Article信州大学理学部紀要 7(1): 7-12(1972)departmental bulletin pape

KISHIMOTO,  KAZUO

English

37262

Shinshu University Institutional Repository

JOUR, FAC. SCI. SHINSHU UNIV., Vol. 7. pp. 7-12, June 1972Note on Quadratic Extensions of Rings Ⅱ              By KAZUO KISHIMOTO       Department of Mathematics, Faculty of Science,                 Shinshu University               (Received April 30 1972)   gntroduction. Throughout the present paper B will mean a ring with anidentity 1, A=B+xB =:B+BxRB an extension ring of B with an identitycoinciding with the jdentity of B.   As an extension of result of [5], T. Nagahara gave characterizations for acommutative ring A to be a Galois extension over B ([7]). The main purpose ofthis note is to extend the above Nagahara's result to some non commutative case.   Let A == B (D xB == Be Bx, dx = xdi + do for each dEB (di, de E B). Thenthe map p:d-di is an automorphism of B and the map D:d-do is a p-derivation of B. Further, if x2 == xbi+bo for some bi, boeB, the map a of Adefined by a(xb' + c') = (xc + b)b' + c'(b, c, b', c' E B) is a B ring epimorphism ofA if and only if there hold followings     (I) cisaunit element of Z) the center of B.     (II) (1-c)D(d) =: db -bp(d) for each d ff B.     (III) cbi == c(p(c)bi + D(c) + b + p(b)).     (IV) bbi + bo = c(p(c)be + D(b)) + b2.   Eor if ff is a B-homomorphism, we obtain   a(dx) = d(a(x)) = d(xc+b) = x,o(d)c + D(d)c + db and o(dx) == a(xp(d) + D(d)) == xcp(d)+ b,o(d) + D(d).   Hence cG Z. Moreover, if a is an epimorphisin, cB=B implies that c is aunit element. Under the assumption that cEU(Z), the validity of (II)-(IV) isequivalent to that a is a homomorphism of A by [2].   Now, we set the condition*) as f･ollowing:   *) If M isaright, as well as left, free A-module of finite rank, then therank is uniquei).   In a!1 that follows, we assume that A satisfies ').   1. Necessary and sufficient conditions for A to be Galois over B.   We shall begin our study from the following   Leinma 1. Let A/B be a Galois extension with a Galois grouP {E;. Then1) If A is commutative, A satisfies *).･8 KAzuo KIsHIMoTo     (a) S is of order 2.     (b) For a( 7k l) G S, x - a(x) is inversible.     (c) {1, x} is a free B-basis for A2).   Proof. Let a(iLl) Ei S. We suppose that x-a(x) is not right inversible. Thenthere exists a proper right ideal r of A such that r ii) x - a(x). On the other hand,since A == B(D xB, (1-6)A == {y-a(y)lyE A} is contained in r. Let {xi, x2,･･･,                                                          nXn; Yi, N2, '･･, y.} be a (S}-Galois coordinate system for A/B with Z                                                            T(Xi)Yi = O"1, .                                                         i=1for each rE (S}. Then we have a contradiction 1= ZI`m.i(x, -a(x,))y, Et. Thusx - 6<x) is right inversible. Since A -- BeBx, the same arguments enable us tosee that x - a(x) is left inversible.   Now, let c' +xb'=:O (resp. c'+b'x == O) for some c', b' ci B. Then O=(c'+xb') - a(c' + xb') == (x - a(x))b' (resp, (c' + b'x) - a(c' + b'x) := b'(x - o(x)) yields c' =bt = O.   Regarding that A <g) BA is a Ieft (resp. right) A-module by a(b' (2b c') = ab' (g> c'(resp. (b' (g) c')a == b'Xc'a) for each a, b', c' E A, Aop BA =A op B(BOBx) =AOA op BBx == A(1 E9 1) + A (1 op x) (resp. A E9 .A = (B exB) op BA = Ae xB op BA=(1 (g) 1)A + (x (E9 1)A) is a free A-module of rank 2. 