Chapter 1 The Category of Graded Rings 1.1 Graded Rings Unless otherwise stated, all rings are assumed to be associative rings and any ring R has an identity 1 ∈ R.IfX and Y are nonempty subsets of a ring R then XY denotes the set of all finite sums of elements of the form xy with x ∈ X and y ∈ Y . The group of multiplication invertible elements of R will be denoted by U(R). Consider a multiplicatively written group G with identity element e ∈ G. AringR is graded of type G or R,isG-graded, if there is a family {Rσ,σ ∈ G} of additive subgroups Rσ of R such that R = ⊕σ∈GRσ and RσRτ ⊂ Rστ , for every σ, τ ∈ G.ForaG-graded ring R such that RσRτ = Rστ for all σ, τ ∈ G,wesaythatR is strongly graded by G. The set h(R)=∪σ∈GRσ is the set of homogeneous elements of R; a nonzero element x ∈ Rσ is said to be homogeneous of degree σ and we write : deg(x)=σ.Anelementr of R has a unique decomposition as ∈ ∈ r = σ∈G rσ with rσ Rσ for all σ G, but the sum being a finite sum i.e. almost all rσ zero. The set sup(r)={σ ∈ G, rσ =0 } is the support of r in G. By sup(R)={σ ∈ G, Rσ =0 } we denote the support of the graded ring R. In case sup(R) is a finite set we will write sup(R) < ∞ and then R is said to be a G-graded ring of finite support. ∩ If X is a nontrivial additive subgroup of R then we write Xσ = X Rσ ∈ for σ G.WesaythatX is graded (or homogeneous) if : X = σ∈G Xσ.In particular, when X is a subring, respectively : a left ideal, a right ideal, an ideal, then we obtain the notions of graded subring, respectively : a graded left ideal, a graded right ideal, graded ideal. In case I is a graded ideal of R then the factor ring R/I is a graded ring with gradation defined by : (R/I)σ = Rσ + I/I, R/I = ⊕σ∈G(R/I)σ. C. Nˇastˇasescuand F. Van Oystaeyen: LNM 1836, pp. 1–18, 2004. c Springer-Verlag Berlin Heidelberg 2004 2 1 The Category of Graded Rings 1.1.1 Proposition Let R = ⊕σ∈GRσ be a G-graded ring. Then the following assertions hold : 1. 1 ∈ Re and Re is a subring of R. 2. The inverse r−1 of a homogeneous element r ∈ U(R) is also homoge- neous. 3. R is a strongly graded ring if and only if 1 ∈ RσRσ−1 for any σ ∈ G. Proof 1. Since ReRe ⊆ Re,weonlyhavetoprovethat1∈ Re.Let1= rσ be ∈ ∈ ∈ the decomposition of 1 withrσ Rσ. Then for any sλ Rλ(λ G), ∈ we have that sλ = sλ.1= σ∈G sλrσ,andsλrσ Rλσ.Consequently sλrσ =0foranyσ = e,sowehavethatsrσ =0foranys ∈ R. In particular for s = 1 we obtain that rσ =0foranyσ = e. Hence 1=re ∈ Re. ∈ ∩ −1 −1 −1 ∈ 2. Assume that r U(R) Rλ.Ifr = σ∈G(r )σ with (r )σ Rσ, −1 −1 ∈ −1 ∈ then 1 = rr = σ∈G r(r )σ.Since1 Re and r(r )σ Rλσ,we −1 −1 have that r(r )σ =0foranyσ = λ .Sincer ∈ U(R)wegetthat −1 −1 −1 −1 (r )σ =0for σ = λ , therefore r =(r )λ−1 ∈ Rλ−1 . 3. Suppose that 1 ∈ RσRσ−1 for any σ ∈ G.Thenforσ, τ ∈ G we have: Rστ = ReRστ =(RσRσ−1 )Rστ = Rσ(Rσ−1 Rστ ) ⊆ RσRτ therefore Rστ = RσRτ , which means that R is strongly graded. The converse is clear. 1.1.2 Remark The previous proposition shows that ReRσ = RσRe = Rσ,provingthatRσ is an Re-bimodule. gr If R is a G-graded ring, we denote by U (R)=∪σ∈G(U(R) ∩ Rσ)thesetof the invertible homogeneous elements. It follows from Proposition 1.1.1 that U gr(R) is a subgroup of U(R). Clearly the degree map deg : U gr(R) → G is a group morphism with Ker(deg) = U(Re). A G-graded ring R is called a crossed product if U(R)∩Rσ = ∅ for any σ ∈ G, which is equivalent to the map deg being surjective. Note that a G-crossed product R = ⊕σ∈GRσ is a strongly graded ring. Indeed, if uσ ∈ U(R) ∩ Rσ, − − 1 ∈ − 1 ∈ − then uσ Rσ 1 (by Proposition 1.1.1), and 1 = uσuσ RσRσ 1 . 