ I λ ! i = X λ ∈ Λ ( I λ ) i for i = 0 , . (ii) The mixed product I · ♯ J := ( I · ♯ J ) ⊕ ( I · ♯ J ) of I and J is a triideal,where ( I · ♯ J ) = I J and ( I · ♯ J ) = I ♯ J . Proof
Clear.
We begin this section by introducing the notion of prime triideals.
Definition 2.1
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. An triideal P = P ⊕ P of R is called a prime triideal if P = R and x y ∈ P ⇒ x ∈ P or y ∈ P , (7) x y ∈ P ⇒ x ∈ P or y ∈ P , (8) x y ∈ P ⇒ x ∈ P or y ∈ P , (9) x ♯ y ∈ P ⇒ x ∈ P or y ∈ P , (10) where x i , y i ∈ R i for i = 0 and . Let ( R = R ⊕ R , + , · ♯ ) be a triring. The set of all prime triideals of R is called the trispectrum of R and denoted by Spec ♯ R . It is clear that Spec ♯ R = Spec ♯ R ∪ Spec ♯ R and Spec ♯ R ∩ Spec ♯ R = ∅ , Spec ♯ R := { P | P ∈ Spec ♯ R and P ⊇ R } is called the even trispectrum of R and Spec ♯ R := { P | P ∈ Spec ♯ R and P R } is called the odd trispectrum of R .Let P = P ⊕ P be a triideal of a triring R . Then P ∈ Spec ♯ R if and onlyif P is a prime ideal of the commutative ring ( R , + , · ). It is also obviousthat P ∈ Spec ♯ R if and only if P is a prime ideal of the commutative ring( R , + , · ), P is a prime ideal of the commutative ring ( R , + , ♯ ), and the (cid:18) R P , R P (cid:19) -bimodule RP is faithful as both left module and right module, wherethe left R P -module action on RP is defined by( x + P )( y + P ) := x y + P for x ∈ P and y ∈ R and the right R P -module action on RP is defined by( y + P )( x + P ) := yx + P for x ∈ P and y ∈ R. Clearly, the even trispectrum
Spec ♯ R of a triring R is not empty. A basicproperty of trirings is that the odd trispectrum Spec ♯ R is always not emptyprovided R = 0. This basic fact is a corollary of the following Proposition 2.1
Let ( R = R ⊕ R , + , · ♯ ) be a triring with R = 0 . If P is a prime ideal of the commutative ring ( R , + , ♯ ) , then there exists primeideal P of the commutative ring ( R , + , · ) such that P := P ⊕ P is a primetriideal of R , and P contains every ideal I of the ring ( R , + , · ) which hasthe property: R I ⊆ P . Proof
Consider the set Ω defined byΩ := { I | I is an ideal of ( R , + , · ) and R I ⊆ P } . Clearly, 1 I if I ∈ Ω. Since 0 ∈ Ω, Ω is not empty. The relation ofinclusion, ⊆ , is a partial order on Ω. Let ∆ be a non-empty totally orderedsubset of Ω. Let J := [ I ∈ ∆ I . Then J ∈ Ω. Thus J is an upper bound for∆ in Ω. By Zorn’s Lemma, the partial order set (Ω , ⊆ ) has a maximal element P . We are going to prove that P := P ⊕ P is a prime triideal of R . Clearly, P = P ⊕ P is a triideal satisfying (10). Let x i , y i ∈ R i for i = 0, 1.6f x y ∈ P and x P , then x ♯ (1 ♯ y ) = x y ∈ P , which implies that1 ♯ y ∈ P . Hence, we have R ( P + R y ) ⊆ R P + R R y ⊆ P + R y = P + R ♯ (1 ♯ y ) ⊆ P + R ♯ P ⊆ P . (11)Using (11) and the fact that P + R y is an ideal of R , we get P + R y ∈ Ω.Since P + R y ⊇ P , we have to have P + R y = P , which implies that y ∈ P . This proves that (9) holds. Similarly, we have that both (7) and (8)hold. Hence, P = P ⊕ P is a prime triideal of R .The following proposition gives another characterization of prime triideals. Proposition 2.2
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. The following areequivalent. (i) P is a prime triideal. (ii) For two triideals I , J of R , I · ♯ J ⊆ P implies that I ⊆ P or J ⊆ P . Proof
Use By Proposition 1.2.