0n the other hand, (b), (c), (d)and (e) of [1], Theorem 1. 3 are equivaient without assumptions that A and B arecommutative3). Therefore A (8} BA is isomorphic to a direct sum of 161 -copies ofA. Consequently we have ISI=2 by ").   Theorem 1. 4) Let A have a relation x2 = xbi + bo for some bo, bi E B. Then A/Bis a Galois extension if and only if there hold that     (a) {1, x} is a free B-basis for A.     (b) there exists an element b of B satisylying     (i) 2D(d) = db - bp(d),     (ii) b+p(b)=2bi,     (iii) bb, == b2 - D(b),     (iv) 2x-b is inversible, where p, D are ･maPs of B deij7ned by d--->dt,       'd--> do resPectively for each dEB with dx == xdt + do (di, do E B),   Moreover, if A is commutative (i), (ii) and (iii) of (b) are needless and (iv) can berePlaced (iv') 2x -bi is inversible. '   Proof. Let A/B be a Galois extension. Then by Lemma 1, 6, the group ofB-automorphisrns of A is {1, a} and {1, x} is a free B-basis for A.    Let a(x) =: xc + b. Then BDx÷ a(x) = x(1+c) +b implies c= -1, and hence,x-a(x) == 2x -b is inversible by Lemma 1. The validity of G), (ii) and (iii) of (b)is a direct consequence of (II), (III) and (IV).2)3)4)A free basis means a free right, as well as, leftNeedless to say a B-algebra homomorphism of [1]Cf. [7], Lemma 1.basis. replace to a B-module homomorphism.                    Note on Quadratic Extensions of Rings ll 9   Conversely, assume that A satisfy (a) and (b). Then by (a) and (i), (ii) and (iii)of (b), the map a defined by xb' + c' -(-x + b)b' + c' (b',c' E B) is a B-automor-phism of A. Let a(xb' + c') == xb' + c'. Then (x - ff(x))b' = (2x - b)b' = O impliesb' =O by (iv) of (b). Thus Av= B. Since (x - o(x))"ix -(x - a(x))-i. a(x)el := 1 and(x - a(x))"o(x) - (x - a(x))-'a(x)a(1) == O, AIB is a Ga!ois extension.   Let A be commutative. Then we have bbi=b2 by (iii) of (b), and the maprp : xb' + c' -(-x + bi)b' + c' is a B-automorphism of A by (I), (II), (III) and (IV).If rp =1 then x= rp(x) -- -x+ bi, and hence 2x =bi = O. On the other hand, since2x -b is inversible by (iv) of (b), we can see that b is inversible, But, this con-tradicts to b2 == bbi. Thus rp == o(; 1) and x - c(x) =: 2x - bi is inversible by Lemma1 (b).   Let T be a ring, P an automorphism of T, E a P-derivation of T. Then byT[X ; P, E] we denote a ring of polynomials {XX'tilti e T} whose multiplicationis defined by the distributive laws and the rule tX = XP(t) + E(t) for each t E T.A monic polynomial f<X) E T[X; P, E] is called a non-vanishing PolNnomial if theright ideal f(X)T[X; P, E] is a two-sided ideal of T[X; P, E], and, an elementt E T is cailed a root of f(X) if f(t) =O and X-t is non-vanishing5).   Corol}ary 1. Let AIB be a Galois extension with x2 == xbi+be (bi, boEB) anddx = xp(d) + D(d) for each d E B. Tlaen the following conditions are equivalent :     (a) 2el=O     (b) x-o(x) is an element of B.     (c) there exists a free B-basis {1, y} for A with a(y) =y- 1.     (d) there exists a free B-basis {1, w} for A such thatwand w+1 are roots        of the Polynomial X2 - X - (zv2 - w) E A[X ; I.]. 6)   Moreover, if A has no ProPer central idemPotents, then the only roots of thePolynonzial X2 -X-(w2 - w) given in (d) are w and w+ 1.   Proof. (a)-(b). Let 2el ::= O. Then x+a(x) == x-a(x) means that x-a(x) EB.   (b)-(c). Let b=x-a(x)EB. Then, by Lemma 1, b is inversible,--Henceif we set y == xb-i, {1, y} is a free B-basis for A and a(y) == (x-b)bTi =: y - 1.   (c) ----> (d). Since dy - yd EB for each dE B, dy = yd + D(d), where D is aderivation of B. Now we shall show that X2-X-(N2-y)EA[X; I.] is therequested polynomial. X(X - y) == (X - y)X, X(X - (y + 1)) = (X - (y + 1))X and                                              'd(X - y) - Xd - dy + D(d) = (X - y)d, d(X - (y + 1)) -= (X - (y + 1))d show that yand y+1 are roots of X2-X-(N2- y).   (d)-(a). Let {1, w} be a free B-basis for A such that w and w+1 areroots of X2 -X- (w2- w). Then O := (w + 1>2 -(zv + 1) -(w2 - w) = 2w shows that5)6)Cf. [4].4. means the inner derivation generated by zv.10 KAzuo KIsHIMoTo2el =' O.    Let A be a ring without proper central idempotents, and let z be a root ofX2 -Xr (zv2 - zv) given in (d). Then X(X - z) = (X - z)X = X(X - z) - D(z) andd(X - 2) = (Xd - d2 + D(d)) == (X - 2)d for each dE B. Hence we have D(z) = zw- zvz == O and dw - wd := dz - zd respectively. Hence w+zE V, the centralizerof B in A. Since 2w == wz, we have w+2Ei C, that is, z=w+c for some cEC. Noting that 2el = O, O ::= z2 - z - (w2 - w) :=: (z + w)2 - (z + w) = c2 + c, c is acentral idempotents, and hence c =7- O or c := 1.    Theorem 2. Let /llB be a Galois extension. Then 2el is inversible if and onlyif there exists an element yEA sach that A=BeyB =BeBy, y2EBand ya(y)= a(y)y for each a E (S} == S(A!B), and if this is the case, y is inversible.   Proof. Let 2el be inversible, and let y=(x-a(x))12. Then y is inversibie,a(y) = -y and y2 (ii U(B). Since y'i/2ey+ y-i12eyel == 1 and N-`12ea(y) ÷ y-i12epteo(1)= O, B[y] == B + yB == B + By =: A by [6, Theorem 2. 3]. By Lemma 1, {1, y}is a free B-basis for A.   Conversely, assume that there exist$ an element y E A such that A =:: B e yB=: Be By, y2 G B and ya(y) = a(y)N for each o E S, Then y(y + a(y)) == y2 + ya(y)ci B yields y+o(y)=O, and hence a<y) = -y. Consequently, we can see that 2yis inversible by Theorem 1. Thus 2el and y are inversible.   Corollary 2. Let A be a Galois extension with x2 E B, and dx == xp(d) -t- D(d) foreach d E B. Then the following conditions are equivalent :     <a) xa(x) ::= a<x)x for each aE(S. ,     (b) D=O and 2el, x are inversible.     (c) p= 'S'HilB and 2el is inversible. '     (d) p can be extended to an automorPhism P of A zvith P(x) == x, x and -x        are distinct roots of X2 - x2 of A[X ; P] in A.   Proof. Firstly, we shall note that if a(x)+x=:]b for some bci B, then bsatlsfies 2D(d) -- db - bp(d) for each d cii B (Theorem. 1 (b), (i>),   (a)-(b), As is shown in the proof of the sufficiency of Theorem 2, a(x)=-x, 2el and x are inversible. Since o(x)+x :=: O, we have D(d>= d(b12) -(b12)p(d)== O for each dEB                 '   (b)--(c), This implication is evident.   (c) ---> (d). If p == hi"i1B then P == hi-` is an automorphism of A with P(x) = x,and X(X ± x) ::= (X ± x)X, d(X ± x) = (X ± x)p(d) are clear.   (d)---> (a). Since d(X - x) = (X- x)p(d) for each d cr B, p= bufim'lB. Hence themap a defined by a(xb'+c')=-xb'+c' (b', c' GB) is a B-automorphism of A.Thus xa(x) = a(x)x for each o E S.   