ON FREE RING EXTENSIONS OF DEGREE N

Nagahara and Kishimoto [1] studied free ring extensions B(x) of degree n for some integer n over a ring B with 1, where xn=b, cx=xρ(c) for all c and some b in B(ρ=automophism of  B), and {1,x…,xn−1} is a basis. Parimala and Sridharan [2], and the author investigated a class of free ring extensions called generalized quaternion algebras in which b=−1 and ρ is of order 2. The purpose of the present paper is to generalize a characterization of a generalized quaternion algebra to a free ring extension of degree n in terms of the Azumaya algebra. Also, it is shown that a one-to-one correspondence between the set of invariant ideals of B under ρ and the set of ideals of B(x) leads to a relation of the Galois extension B over an invariant subring under ρ to the center of B.

ring B with i with a basis {i, x} such that x xa + b for some a and b in B, and cx x0(c) for each c in B, where p is a ring automorphism of B of order 2. (2) B(x), a free ring extension of degree n > 2 is similarly defined with a basis n-i x n {i, x, x }, and b for some b in B and cx xp(c) for each c in B, where 0 is of order n.Some special free ring extensions called generalized quaternion algebras were investigated by Parimala and Sridharan [2] and the author Szeto ([4] [5]).One of their results is a characterization of the Galois extension of B over a subring ([2], Proposition 1.1): Let B(x) be a generalized quaternion algebra (x 2 -I) over a cummutative ring B with 2 a unit in B. Then B is Galois over A (={a in B/0(a) a for an automorphism 0 of order 2}) if and only if BAB(X) is a matrix algebra of order 2. The above characterization was generalized to a free ring extension of degree n, B(x) with x n -I ([4], Theorems 3.4 and 3.5}.Te purpose of the present paper Is to continue the above general- ization to a free ring extension.Also, we shall show that there is a one-to-one correspondence between the set of invariant ideals of B under O and the set of ideals of B(x).This correspondence will lead to a relation of the Galois extension B over the invariant subring A under 0 to the center Z of B over A.

PRELIMINARIES.
Throughout, we assume that B is a ring (not necessarily cummutative) with i, p an automorphism of B of order n for some positive integer n, A {a in B/0(a) a}, n-l} and B(x) a free ring extension over B with a basis {i, x, Let T be a ring containing a subring R with i.Then T is called a separable extension over R if there exist elements (ui, v i / i i, m for some integer m} such that a (uivl) (luivi)a for all a in T where @ is over R and .uivi 1 ([6], [7]).Such an element luivi is called a separable idempotent for T. If R is in the center of T, the separable extension T is called a separable R-algebra.
In particular, if R is the center of T, the separable R-algebra T is called an Azumaya R-a!geb.r a (16], [7]).A commutative ring extension S of R is called a splitting ri for the Azumaya R-algebra T if ST HOms(P,P for a progenerator S-module P ([6], [7]).The ring extension T over R is called a Galois extension with a finite automorphism group G (Galois group) if (I) R {a in T / u(a) a for all u in G}, and (2) there exist elements {ui, v i in T / i I,..., m for some integer m} such that (i) .uivi i and (2) .uiu(vi) 0for each u the identity of G ([7], [8]).
In this section, we shall generalize the Parimala-Sridharan [2] theorem to a free ring extension B(x) of degree n for an integer n such that x n b and ax xp(a) for some b and all a in B where p is an automorphism of B of order n.
We note that if B(x) is separable over B then b is a unit ([i], Proposition 2.4).
The converse holds if n is also a unit: LEMMA 3.1.If n and b are units, then B(x) is a separable extension over B. n PROOF.Since b is in A ([I], p. 20) and s-ince p the Identity, it is straightforward to verify that the element u b-i n-i (i=ox.n-iixn-i) satisfies the equations: au ua for all a in B(x), and b -I n -I (xix n-i) i, where @ is over B.
We remark here that there are separable extensions with n (--2) not a unit ([4], Theorem 4.2).With the same proof as given for Proposition 1.2 in [7] we have a characterization for Galois extensions of non-commutative rings: LEMMI 3.2.Let B be a ring extension of A with a finite automorphism group a for each in G}).Then B is Galois over A if and only if the left Ideal generated by {a-u(a) / for a in B} B for any the identity of G.
THEOREM 3.3.Let n and b be units in B. If B is Galois over A which is con- n-i tained in the center, Z of B with a Galois group {i, 0, .,0} of order n, n then the free ring extension B(x) of degree n is an Azumaya A-algebra, where x b and cx xp(c) for each c in B.
PROOF.By Lemma 3.1, B(x) is separable over B. Since B is Galois over A, B is separable over A. Hence B(x) is separable over A ( [4], the proof of Theorem 3.4).