Let R = R ⊕ R be a triring with the local identity 1 ♯ . For α ∈ R , the local n th power α ♯n is defined by: α ♯n := ♯ , if n = 0; α ♯ α ♯ · · · ♯ α | {z } n , if n is a positive integer.The products ( x m )( α ♯n ) and ( α ♯n )( x m ) will be denoted by x m α ♯n and α ♯n x m respectively, where x ∈ R and α ∈ R . Proposition 3.1
Let ( R = R ⊕ R , + , · , ♯ ) be a triring with the local identity ♯ . If x , y ∈ R , α , β ∈ R and m ∈ Z > , then ( xα ) ♯ ( yβ ) = ( xy )( α ♯ β ) , ( αx ) ♯ ( βy ) = ( α ♯ β ) xy (12) and ( xα ) ♯m = x m α ♯m , ( αx ) ♯m = α ♯m x m . (13)7 roof By the triassociative law, we have( xα ) ♯ ( yβ ) = x ( α ♯ ( yβ )) = x (( yβ ) ♯ α ) = ( xy )( β ♯ α ) = ( xy )( α ♯ β )and ( αx ) ♯ ( βy ) = (( αx ) ♯ β )) y = (( β ) ♯ ( αx )) y = ( β ♯ α ) xy = ( α ♯ β ) xy. Hence, (12) holds. Clearly, (13) follows from (12).
Definition 3.1
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. (i) An element x of R is said to be trinilpotent if x m = 0 and x ♯ n = 0 for some m , n ∈ Z > , (14) where x and x are the even component and the old component of x respectively. (ii) The set of all trinilpotent elements of R is called the trinilradical of R and denoted by nilrad ♯ ( R ) or ♯ √ . The ordinary nilradical of a ring ( A, + , · ) is denoted by nilrad ( A ) or nilrad ( A, + , · ); that is, nilrad ( A ) := { x | x m = 0 for some m ∈ Z > } . If ( R = R ⊕ R , + , · , ♯ ) is a triring, then nilrad ( R, + , · ) = nilrad ( R , + , · ) ⊕ R and nilrad ♯ R = nilrad ( R , + , · ) ⊕ nilrad ( R , + , ♯ ) . (15)Hence, nilrad ( R, + , · ) ⊇ nilrad ♯ R , i.e., the trinilradical of a triring R is smallerthan the ordinary nilradical of the ring ( R, + , · ). Proposition 3.2
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. (i) The trinilradical nilrad ♯ ( R ) is a triideal of R . (ii) nilrad ♯ (cid:18) Rnilrad ♯ ( R ) (cid:19) = 0 . Proof (i) By the definition of trinilradicals, we have( nilrad ♯ R ) ∩ R = nilrad ( R , + , · ) and ( nilrad ♯ R ) ∩ R = nilrad ( R , + , ♯ ) . { xa, ax } ⊆ nilrad ♯ R for x ∈ R and a ∈ nilrad ♯ R. (16)Let x = x + x ∈ R and a = a + a ∈ nilrad ♯ R , where x i , a i ∈ R i for i = 0, 1. Then we have a m = a ♯ m = 0 for some m ∈ Z > . Since xa = ( x + x )( a + a ) = x a + ( x a + x a ) , (17)( x a ) m = x m a m = x m , ( x a ) ♯m = x m a ♯m = x m , ( x a ) ♯m = ( x a ) ♯m = x ♯m a m = x ♯m , we get x a ∈ nilrad ( R , + , · ) and x a + x a ∈ nilrad ♯ ( R , + , ♯ ) . (18)It follows from (17) and (18) that xa ∈ nilrad ♯ R . Similarly, we have ax ∈ nilrad ♯ R . This proves (i).(ii) follows from (i).If I is a triideal of a triring R = R ⊕ R , then the trinilradical ♯ √ I of I isdefined by ♯ √ I := { x ∈ R | x m ∈ I and x ♯ n ∈ I for some m , n ∈ Z > } , where x = x + x , x i ∈ R i and I i = I ∩ R i for i = 0, 1. Since nilrad ♯ (cid:18) RI (cid:19) = ♯ √ II , ♯ √ I is a triideal of R . A triideal I of a triring R is called a radical triideal if ♯ √ I = I .We now characterize the trinilradical of a triring by using prime triideals. Proposition 3.3