Let A be a ring without proper central idempotents, and let z be a root ofX2 - x2 given in (d). Then X(x - z) == (X - z)X and d(X - z) = (X - z)p(d) for each                    Note on Quadratic Extensions of Rings ll 11dcii B. Hence we have xz=zx, dz=zp(d) respectively. Hence z=: xc for somec E U(C) with c2 == 1. Since C is a commutative ring without proper idempotents,c= ±1 by [3, Corollary 2. 5].   Tfie following will be easily seen from Theorem 2 and Coro!lary 2.   Corollary 3.7) Let A have a relation x2 c!i B. Then AIB is a Galois extensionwith xa(x) == a(x)x for each o E S if and only if there holds that     (a) {1, x} is a free B-basis for A.     (b) 2el and x are inversible.     (c) D == O zvhere D is the map dofned by dx =xp(d)+ D(d) foreach dEB.   2. Structure of the centralizer.   In the rest, we shall determine the structure of the centralizer of a quadraticextension.   Let A=Bes xB=BOBx be a Cl3 =:: {1, a} Galois extension, and let V bethe centralizer of B in A. Then we may assume that x2 E U(B), o(x) = -x anddx == xp(d) for some automorphism p of B if 2el is inversible for each dG B, anddx = xd + D(d), a(x) := x+1 for some derivation D of B if 2el=O for each dEB.   Theorem 4. Let 2el be inversible or 2el=O. Then V=C[Z], the comPosite ofthe center C of A and the center Z of B. More Precisely, Y [= CG) Zb, where Z} =Zn 1} and 1} == {a e Aiay = a(y)a for each y E AiJ,   Proof. It is evident that V=Z if o=D for some vG V. Hence we considerthe case a iLD for each vG U(V). Firstly, we note that V:= COL.   case 2el = O. Let v == xb + c (b, c E B). Then dv = vd for each d er B implyxdb + D(d)b + dc = xbd + cd and henceand D(d)b = cd - dc.   Thus,            D(b)b-O , (2)   Next, let us assume that v E 1}. Then 1} i a (v) - v = b yields bx = a(x)b ==(x + 1)b, and hence   By (2) and (3), we have b2 == O. Then 1+bE U(Z) by (1).   On the other hand, since o;-v for each ve U(V), U(Z)EC. Thus we obtainO=D(1+b)= D(b)=b. Therefore v=cEBnV==Z means that LEZ. Thus7) Cf. [7], Lemma 2.12 KAzuo KIsHIMoToV-Ce4-C[Z].    case 2el is inver sible. Let v = scb + c(b, c E B), Then dv -- vd for each d E Bimplies xp(d)b + dc = xbd ÷ cd, and hence                                                                    '             p(d)b ::= bd, cEZ ' (1)                                           'Thus             p(b)b-b2 (2)    Next, let us assume that vEL. Then 1} g 112(a (v) - v) == xb and xbx=x2p(b)=a(x)xb=-x2b, and hence .             p(b) -= -b (3)    By (2) and (3), we have p(b)b=b2=O. Thus (xb)2 :=: x2p(b)b=O, and hence 1- xb E U( V). Since U( V) g U(C), we have xb E 1} n C = O. Consequently, V == Ce 4 - C[Z].                                 References[1] S. U. CHAsE, D. K. HARRisoN and A. RosENBERG:Galois theory and Galois cohomo-   logy of commutative rings, Mem. Amer. Math. Soc., No. 52 (l965),[2] s3:lslll6. CO}IN : QUadratiC extensions of sl<ew fields, proc. London Math. soc., 11 (lg61),[3] G. J. JANusz:Separable algebras over commutative rings, Trans. Amer. Math. Soc.,   122 (l966), 461-479.[4] K. KismMoTo:Zeros of polynomials and Galois extensions of simple rings, J. Fac.   Sci. Shinshu Univ., 2 (1967), 117-122.[5] :Note on quadratic extensions of rings, J. Fac. Sci. Shinshu Univ.,5<1970),   25-28.[6] Y. MiyAsHiTA:Finite outer Galois theory of non-commutative rings, J. Fac. Sci.   Hokkaido Univ., I9 (1966), 114-134.[7] T, NAGAHARA : A quadratic extension, Proc. Jap. Acad., 47 (1971), 6-7.