G. SZETO n-i
So, lt suffices to show that the center of B(x) is A. Let u a lx be in the i--O n-i center.Then xu ux.Noting that {l,x ,x is a basis for B(x) over B, we have that a i are in A. Also, au ua for all a in B, so a i(a-pi(a)) 0 for each i O. Hence the central elements a i are in the left annihilators of the left ideal generated by {a-pi(a) / a n B} for i # O.By hypothesis, B is Galols over A, so a_ 0 for each i # 0 by Lemma 3.2.Thus u a in A. Clearly, A is in the cen- o ter of B(x).Therefore, A the center of B(x).
By the Parimla-Sridharan theorem ([3], Proposition i.I), let B(x) be a gen- eralized quaternion algebra (x 2 =-i) over a commutative ring B. Then, B is Galois over A (= {a in B / (a) a for an automorphism 0 of order 2}) if and only if BHAB(X) is a matrix algebra of order 2 over B. Hence, Theorem 3.3 generalizes the necessity of the Parlmala-Sridharan theorem.
For the sufficiency, we first give a one-to-one correspondence between the sets of ideals of B, of B(x), of A, and the center Z of B. An ideal I of B is called a G-ideal if (I) Z, a G-ideal J of Z is similarly defined, where G i,0,...,0 }.THEOREM 3.4.Let B(x) be an Azumaya A-algebra.Then there exists a one-to- one correspondence between (i) the set of G-ideals of B, (2) the set of ideals of B(x), and (3) the set of ideals of A.
PROOF.At first, we want to give a structure of a G-ideal I of B. Since -i I, xlB(x> c 0 (1)B(x) IB(x).Hence IB(x) is an ideal of B(x).By hy- pothesls, B(x) is an Azumaya A-algebra, so IB(x) I B(x) where I IB(x) n A o o n-i ( [7], Corollary 3.7, p. 54).Noting that {l,x,...,x } is a basis for B(x) over B, we have I I B and I I n A. Next, it is easy to see that J B is a G-ideal o o o of B for any ideal J of A. Thus the set of G-ideals of B are in one-to-one corre- o spondence with the set of ideals of A from the above representation I B of a G- o ideal I of B. By hypothesis again, B(x) is an Azumaya A-algebra, so the set of ideals of B(x) and the set of ideals of A are in one-to-one correspondence under IoB(X)++l.for an ideal I of A. Thus the theorem is proved.
o o COROLLARY 3.5.Let n and b be units in B. Suppose B is Galols over A which is contained in Z. Then there exists a one-to-one correspondence between the set of G-ideals of Z and the set of ideals of B(x). ,} obtained from G (= {i , }) by taking m as the minimal integer such that 0m the identity on Z.We denote the ideal generated by {a-oi(a) / a in Z} by I i for i l,...,m-l.It is easy to see that each I i is a G-ideal such that Ira_ 1 c Ira_ 2 c c I I.We shall show that the chain of li's characterizes the Galols extension of Z over A. That is: THEOREM 3.6.If B(x) is an Azumaya A-algebra such that I 1 12 Ira_l, then Z is Galois over A with a Galols group G'.
PROOF.In case Z A the theorem is trivial.Let Z # A. Then m # 0.
Clearly, A BG Z G'.Now we assume Z is not Galols over A. Then the ideal I 1 of Z is not Z ([7], Proposition 1.2, p. 80) since I 1 12 Ira_ 1 by hypothesis.Since I 1 is a G-ideal, I 1 IZ for some ideal I of A by Theorem 3.4.