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. The trinilradical of R is the intersection of the prime triideals of R . Proof
Let x = x + x be any element of nilrad ♯ ( R ), where x ∈ R and x ∈ R . Then x m = 0 and x ♯ n = 0 for some m , n ∈ Z > . Let P = P ⊕ P beany prime triideal of R . Since x m = 0 ∈ P , we have x ∈ P (19)by (7). 9f P R , then P is a prime ideal of the commutative ring ( R , + , ♯ ).Using this fact and x ♯ n = 0, we get x ∈ P . (20)If P ⊇ R , then (20) is obviously true. It follows from (19) and (20) that x = x + x ∈ P . This proves that nilrad ♯ ( R ) ⊆ \ P ∈ Spec ♯ R P. (21)Conversely, we prove that z nilrad ♯ ( R ) ⇒ z \ P ∈ Spec ♯ R P. (22) Case 1: z m = 0 for all m ∈ Z > , in which case, z m R for all m ∈ Z > .Hence, z + R is not a nilpotent element of the commutative ring RR ; that is, z + R = nilrad (cid:18) RR (cid:19) = \ IR ∈ Spec (cid:0) RR (cid:1) (cid:18) IR (cid:19) , where Spec (cid:18) RR (cid:19) is the ordinary spectrum of the commutative ring RR . Hence,there exists a prime ideal IR of the commutative ring RR such that z I . Since I is a prime triideal of R , (22) holds in this case. Case 2: z m = 0 for some m ∈ Z > . Let z = z + z with z ∈ R and z ∈ R . Then 0 = z m = z m + z m − r for some r ∈ R by (4). Thus, z m = 0,which implies that z ♯ n = 0 for all n ∈ Z > in this case. We now consider thefollowing set T := n J (cid:12)(cid:12)(cid:12) J is a triideal of R and z ♯ n J for all n ∈ Z > o . Since { } ∈ T , T is nonempty. Clearly, ( T, ⊆ ) is a partially order set, where ⊆ is the relation of set inclusion. If { J λ | λ ∈ Λ } is a nonempty totally orderedsubset of T , then S λ ∈ Λ J λ is an upper bound of { J λ | λ ∈ Λ } in T . By Zorn’sLemma, the partially ordered set ( T, ⊆ ) has a maximal element P . We aregoing to prove that P is a prime triideal of R .Let x = x + x and y = y + y be two elements of R , where x i , y i ∈ R i for i = 0, 1. First, if x P and y P , then P ⊂ P + Rx and P ⊂ P + Ry . (23)10ince both P + Rx and P + Ry are triideals of R by Proposition 1.2, (23)implies that z ♯ u ∈ P + Rx and z ♯ v ∈ P + Ry or z ♯ u ∈ P + R x and z ♯ v ∈ P + R y for some u , v ∈ Z > .Thus, we have z ♯ ( u + v )1 = z ♯ u ♯ z ♯ v ∈ ( P + R x ) ♯ ( P + R y ) ⊆ P ♯ P + P ♯ ( R y ) + ( R x ) ♯ P | {z } This is a subset of P +( R x ) ♯ ( R y ) ⊆ P + R x y , which implies that x y P . This proves that x P and y P ⇒ x y P . (24)Similarly, we have x P and y P ⇒ x y P , (25) y P and x P ⇒ y x P (26)and x P and y P ⇒ x y P . (27)By (24), (25), (26) and (27), P is a prime triideal. Since z P , (22) alsoholds in Case 2.It follows from (21) and (22) that Proposition 3.3 is true.The next proposition is a corollary of Proposition 3.3. Proposition 3.4 If I is a triideal of a Hu-Liu triring R and I = R , then ♯ √ I = \ P ∈ Spec ♯ R and P ⊇ I P.
