Generalizations of the Primitive Element Theorem

In this paper we gemeralize the primitive element theorem to the generation of separable algebras over fields and rings. We prove that any finitely generated separable algebra over an infinite field is generated by two elements and if the algebra is commutative it can be generated by one element. We then derive similar results for finitely generated separable algebras over semilocal rings. A) PASES: Generation of algebras, separable algebras, semilocal rings 1980 stec’c ict.i c]e: 16,17 i. fRXRZTIC. It is a well known result (Nagahara, i] that any finitely generated separable simple algebra A over a field F is generated over F by two conjugate elements of A. It is also known that if x is an element of A which does not belong to the center of A, then there exists a unit xI in A such that A is generated over F by x and xI (Nagahara, [i] ). We present a proof of some of these results in section 2. In section 3, we examine the problem of generating separable finitely generated algebras, not necessarily simple, over infinite fields and local or semilocal rings. Namely, we show that a finitely generated separable algebra over an infinite field F is generated by two elements over F. In the case the algebra is commutative it can be generated by one element. We give a counter example to shc that the condition, that the ground field is inifinite, is necessary. In section 4, we examine algebras over semilocal rings and we show that a finitely generated central separable algebra over a semilocal ring can also be generated by two elements over the ring and one element if the algebra is commutative. The condition that the algebra is oentral can be eliminated and the theorem still holds if the local fields of the ring, i.e. the ring modulo its maximal ideals, are infinite. 464 C. NIKOLOPOULOS AND P. NIKOLOPOULOS Let A Z D eij be a simple ring finite over a simple subring B as left i,j=l B-module where the eij are matrix units and D CA({eij}) is a division ring. CA(S is the centralizer of the subset S of A in A. Let B1 B[{eij}] n D1 eij, where D1 CBl({eij} is a division ring. We prove i,j=l basic results frcm [I], using a scmewhat different approach. here some the If D is a finitely generated separable division algebra over then there exist units u,d in D such that D F [u, d u d-l]. Proof. Let M be a separable maximal subfield of D over F, then M F[u] with some u (see [I] Corollary 7.9). Let C be the center of D. Then M C[u] and D is finitely generated, central division algebra over C, with maximal subfield M. Therefore if [D:C] n2, (by [3], theorem VII ii.3), we know that there exists a unit d in D such that the set {ui d uJ/i,j 0, i, n-l} is basis for D over C. Consequently, D ui d uJ C ui(dud-I)jC C[u, dud-1 i,j i,j C[u] [dud-1] F[u,dud-l], which completes the proof. T 2.2: Let A, B be as in the conventions. If D Dl[X,y with some conjugate x,y in D, then A B[u,v] for some conjugate u,v units of A. Proof. Since A r. D eij if n=l the leamm is trivially true. Let n>l i,j=l n n n and w 7 ei,n_i+l w-1, uI 7. ei_l, i and Vl= 7. ei, i_l WUlw-l. Since i=l i=2 x and y are conjugates, we have y dxd-I for some d element of D. Also, eij=vli -luln-lvln-lul j-i for i,j=l n. In case D D1 we have B[l-Ul, w(l-Ul)w-l] B[Ul,Vl] B [{eij ]= A, and l-uI is a unit. Hence, assume that D # DI. Exaraine two cases: (i) if xy # 1 and (ii) if xy i. In the first case, define u as Ul+xenl v as vl+Yeln. Then v=dwu(dw) -I and u--l=vl+x-lelniS an element of B[u]. Therefore (x-l-y)eln u-l-v is an element of B[u,v]. Hence (l-xy) ann u(x-l-y)eln is an element of B[u,v] and (l-xy)ein un-i(l-xy)enn and (l-xy)enj (l-xy)ennvn-j are elements of B[u,v], for every i,j =I n. Hence for every i,j I,..., n, we have (l-xy) 2 eij is in B[u,v]. Since (l-xy) 2 is a unit in B[u,v], we get {eij is a subset of B[u,v]. n Also x un, y r. eil v eni are in B[u,v] and therefore B[u,v] B[x,y,{eij}] i=l GENERALIZATIONS OF THE PRIMITIVE ELEMENT THEOREM 465 D1 [x,y, {eij}] D[{eij}] A. In the second case, (ii) if xy=l then D=D1 [x,y]=Dl[X and since D # DI we get x # _+i or #i. We can nw apply the first case for y=x to cmplete the proof. 2.1: Any separable simple algebra finitely generated over a field is generated over the ground field by two conjugate invertible elements. Proof. If A is separable simple finitely generated over a field F and A n Deij then Lemma 2.1 implies D=F[x, d x d-l], where x is a generating i,j=l element over F of a maximal subfield of D. Therefore, by Lemma 2.2 we have A F[u,v] for sme conjugate u,v units in A, which completes the proof of the theorem. T 2.3: Let E a proper division subring of a division ring D and a in D with ab # ba for sce b in D E. Let C the center of D. Then: (i) There exist at most two elements ci, i 1,2 in C N E such that (b+ci) a (b+ci) -I e E. (2) If a g E, then there exists at most one element c of the centralizer of a in E such that (b+c) a (b+c) -i E. Proof. Suppose that there exist Cl, c2, c3 three different elements in CA E such that ai=(b+ci)a(b+ci)-I is contained in E, for i=i,2,3. Then ba+cia= aib+aici for i=i,2,3 hence (Cl-C2)a=(al-a2)b+(alcl-a 