Note on Quadratic Extensions of Rings II

KISHIMOTO  KAZUO

JOUR, FAC. SCI. SHINSHU UNIV., Vol. 7. pp. 7-12, June 1972Note on Quadratic Extensions of Rings Ⅱ              By KAZUO KISHIMOTO       Department of Mathematics, Faculty of Science,                 Shinshu University               (Received April 30 1972)   gntroduction. Throughout the present paper B will mean a ring with anidentity 1, A=B+xB =:B+BxRB an extension ring of B with an identitycoinciding with the jdentity of B.   As an extension of result of [5], T. Nagahara gave characterizations for acommutative ring A to be a Galois extension over B ([7]). The main purpose ofthis note is to extend the above Nagahara's result to some non commutative case.   Let A == B (D xB == Be Bx, dx = xdi + do for each dEB (di, de E B). Thenthe map p:d-di is an automorphism of B and the map D:d-do is a p-derivation of B. Further, if x2 == xbi+bo for some bi, boeB, the map a of Adefined by a(xb' + c') = (xc + b)b' + c'(b, c, b', c' E B) is a B ring epimorphism ofA if and only if there hold followings     (I) cisaunit element of Z) the center of B.     (II) (1-c)D(d) =: db -bp(d) for each d ff B.     (III) cbi == c(p(c)bi + D(c) + b + p(b)).     (IV) bbi + bo = c(p(c)be + D(b)) + b2.   Eor if ff is a B-homomorphism, we obtain   a(dx) = d(a(x)) = d(xc+b) = x,o(d)c + D(d)c + db and o(dx) == a(xp(d) + D(d)) == xcp(d)+ b,o(d) + D(d).   Hence cG Z. Moreover, if a is an epimorphisin, cB=B implies that c is aunit element. Under the assumption that cEU(Z), the validity of (II)-(IV) isequivalent to that a is a homomorphism of A by [2].   Now, we set the condition*) as f･ollowing:   *) If M isaright, as well as left, free A-module of finite rank, then therank is uniquei).   In a!1 that follows, we assume that A satisfies ').   1. Necessary and sufficient conditions for A to be Galois over B.   We shall begin our study from the following   Leinma 1. Let A/B be a Galois extension with a Galois grouP {E;. Then1) If A iscommutative, A satisfies *).･8 KAzuo KIsHIMoTo     (a) S is of order 2.     (b) For a( 7k l) G S, x - a(x) is inversible.     (c) {1, x} is a free B-basis for A2).   Proof. Let a(iLl) Ei S. We suppose that x-a(x) is not right inversible. Thenthere exists a proper right ideal r of A such that r ii) x - a(x). On the other hand,since A == B(D xB, (1-6)A == {y-a(y)lyE A} is contained in r. Let {xi, x2,･･･,                                                          nXn; Yi, N2, '･･, y.} be a (S}-Galois coordinate system for A/B with Z                    T(Xi)Yi = O"1, .                                                         i=1for each rE (S}. Then we have a contradiction 1= ZI`m.i(x, -a(x,))y, Et. Thusx - 6<x) is right inversible. Since A -- BeBx, the same arguments enable us tosee that x - a(x) is left inversible.   Now, let c' +xb'=:O (resp. c'+b'x == O) for some c', b' ci B. Then O=(c'+xb') - a(c' + xb') == (x - a(x))b' (resp, (c' + b'x) - a(c' + b'x) := b'(x - o(x)) yields c' =bt = O.   Regarding that A <g) BA is a Ieft (resp. right) A-module by a(b' (2b c') = ab' (g> c'(resp. (b' (g) c')a == b'Xc'a) for each a, b', c' E A, Aop BA =A op B(BOBx) =AOA op BBx == A(1 E9 1) + A (1 op x) (resp. A E9 .A = (B exB) op BA = Ae xB op BA=(1 (g) 1)A + (x (E9 1)A) is a free A-module of rank 2. 0n the other hand, (b), (c), (d)and (e) of [1], Theorem 1. 