46). But
() in B(x)/llB(X for each a in Z, so R '--.This implies that Z is contained in the center A/I of the Azumaya A/l-algebra A/IQAB(X).This is impossible since Z is not contained in A. Thus Z is Galois over A. COROLLARY 3.7.By keeping the notations of Theorem 3.6, if B is Galois over n-i A with a Galols group G (= {l,p,...,o }) such that I 1 12 Ira_l, then Z is Galois over A with a Galols group G', where b and n are units in B.
PROOF.Theorem 3.3 implies that B(x) is an Azumaya A-algebra, so the corollary is a consequence of Theorem 3.6.
As given in Theorem 3.6, let B(x) be an Azumaya A-algebra.If B is commutative B Z. Now assume B is not Galois over A. Then there is an I i for some i l,...,m-I such that I i # Z.One can show as given in Theorem 3.6 that A/IIQAB(x is an Azu- maya algebra such that x i is in the center A/I i.
Thus we have a contradfctlon.
This proves that B is Galols over A. So, Theorem 3.6 generalizes Theorems 3.4 and and 3.5 in 4]. 4.

SPLITTING RINGS.
In this section, we shall show that if B(x) is an Azumaya A-algebra in which b and n are units, then A(x) is a splitting ring for B(x) such that A(x) is a chain of Galois extensions of degree 2 (that is, A(x) A(x2) . A(xn) A, such that A(xi) is Galois over A(x21)).PROOF.We define a mapping =: A(x)/A(x) by e(x) -x and a(alxl )" lai(e(x))i for i 0,i, n-l.Then it is straightforward to check that s is an automorphism of A(x) of order 2 such that (A(x)) A(x2).Since m-i 2 n (= 2 m 2.2m-l) and b (= x n (x2) 2 are units in A, 2 and x are units in A(x2).Now we claim that A(x) is Galols over A(x2) with a Galols group {I,}.
In fact, let a.] (2x2)-ix, a 2 2-1, b I x and b 2 I. Then we have albl+a2b2 i and al(bl)+a2(b2) 0. Thus A(x) is Galois over A(x2) of degree 2. Similarly, we can show that A(x2) is Galois over A(x4) with a Galois group {1,8} with (x 2 2 8 =-x of order 2. Therefore, an induction argument concludes the existence of a chain of Galois extensions of degree 2. For the class of free ring extensions B(x) of degree n as given in [i], Sec- i) n tlon 2 such that c and (l-c are units in A where c i and i 1,2,...,n-l, we have: THEOREM 4.2.Let A be a commutative ring with i, x n b which is a unit in A, and ax xa for each a in A. If there is an c in A such that n and (l-ci) are units in A for i l,...,n-i with c n i, then A(x) is Galols over A.
PROOF.We define a mapping : A(x)A(x) by u(x) cx and s([.alxl [ai(cx) i.
Then one can check that (A(x)) A and that s is an automorphlsm of A(x) of order i n (for l-c are units in A for i 1,2,...,n-l).Moreover, since (l-c) is a unit in A, (x-u(x)) x-cx (l-c)x is also a unit (for x is also a unit).Therefore, A(x) is Galois over A with a Galois group {i, ,... }( [7], Proposition 1.2, p. 80).
for some b and all a in B (hence o(b) b ([i], p. 20)).
PROOF.Since B is Galols over A, B is a separable A-algebra.Hence B is Azumaya over its center Z ([7], Theorem 3.8, p. 55).Thus the set of G-ideals of B and the set of G-ideals of Z are in one-to-one correspondence; and so Theorem 3.4 implies the corollary.Now we show a generalization of the sufficiency of the Parimala-Srldharan theorem.The set {a in B / p(a) a} is denoted by B p. Let G' be an automorphlsm

THEOREM 4 .
1. Let A be a commutative ring with I, x n b in A, and ax xa for each a in A. If b and n are units in A with n a power of 2 (-2 m for some m), then A(x) is a chain of Galols extensions of degree 2.