Proof
By Proposition 3.3, we have x ∈ ♯ √ I ⇔ x + I ∈ nilrad ♯ (cid:18) RI (cid:19) = \ PI ∈ Spec ♯ (cid:0) RI (cid:1) PI ⇔ x ∈ \ P ∈ Spec ♯ R and P ⊇ I P. Extended Zariski Topology
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. For a triideal I of R , we define asubset V ( I ) of Spec ♯ R by V ♯ ( I ) := { P | P ∈ Spec ♯ R and P ⊇ I } . (28) Proposition 4.1
Let R be a triring. (i) V ♯ (0) = spec ♯ R and V ♯ ( R ) = ∅ . (ii) V ♯ ( I ) ∪ V ♯ ( J ) = V ♯ ( I ∩ J ) = V ♯ ( I · ♯ J ) , where I and J are two triidealsof R . (iii) \ λ ∈ Λ V ♯ (cid:0) I ( λ ) (cid:1) = V ♯ X λ ∈ Λ I ( λ ) ! , where (cid:8) I ( λ ) (cid:12)(cid:12) λ ∈ Λ (cid:9) is a set of triidealsof R . Proof
Since (i) and (iii) are clear, we need only to prove (ii).Since I · ♯ J ⊆ I ∩ J ⊆ I , we get V ♯ ( I · ♯ J ) ⊇ V ♯ ( I ∩ J ) ⊇ V ♯ ( I ). Similarly,we have V ♯ ( I · ♯ J ) ⊇ V ♯ ( I ∩ J ) ⊇ V ♯ ( J ). Thus, we get V ♯ ( I ) ∪ V ♯ ( J ) ⊆ V ♯ ( I ∩ J ) ⊆ V ♯ ( I · ♯ J ) . (29)Conversely, if P ∈ V ♯ ( I · ♯ J ), then I · ♯ J ⊆ P . By Proposition 2.2, we get I ⊆ P or J ⊆ P . Hence, P ∈ V ♯ ( I ) ∪ V ♯ ( J ). This proves that V ♯ ( I · ♯ J ) ⊆ V ♯ ( I ) ∪ V ♯ ( J ) . (30)It follows from (29) and (30) that (ii) is true.Let ( R = R ⊕ R , + , · , ♯ ) be a triring. By Proposition 4.1, the collection V ♯ := { V ♯ ( I ) | I is a triideal of R } of subsets of Spec ♯ R satisfies the axioms for closed sets in a topological space.The topology on Spec ♯ R having the elements of V ♯ as closed sets is called the extended Zariski topology . The collection D ♯ := { D ♯ ( I ) | I is a triideal of R } consists of the open sets of the extended Zariski topology on Spec ♯ R , where D ♯ ( I ) := Spec ♯ R \ V ♯ ( I ) = { P | P ∈ Spec ♯ R and P I } . x i ∈ R i with i ∈ { , } , both Rx and R ♯ x are triideals of R byProposition 1.2. Let D ♯ ( x ) := D ♯ ( Rx ) , D ♯ ( x ) := D ♯ ( R ♯ x ) . If I and I are the even part and odd part of an triideal I , then D ♯ ( I ) = [ x i ∈ I i , i =0 , D ♯ ( x i ) . Thus, { D ♯ ( x i ) | x i ∈ R i with i = 0 , } forms an open base for the extendedZariski topology on Spec ♯ R . Each D ♯ ( x i ) is called a basic open subset of Spec ♯ R . Clearly, D ♯ (0) = ∅ , D ♯ (1) = Spec ♯ R , D ♯ (1 ♯ ) = Spec ♯ R , and D ♯ ( x ) ⊆ Spec ♯ R for x ∈ R . Proposition 4.2
Let I and J be triideals of a Hu-Liu triring. (i) V ♯ ( I ) ⊆ V ♯ ( J ) if and only if ♯ √ J ⊆ ♯ √ I . (ii) V ♯ ( I ) = V ♯ ( ♯ √ I ) . Proof (i) follows from Proposition 3.4, and (ii) follows from (i).
