By I. Kaplansky

ISBN-10: 0821816500

ISBN-13: 9780821816509

An algebraic prelude Continuity of automorphisms and derivations $C^*$-algebra axiomatics and simple effects Derivations of $C^*$-algebras Homogeneous $C^*$-algebras CCR-algebras $W^*$ and $AW^*$-algebras Miscellany Mappings protecting invertible parts Nonassociativity Bibliography

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We have αφ(J ) ⊆ αm ⊆ m ⊆ R for any α ∈ m : K m, and hence α ∈ R(J −1 ). Thus m : K m ⊆ R(J −1 ). But since R(J −1 ) ⊇ m :k m⊃ R, we − have R = R(J −1 )⊃ R, which is a contradiction. 3 Let J be a divisorial integral ideal of R. If R(J −1 ) = R, then there exists a nonzero element α ∈ K such that J = Iα . Proof We first show the following claim: If p ∈ Dp1 (R) contains J , then p is a prime divisor of J . 2. As p R p is a prime divisor of a principal ideal a R p = J p (cf. √ [Y]), p is a prime divisor of J .

Now we consider a certain over-ring of R, which is seen in Chapter 1. Let J be a fractional ideal of R. Recall R(J ) := J : K J, which is an over-ring of R. 9 Let J be a divisorial ideal of R. Then R(J ) = R if and only if R(J −1 ) = R. Proof Since J is divisorial, we have (J −1 )−1 = J. So it suffices to prove one of the implications. Assume that R(J ) = R. The implication R(J −1 ) ⊇ R is obvious. Take λ ∈ R(J −1 ). Then λJ −1 ⊆ J −1 . Thus R : K λJ −1 ⊇ R : K J −1 = (J −1 )−1 = J. On the other hand, we have R : λJ −1 = λ−1 R : K J −1 = λ−1 (R : K J −1 ) = λ−1 (J −1 )−1 = λ−1 J.

Let α be an element of an algebraic field extension L of the quotient field K of R and let π : R[X ] −→ R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with deg ϕα (X ) = d d (R : R ηi ). For and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Let I[α] := i=1 f (X ) ∈ R[X ], let c( f (X )) denote the ideal generated by the coefficients of f (X ). Let J[α] := I[α] c(ϕα (X )), which is an ideal of R and contains I[α] . The element α is called an anti-integral element of degree d over R if Ker(π ) = I[α] ϕα (X )R[X ].

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