Algebra II 13ed. by Armando Rojo

By Armando Rojo

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If a was left invertible, one would have e = a−1 a ∈ I, so that I would not be proper. Conversely, assume that a is not left invertible. Consider the left ideal I given by I := {ba : b ∈ A}. Then e ∈ / I, so that a lies in the proper left ideal I. (iii) follows from the fact that the set of left invertible elements of a unital Banach algebra is an open neighbourhood of the unit, cf. 4). (iv) follows from (iii). 7). Ä ÑÑ º A proper left ideal in a unital algebra is contained in a maximal left ideal.

Since e + a and e + b are invertible, we may define c := a(e + a)−1 , d := b(e + b)−1 . The Rational Spectral Mapping Theorem yields rλ (c) < 1 and rλ (d) < 1. The preceding proposition then gives rλ (cd) < 1, so e − cd is invertible. We now have (e + a)(e − cd)(e + b) = e + a + b, so that e + a + b is invertible. 8). ÈÖÓÔÓ× Ø ÓÒº Let A be a Hermitian Banach ∗-algebra. Let a = a∗ , b = b∗ be Hermitian elements of A. We then have rλ (a + b) ≤ rλ (a) + rλ (b), whence also | rλ (a) − rλ (b) | ≤ rλ (a − b) ≤ | a − b |.

Let J := C. It shall be shown that J ∈ Z. The left ideal J is proper. Indeed one notes that J does not contain the unit of A, because otherwise some element of Z would contain the unit. Now Zorn’s Lemma yields a maximal element of Z. 8). Ä ÑÑ º Let I be a proper two-sided ideal in a unital algebra A. The unital algebra A/I then is a division ring if and only if I is both a maximal left ideal and a maximal right ideal. § 14. MULTIPLICATIVE LINEAR FUNCTIONALS 45 Proof. Please note first that A/I is a unital algebra since the twosided ideal I is proper, cf.

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