3 are equivaient without assumptions that A and B arecommutative3). Therefore A (8} BA is isomorphic to a direct sum of 161 -copies ofA. Consequently we have ISI=2 by ").   Theorem 1. 4) Let A have a relation x2 = xbi + bo for some bo, bi E B. Then A/Bis a Galois extension if and only if there hold that     (a) {1, x} is a free B-basis for A.     (b) there exists an element b of B satisylying     (i) 2D(d) = db - bp(d),     (ii) b+p(b)=2bi,     (iii) bb, == b2 - D(b),     (iv) 2x-b is inversible, where p, D are ･maPs of B deij7ned by d--->dt,       'd--> do resPectively for each dEB with dx == xdt + do (di, do E B),   Moreover, if A is commutative (i), (ii) and (iii) of (b) are needless and (iv) can berePlaced (iv') 2x -bi is inversible. '   Proof. Let A/B be a Galois extension. Then by Lemma 1, 6, the group ofB-automorphisrns of A is {1, a} and {1, x} is a free B-basis for A.    Let a(x) =: xc + b. Then BDx÷ a(x) = x(1+c) +b implies c= -1, and hence,x-a(x) == 2x -b is inversible by Lemma 1. The validity of G), (ii) and (iii) of (b)is a direct consequence of (II), (III) and (IV).? ? ?A free basis means a free right, as well as, leftNeedless to say a B-algebra homomorphism of [1]Cf. [7], Lemma 1.basis. eplace toa B-module homomorphism.                    Note on Quadratic Extensions of Rings ll 9   Conversely, assume that A satisfy (a) and (b). Then by (a) and (i), (ii) and (iii)of (b), the map a defined by xb' + c' -(-x + b)b' + c' (b',c' E B) is a B-automor-phism of A. Let a(xb' + c') == xb' + c'. Then (x - ff(x))b' = (2x - b)b' = O impliesb' =O by (iv) of (b). Thus Av= B. Since (x - o(x))"ix -(x - a(x))-i. a(x)el := 1 and(x - a(x))"o(x) - (x - a(x))-'a(x)a(1) == O, AIB is a Ga!ois extension.   Let A be commutative. Then we have bbi=b2 by (iii) of (b), and the maprp : xb' + c' -(-x + bi)b' + c' is a B-automorphism of A by (I), (II), (III) and (IV).If rp =1 then x= rp(x) -- -x+ bi, and hence 2x =bi = O. On the other hand, since2x -b is inversible by (iv) of (b), we can see that b is inversible, But, this con-tradicts to b2 == bbi. Thus rp == o(; 1) and x - c(x) =: 2x - bi is inversible by Lemma1 (b).   Let T be a ring, P an automorphism of T, E a P-derivation of T. Then byT[X ; P, E] we denote a ring of polynomials {XX'tilti e T} whose multiplicationis defined by the distributive laws and the rule tX = XP(t) + E(t) for each t E T.A monic polynomial f<X) E T[X; P, E] is called a non-vanishing PolNnomial if theright ideal f(X)T[X; P, E] is a two-sided ideal of T[X; P, E], and, an elementt E T is cailed a root of f(X) if f(t) =O and X-t is non-vanishing5).   Corol}ary 1. Let AIB be a Galois extension with x2 == xbi+be (bi, boEB) anddx = xp(d) + D(d) for each d E B. Tlaen the following conditions are equivalent :     (a) 2el=O     (b) x-o(x) is an element of B.     (c) there exists a free B-basis {1, y} for A with a(y) =y- 1.     (d) there exists a free B-basis {1, w} for A such thatwand w+1 are roots        of the Polynomial X2 - X - (zv2 - w) E A[X ; I.]. 6)   Moreover, if A has no ProPer central idemPotents, then the only roots of thePolynonzial X2 -X-(w2 - w) given in (d) are w and w+ 1.   Proof. (a)-(b). Let 2el ::= O. Then x+a(x) == x-a(x) means that x-a(x) EB.   (b)-(c). Let b=x-a(x)EB. Then, by Lemma 1, b is inversible,--Henceif we set y == xb-i, {1, y} is a free B-basis for A and a(y) == (x-b)bTi =: y - 1.   (c) ----> (d). Since dy - yd EB for each dE B, dy = yd + D(d), where D is aderivation of B. Now we shall show that X2-X-(N2-y)EA[X; I.] is therequested polynomial. X(X - y) == (X - y)X, X(X - (y + 1)) = (X - (y + 1))X and                                              'd(X - y) - Xd - dy + D(d) = (X - y)d, d(X - (y + 1)) -= (X - (y + 1))d show that yand y+1 are roots of X2-X-(N2- y).   (d)-(a). Let {1, w} be a free B-basis for A such that w and w+1 areroots of X2 -X- (w2- w). Then O := (w + 1>2 -(zv + 1) -(w2 - w) = 2w shows that? ?Cf. [4].4. means the inner derivation generated by zv.10 KAzuo KIsHIMoTo2el =' O.    Let A be a ring without proper central idempotents, and let z be a root ofX2 -Xr (zv2 - zv) given in (d). Then X(X - z) = (X - z)X = X(X - z) - D(z) andd(X - 2) = (Xd - d2 + D(d)) == (X - 2)d for each dE B. Hence we have D(z) = zw- zvz == O and dw - wd := dz - zd respectively. Hence w+zE V, the centralizerof B in A. Since 2w == wz, we have w+2Ei C, that is, z=w+c for some cEC. Noting that 2el = O, O ::= z2 - z - (w2 - w) :=: (z + w)2 - (z + w) = c2 + c, c is acentral idempotents, and hence c =7- O or c := 1.    Theorem 2. Let /llB be a Galois extension. Then 2el is inversible if and onlyif there exists an element yEA sach that A=BeyB =BeBy, y2EBand ya(y)= a(y)y for each a E (S} == S(A!B), and if this is the case, y is inversible.   Proof. Let 2el be inversible, and let y=(x-a(x))12. Then y is inversibie,a(y) = -y and y2 (ii U(B). Since y'i/2ey+ y-i12eyel == 1 and N-`12ea(y) ÷ y-i12epteo(1)= O, B[y] == B + yB == B + By =: A by [6, Theorem 2. 3]. By Lemma 1, {1, y}is a free B-basis for A.   Conversely, assume that there exist$ an element y E A such that A =:: B e yB=: Be By, y2 G B and ya(y) = a(y)N for each o E S, Then y(y + a(y)) == y2 + ya(y)ci B yields y+o(y)=O, and hence a<y) = -y. Consequently, we can see that 2yis inversible by Theorem 1. Thus 2el and y are inversible.   Corollary 2. Let A be a Galois extension with x2 E B, and dx == xp(d) -t- D(d) foreach d E B. Then the following conditions are equivalent :     <a) xa(x) ::= a<x)x for each aE(S. ,     (b) D=O and 2el, x are inversible.     (c) p= 'S'HilB and 2el is inversible. '     (d) p can be extended to an automorPhism P of A zvith P(x) == x, x and -x        are distinct roots of X2 - x2 of A[X ; P] in A.   Proof. Firstly, we shall note that if a(x)+x=:]b for some bci B, then bsatlsfies 2D(d) -- db - bp(d) for each d cii B (Theorem. 1 (b), (i>),   (a)-(b), As is shown in the proof of the sufficiency of Theorem 2, a(x)=-x, 2el and x are inversible. Since o(x)+x :=: O, we have D(d>= d(b12) -(b12)p(d)== O for each dEB                 '   (b)--(c), This implication is evident.   (c) ---> (d). If p == hi"i1B then P == hi-` is an automorphism of A with P(x) = x,and X(X ± x) ::= (X ± x)X, d(X ± x) = (X ± x)p(d) are clear.   (d)---> (a). Since d(X - x) = (X- x)p(d) for each d cr B, p= bufim'lB. Hence themap a defined by a(xb'+c')=-xb'+c' (b', c' GB) is a B-automorphism of A.Thus xa(x) = a(x)x for each o E S.   Let A be a ring without proper central idempotents, and let z be a root ofX2 - x2 given in (d). Then X(x - z) == (X - z)X and d(X - z) = (X - z)p(d) for each                    Note on Quadratic Extensions of Rings ll 11dcii B. Hence we have xz=zx, dz=zp(d) respectively. Hence z=: xc for somec E U(C) with c2 == 1. Since C is a commutative ring without proper idempotents,c= ±1 by [3, Corollary 2. 