Definition 4.1
Let X be a topological space. (i) A closed subset F of X is reducible if F = F (1) ∪ F (2) for proper closedsubsets F (1) , F (2) of X . We call a closed subset F irreducible if it is notreducible. (ii) X is quasicompact if given an arbitrary open covering { U ( i ) | i ∈ I } of X , there exists a finite subcovering of X , i.e., there exist finitely manymembers U ( i ) , . . . , U ( i n ) of { U ( i ) | i ∈ I } such that X = U ( i ) ∪ · · · ∪ U ( i n ) . Using the topological concepts above, we have the following
Proposition 4.3
Let ( R = R ⊕ R , + , · , ♯ ) be a triring. (i) The trispectrum
Spec ♯ R is quasicompact. (ii) Both
Spec ♯ R and Spec ♯ R are quasicompact subsets of Spec ♯ R . (iii) If I is a triideal of R , then the closed subset V ♯ ( I ) of Spec ♯ R is irreducibleif and only if ♯ √ I is a prime triideal. roof (i) Let { D ♯ ( I ( i ) ) | i ∈ ∆ } be an open covering of Spec ♯ R , where I ( i ) = I ( i )0 ⊕ I ( i )1 is a triideal with the even part I ( i )0 and the odd part I ( i )1 for each i ∈ ∆. Thus, Spec ♯ R = S i ∈ ∆ D ♯ ( I ( i ) ) = D ♯ ( P i ∈ ∆ I ( i ) ). Hence, V ♯ ( P i ∈ ∆ I ( i ) ) = ∅ .If 1 ( P i ∈ ∆ I ( i ) ) = P i ∈ ∆ I ( i )0 . Then ( P i ∈ ∆ I ( i ) ) is a proper ideal of thecommutative ring ( R , + , · ). Hence, there exists a maximal ideal M of thering ( R , + , · ) such that ( P i ∈ ∆ I ( i ) ) ⊆ M . Since M ⊕ R is a prime triidealof R and P i ∈ ∆ I ( i ) ⊆ M ⊕ R , we get that M ⊕ R ∈ V ♯ ( P i ∈ ∆ I ( i ) ) = ∅ ,which is impossible. Therefore, 1 ∈ ( P i ∈ ∆ I ( i ) ) = P i ∈ ∆ I ( i )0 , which impliesthat x ( i )0 + x ( i )0 + · · · + x ( i n )0 = 1 for some positive integer n and x ( i k )0 ∈ I ( i k )0 with i k ∈ ∆ and n ≥ k ≥
1. It follows that D ♯ ( x ( i k )0 ) ⊆ D ♯ ( I ( i k ) ) and Spec ♯ R = [ i ∈ ∆ D ♯ ( I ( i ) ) ⊇ n [ k =1 D ♯ ( I ( i k ) ) ⊇ n [ k =1 D ♯ ( x ( i k )0 )= n [ k =1 D ♯ ( Rx ( i k )0 ) = D ♯ ( n X k =1 Rx ( i k )0 ) = D ♯ ( R ) = Spec ♯ R, which implies that Spec ♯ R = S nk =1 D ♯ ( I ( i k ) ).(ii) Note that a closed subset of a quasicompact topological space is a quasi-compact subset. Since Spec ♯ R = V ♯ ( R ) is a closed subset of the quasicompacttopological space Spec ♯ R , Spec ♯ R is a quasicompact subset.Note that a subset C of a topological space X is a quasicompact subset of X if and only if every covering of C by open subsets of X has a finite subcovering.Hence, in order to prove that Spec ♯ R is a quasicompact subsets of Spec ♯ R , itsuffices to prove that if Spec ♯ R = S j ∈ Γ D ♯ ( J ( j ) ) for triideals J ( j ) of R , thenthere exists a positive integer m such that Spec ♯ R = S mk =1 D ♯ ( J ( j k ) ) for some j , · · · , j k ∈ Γ.Since
