SPLITTINGS OF EXTENSIONS OF BANACH ALGEBRAS 13

1.8. THEOREM. Let A be a [commutative] Banach algebra,

(i) Let m G N. Suppose that every [commutative] singular extension of A splits

strongly. Then every [commutative] nilpotent extension of A splits strongly.

(ii) Suppose that every [commutative] singular extension of dimension at most m splits

strongly Then every [commutative] extension of dimension at most m splits strongly

PROOF: (i) By Theorem 1.7, it suffices to prove that every nilpotent [commutative] ex-

tension of dimension at most m splits strongly.

The proof of this statement is by induction on the minimum index n G N such that

In = 0. The hypothesis gives the case where n = 2; we suppose that *%2(%L\I) is an

extension of A such that / has index k 3, and assume that the result holds for each

nilpotent [commutative] extension of dimension at most m and index at most k — 1. Set

J = I2 , a closed ideal of 21. Then the inductive hypothesis applies to each [commutative]

extension of A by J and to the extension 52 (21/J; I/J), and so each of these extensions

splits strongly. By Proposition 1.5(i), every [commutative] extension of A by I splits

strongly, continuing the induction.

(ii) This is the same proof, save that now we set J = T2, and note that Jk~l = 0. •

Note that we have no result which proceeds from 'singular extension' to 'nilpotent

extension' in the case of algebraic splittings of extensions of a general Banach algebra; we

shall obtain a result in this direction in the case where A is commutative in §4. However,

we do have the following result, which will be used in §4.

1.9. THEOREM. Let A be a Banach algebra, and let 52(^5 -0 be an extension of A.

Suppose that I is the direct sum I = J®K of two closed ideals J and K in 21, and that

the extensions £(21/J; I/J) and £(2l/if; I/K) both split algebraically Then £(21; I)

splits algebraically.

PROOF: There exist subalgebras £j and CK , respectively, such that

21/ J = £jQK and 2l/if = ^ 0 J .

Define 03j = {a G 21 : a + J € €j} and 93K = {a € 21 : a + K G €K}, so that Q3j and

03K are subalgebras of 21. Then define

B = BjnB

K

,

so that 03 also a subalgebra of 21.

Take a €21. Then

a + J = 6 + J = c + I

for some b G 03j and c G 03K • Since b — c G / , there exist x G J and y G K with

b-c = x + y. We have 6 - x = c + y G 0 3 j n 03K = 03, and so 21 = 03 4- / .

Let x G 03 fl / . Then x + J G £j D K = {0}, and so x G J. Similarly, x G K, and so

x = 0. Thus 0 3 n / = 0.

We have shown that 21 = 03 © / , and so XX^5 I) splits algebraically. •