2c2 and (Cl-C3)a=(aI -a3) b+(alcl-a3c3) so a=(cl-c2)-l(al-a2)b+(Cl-C2)-l(alcl-a2c2 and a=c (ci-c3)-I (al-a3)b+(Cl-C3)-l(alcl-a3c3) Subtracting those two, by elementary calculation we get a2=a3 which oontradicts the fact that if c2 then (b+c2)a(b+c2)-i #(b+)a(b) -I, for if (b) a(b) -I =(b) a(b)--a then (c2-c3) a= a’(c2 -c3) which gives a’. But (b+c2)a(b+c2)--a leads to a contradiction ba=ab. To prove the second assertion suppose there are two, elements cI and c2 in CR({a}) with cI c2 and ai (b+ci)a(b+ci)-I an element of F, for i=i,2. Then b (a2-al)-i (alCl-Cl a)-(a2 c2-c2a E and this is a contradiction, which proves the lemmm. Using the above lena we can prove the follwing: 2.2: If D is sepaz-able division algebra finitely generated over field F and a is an element of D such that a is not contained in the center of D, then D F[a, al] for sme aI in D. Proof. Consider M the maximal separable subfield of D. If M F[x], then theorem 2.1 implies D=[x,y], for sme y. By the fundamental theorem of Galois 466 C. NIKOLOPOULOS AND P. NIKOLOPOULOS theory for simple rings (see [i], Theorem 7.7), since M CD(M), we have that the number of intermediate fields between M and C is equal to the number of intermediate rings between D and M. Therefore, the number of intermediate rings between D and M is finite, say {A1 An}. Now y is not contained in Ai for any i, since Fix,Y] D. We examine two cases: (i) If ay ya then a(x+y) # (x+y)a, since a is not in the center of D, so D F[x,x+y] and we can apply (ii) for x, x+y. (ii) If ay # ya, from lemma 2.3 we have that for every i, there exist at most two elements Cl, c2 in F such that (Y+Cl)a(y+cl)-I is an element of Ai, and (Y+c2)a(y+c2)-I is an element of Ai. Let y0=Y+c0where co in F such that Y0aY0 -I is not contained in Ai for any i. Then D=M[Y0aY0-1]=F[Y0aY0 -l,u]=F[a,Y0-1 uY0]. 2.3: Let A be a separable finitely generated simple algebra over a field F and a an element of A not contained in the center of A. Then A F[a, al] for some aI unit in A. Proof. See i], Theorem 12. I. T 3.1: If S is a commutative algebra over a field F, then S is separable over F if and only if S is the direct sum of separable field extensions of F. Prxf. The proof follows since cfmmtative separable algebras over fields are semisimple, hence S is the direct sum of field extensions of F which are separable over F since S is. We now prove a generalization of the primitive element theorem for finitely generated conmfcative separable algebras. 3.1: If S is a finitely generated commutative separable algebra over an infinite field F, then S is generated over F by only one element, i.e. S F[a], for some a in S. Proof. By Lemma 3.1 we have S=FIF2+... @Fn, with Fi a separable finite field extension of F. By the primitive element theorem, Fi F[xi] for some xi in S, which gives S=F[Xl, Xn]. Let n=2. Then S=F1 F2=F[Xl]F[x2]. Now, since Fi=F[xi] is a separable field extension of F, there are a finite number of fields between Fi and F. If A is a commutative separable subalgebra of S, then A=F’ 1F’ 2 with F’ i subfield of Fi and therefore there is only a finite number of commutative separable subalgebras of S. If X=Xl+ax2, where a in F, then GENERALIZATIONS OF THE PRIMITIVE ELEMENT THEOREM 467 F[x] is a cmutative separable subalgebra of S and since F is infinite but there can onlybea finite number of F[x]’s, there exist twodifferent elements x’ x" of S, x’ ---x+alx2 and x"=xl+a2x2, with ai in F, such that F[x’] F [x’’]. Consequently, x’ and x" are eontained in F[x’], which implies x’x"= (al-a2)x2 also belong to F[x’]. Since aI a2 is invertible, x2 and Xl=X’ -alx2 belong to F[x’], which gives that S F[Xl, x2] is contained in F[x’]. Henoe, S F[Xl, x2] [x’]. An easy induction on n gives the result. We rm4 prove that if the algebra is not cmmtative we can the algebra fr=m just two elements. still generate 3.2: Let S be a finitely generated separable algebra over an infinite field F. Then S Fix,y], for sme x,y in S. Proof. S separable implies S is semisimple so S [email protected] with Si simple and separable over F. By Theorem 2.3 we have Si F[xi, Yi], xi, Yi contained in S. let Zi denote the extension F[xi] of F for i=l, n. Then Zi is separable over F and if Z [email protected] then Z is a cmmtative separable algebra over F. Nw Theorem 3.1 implies that Z F[x] for sme x in Z. Let y Yl +’" "+Yn" Then we have xk is an element of Fix], for hence xk is an element of F[x,y] for every k and xki yJ (xkiek) yJ xki(ekyJ) xki(eky)J xki(ekYk)J =XkkJ where ek is the identity in F[Xk]. So Xkk is contained in F[x,y] hence Yk is contained in F[x,y] so F[Xk, Yk] is contained in F[x,y] for every k. n Therefore S =F[Xk, Yk] is contained in F[x,y], from which S F[x,y]. k=-i This proves the theorem. The condition that the field has to be infinite in Theorems 3.1 and 3.2 is necessary as is shcn by the next example. v If F is the field with q elements, F {al aq} and S Fi, where i=l Fi F for every i, F F[x]/(x-ai). let v>q. Then S is separable over F but cannot be generated by less than two elements since F1 F...Fq+1 = F+F[x]/ q q H (x-ai) F[x]/(x-ai) F[x]/ (x-ai) which cannot be of the form F[x]/