Spec ♯ R = S j ∈ Γ D ♯ ( J ( j ) ) = D ♯ ( P j ∈ Γ J ( j ) ), we have V ♯ ( P j ∈ Γ J ( j ) ) = Spec ♯ R \ D ♯ ( P j ∈ Γ J ( j ) ) = Spec ♯ R . If ( P j ∈ Γ J ( j ) ) = R , then there exists amaximal ideal N of the commutative ring ( R , + , ♯ ) such that ( P j ∈ Γ J ( j ) ) ⊆ N . By Proposition 2.1, there exists an ideal N of the commutative ring( R , + , · ) such that N ⊇ ( P j ∈ Γ J ( j ) ) and N ⊕ N is a prime triideal of R . Thus, N ⊕ N ∈ V ♯ ( P j ∈ Γ J ( j ) ) = Spec ♯ R , which is impossible because N = R . This proves that ( P j ∈ Γ J ( j ) ) = R . Hence, we have y ( j )1 + y ( j )1 + · · · + y ( j m )1 = 1 ♯ for some positive integer m and y ( j k )1 ∈ J ( j k )1 with j k ∈ Γ and m ≥ k ≥
1. It follows that D ♯ ( y ( j k )1 ) ⊆ D ♯ ( J ( j k ) ) and Spec ♯ R = [ j ∈ Γ D ♯ ( J ( j ) ) ⊇ m [ k =1 D ♯ ( y ( j k )1 ) ⊇ = ⊇ m [ k =1 D ♯ ( R ♯ y ( j k )1 ) ⊇ = D ♯ ( m X k =1 R ♯ y ( j k )1 ) = D ♯ ( R ) = Spec ♯ R, Spec ♯ R = S mk =1 D ♯ ( y ( j k )1 ).(iii) By Proposition 4.2, we may assume I = ♯ √ I in the following proof. First,we prove that if V ♯ ( I ) is irreducible , then I is a prime triideal.Suppose that a b ∈ I for some a , b ∈ R . Let J (1) = I + Ra = ( I + R a ) ⊕ ( I + R a )and K (1) = I + Rb = ( I + R b ) ⊕ ( I + R b ) . Then both J (1) and K (1) are triideals of R and J (1) · ♯ K (1) = ( I + R a )( I + R b ) + ( I + R a ) ♯ ( I + R b ) ⊆ I + R a R b + ( R a ) ♯ R b ) ⊆ I + R a b + R a b ⊆ I. Hence, we get V ♯ ( J (1) ) ∪ V ♯ ( K (1) ) = V ♯ ( J (1) · ♯ K (1) ) ⊇ V ♯ ( I ) by Proposition 4.1(ii). It is clear that V ♯ ( J (1) ) ⊆ V ♯ ( I ) and V ♯ ( K (1) ) ⊆ V ♯ ( I ). Hence, we getthat V ♯ ( J (1) ) ∪ V ♯ ( K (1) ) ⊆ V ♯ ( I ). Thus we have V ♯ ( J (1) ) ∪ V ♯ ( K (1) ) = V ♯ ( I ).Since V ♯ ( I ) is irreducible, V ♯ ( J (1) ) = V ♯ ( I ) or V ♯ ( K (1) ) = V ♯ ( I ), which implythat I = ♯ p J (1) ⊇ J (1) ∋ a or I = ♯ p K (1) ⊇ K (1) ∋ b . This proves that a b ∈ I = ⇒ a ∈ I or b ∈ I for a , b ∈ R . (31)Similarly, we have a b ∈ I = ⇒ a ∈ I or b ∈ I for a ∈ R and b ∈ R , (32) a b ∈ I = ⇒ a ∈ I or b ∈ I for a ∈ R and b ∈ R (33)and a ♯ b ∈ I = ⇒ a ∈ I or b ∈ I for a , b ∈ R . (34)By (31), (32), (33) and (34), I is a prime triideal.Next, we prove that if I is a prime triideal, then V ♯ ( I ) is irreducible. Sup-pose that V ♯ ( I ) = V ♯ ( J ) ∪ V ♯ ( K ), where J and K are triideals of R . UsingProposition 4.2 (ii), we can assume that ♯ √ J = J and ♯ √ K = K . In this case,we have V ♯ ( J ) ⊆ V ♯ ( I ) = ⇒ I = ♯ √ I ⊆ ♯ √ J = J and V ♯ ( K ) ⊆ V ♯ ( I ) = ⇒ I = ♯ √ I ⊆ ♯ √ K = K. By Proposition 4.1 (ii), we have V ♯ ( I ) = V ♯ ( J ) ∪ V ♯ ( K ) = V ♯ ( J · ♯ K ). Thisfact and Proposition 4.2 (i) give J · ♯ K ⊆ ♯ √ I = I . Since I is a prime triideal,we get J ⊆ I or K ⊆ I by Proposition 2.2. Hence, I = J or I = K . Thus, V ♯ ( I ) = V ♯ ( J ) or V ♯ ( I ) = V ♯ ( K ). This proves that V ♯ ( I ) is irreducible.15 eferences [1] M. F. Atiyah & I. G. MacDonald, Introduction to Commutative Algebra ,Addison-Wesley Publishing Company, 1969.[2] David Eisenbud & Joe Harris,
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