5].   Tfie following will be easily seen from Theorem 2 and Coro!lary 2.   Corollary 3.7) Let A have a relation x2 c!i B. Then AIB is a Galois extensionwith xa(x) == a(x)x for each o E S if and only if there holds that     (a) {1, x} is a free B-basis for A.     (b) 2el and x are inversible.     (c) D == O zvhere D is the map dofned by dx =xp(d)+ D(d) foreach dEB.   2. Structure of the centralizer.   In the rest, we shall determine the structure of the centralizer of a quadraticextension.   Let A=Bes xB=BOBx be a Cl3 =:: {1, a} Galois extension, and let V bethe centralizer of B in A. Then we may assume that x2 E U(B), o(x) = -x anddx == xp(d) for some automorphism p of B if 2el is inversible for each dG B, anddx = xd + D(d), a(x) := x+1 for some derivation D of B if 2el=O for each dEB.   Theorem 4. Let 2el be inversible or 2el=O. Then V=C[Z], the comPosite ofthe center C of A and the center Z of B. More Precisely, Y [= CG) Zb, where Z} =Zn 1} and 1} == {a e Aiay = a(y)a for each y E AiJ,   Proof. It is evident that V=Z if o=D for some vG V. Hence we considerthe case a iLD for each vG U(V). Firstly, we note that V:= COL.   case 2el = O. Let v == xb + c (b, c E B). Then dv = vd for each d er B implyxdb + D(d)b + dc = xbd + cd and henceand D(d)b = cd - dc.   Thus,            D(b)b-O , (2)   Next, let us assume that v E 1}. Then 1} i a (v) - v = b yields bx = a(x)b ==(x + 1)b, and hence   By (2) and (3), we have b2 == O. Then 1+bE U(Z) by (1).   On the other hand, since o;-v for each ve U(V), U(Z)EC. Thus we obtainO=D(1+b)= D(b)=b. Therefore v=cEBnV==Z means that LEZ. Thus7) Cf. [7],Lemma 2.12 KAzuo KIsHIMoToV-Ce4-C[Z].    case 2el is inver sible. Let v = scb + c(b, c E B), Then dv -- vd for each d E Bimplies xp(d)b + dc = xbd ÷ cd, and hence                                                                    '             p(d)b ::= bd, cEZ ' (1)                                           'Thus             p(b)b-b2 (2)    Next, let us assume that vEL. Then 1} g 112(a (v) - v) == xb and xbx=x2p(b)=a(x)xb=-x2b, and hence .             p(b) -= -b (3)    By (2) and (3), we have p(b)b=b2=O. Thus (xb)2 :=: x2p(b)b=O, and hence 1- xb E U( V). Since U( V) g U(C), we have xb E 1} n C = O. Consequently, V == Ce 4 - C[Z].                                 References[1] S. U. CHAsE, D. K. HARRisoN and A. RosENBERG:Galois theory and Galois cohomo-   logy of commutative rings, Mem. Amer. Math. Soc., No. 52 (l965),[2]s3:lslll6. CO}IN : QUadratiC extensions of sl<ew fields, proc. London Math. soc., 11 (lg61),[3] G. J. JANusz:Separable algebras over commutative rings, Trans. Amer. Math. Soc.,   122 (l966), 461-479.[4] K. KismMoTo:Zeros of polynomials and Galois extensions of simple rings, J. Fac.   Sci. Shinshu Univ., 2 (1967), 117-122.[5] :Note on quadratic extensions of rings, J. Fac. Sci. Shinshu Univ.,5<1970),   25-28.[6] Y. MiyAsHiTA:Finite outer Galois theory of non-commutative rings, J. Fac. Sci.   Hokkaido Univ., I9 (1966), 114-134.[7] T, NAGAHARA : A quadratic extension, Proc. Jap. Acad., 47 (1971), 6-7.

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Note on Quadratic Extensions of Rings II

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