1.2 The Category of Graded Rings 3 1.2 The Category of Graded Rings The category of all rings is denoted by RING. If G is a group, the category of G-graded rings, denoted by G-RING, is obtained by taking the G-graded rings for the objects and for the morphisms between G-graded rings R and S we take the ring morphisms ϕ : R → S such that ϕ(Rσ) ⊆ Sσ for any σ ∈ G. Note that for G = {1} we have G-RING=RING. If R is a G-graded ring, and X is a non-empty subset of G,wedenoteRX = ⊕x∈XRx.Inparticular, if H ≤ G is a subgroup, RH = ⊕h∈H Rh is a subring of R.InfactRH is an H-graded ring. If H = {e},thenRH = Re. Clearly the correspondence R → RH defines a functor (−)H : G − RING → H − RING. 1.2.1 Proposition The functor (−)H has a left adjoint. Proof Let S ∈ H − RING, S = ⊕h∈H Sh. We define a G-graded ring S as follows: S = S as rings, and Sσ = Sσ if σ ∈ H,andSσ = 0 elsewhere. Then the correspondence S → S defines a functor which is a left adjoint of (−)H . We note that if S ∈ RING = H − RING for H = {1}, then the G-graded ring S is said to have the trivial G-grading.LetH ¢ G be a normal subgroup. Then we can consider the factor group G/H.IfR ∈ G − RING, then for any class C ∈ G/H let us consider the set RC = ⊕x∈CRx. Clearly R = ⊕C∈G/HRC ,andRCRC ⊆ RCC for any C, C ∈ G/H. Therefore R has a natural G/H-grading, and we can define a functor UG/H : G − RING → G/H − RING, associating to the G-graded ring R the same ring with the G/H-grading described above. If H = G,thenG/G − RING = RING, and the functor UG/G is exactly the forgetful functor U : G − RING → RING, which associates to the G-graded ring R the underlying ring R. 1.2.2 Proposition The functor UG/H : G − RING → G/H − RING has a right adjoint. Proof Let S ∈ G/H − RING. We consider the group ring S[G], which is a G-graded ring with the natural grading S[G]g = Sg for any g ∈ G.SinceS = ⊕C∈G/HSC , we define the subset A of S[G]byA = ⊕C∈G/HSC[C]. If g ∈ G, there exists a unique C ∈ G/H such that g ∈ C; define Ag = SC g. Clearly the Ag’s define a G-grading on A, in such a way that A becomes a G-graded subring of S[G]. We have defined a functor F : G/H − RING → G − RING, associating to S the G-graded ring A. This functor is a right adjoint of the 4 1 The Category of Graded Rings functor UG/H . Indeed, if R ∈ G − RING and S ∈ G/H − RING, we define a map ϕ :HomG/H−RING(UG/H (R),S) → HomG−RING(R, F(S)) in the following way: if u ∈ HomG/H−RING(UG/H (R),S), then ϕ(u)(rg)= u(rg)g for any rg ∈ Rg.Thenϕ is a natural bijection; its inverse is defined by −1 ◦ ◦ ∈ → ϕ (v)=ε i v for any v HomG−RING(R, A), where i : A S[G]isthe → inclusion map, and ε : S[G] S is the augmentation map, i.e. ε( g∈G sgg)= g∈G sg.IncaseS is a strongly graded ring (resp. a crossed product, then the ring A, constructed in the foregoing proof, is also strongly graded (resp. a crossed product). Clearly if H ≤ G and R is a G-strongly graded ring (respectively a crossed product), then RH is an H-strongly graded ring (respectively a crossed prod- uct). Moreover, if H ¢ G is a normal subgroup, then UG/H(R)isaG/H- strongly graded ring (respectively a crossed product). 1.2.3 Remark The category G-RING has arbitrary direct products. Indeed, if (Ri)i∈I is ⊕ a family of G-graded rings, then R = σ∈G( i(Ri)σ)isaG-graded ring, which is the product of the family (Ri)i∈I in the category G-RING. Note that R is a subring of i∈I Ri, the product of the family in the category RING. gr The ring Ris denoted by i∈I Ri.IfG is finite or I is a finite set, we have gr i∈I Ri = i∈I Ri. 1.2.4 Remark o Let R = ⊕σ∈GRσ be a G-graded ring. We denote by R theoppositeringof R, i.e. Ro has the same underlying additive group as R, and the multiplication o defined by r ◦ r = r r for r, r ∈ R. The assignment (R )σ = Rσ−1 makes R into a G-graded ring. The association R → Ro defines an isomorphism between the categories G − RING and G − RING. 1.3 Examples 1.3.1 Example The polynomial ring If A is a ring, then the polynomial ring R = A[X]isaZZ-graded ring with n the standard grading Rn = AX for 0 ≤ n,andRn =0forn<0.