By Susumu Oda

ISBN-10: 1584888512

ISBN-13: 9781584888512

Even if there are various forms of ring extensions, uncomplicated extensions haven't begun to be completely explored in a single e-book. protecting an understudied point of commutative algebra, basic Extensions with the minimal measure kin of critical domain names offers a entire remedy of assorted easy extensions and their houses. particularly, it examines numerous houses of easy ring extensions of Noetherian necessary domain names. As specialists who've been learning this box for over a decade, the authors current many arguments that they've constructed themselves, frequently exploring anti-integral, super-primitive, and ultra-primitive extensions. inside this framework, they research sure houses, reminiscent of flatness, integrality, and unramifiedness. the various themes mentioned comprise Sharma polynomials, vanishing issues, Noetherian domain names, denominator beliefs, unit teams, and polynomial jewelry. featuring a whole therapy of every subject, uncomplicated Extensions with the minimal measure kinfolk of imperative domain names serves as a fantastic source for graduate scholars and researchers all in favour of the realm of commutative algebra.

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**Extra resources for Simple Extensions with the Minimum Degree Relations of Integral Domains**

**Example text**

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 ].

### Simple Extensions with the Minimum Degree Relations of Integral Domains by Susumu Oda

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