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Math Help - integral extensions

  1. #1
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    integral extensions

    Hi! Let A be a subring of a ring B which is integral over A. Then the following assertions hold:

    (i) B[X] is integral over A[X]
    (ii) Let b be an ideal of B and define a:=A \cap b, then B/b is integral over A/a.

    ad (i)
    Let f \in B[X] (polynomial with degree n). I want to show that (B[X])[f] is a finitely generated A[X]-module. I guess that 1,f,f^2,...,f^(n-1) generates (B[X])[f]. Is this the right way? If yes, is it the right execution?

    ad (ii)
    Here, i have the following idea: Let \overline{x} \in B/b and be x any representative of this equivalence class. As B is integer over A, there exists f \in A[X], so that f(x)=0. Let \pi be the canonical projection B \rightarrow B/b, then it follows that \pi(f(x))=\overline{0}, therefore \pi(f(x))=\sum_{i=1}^n \overline{a_i} \cdot \overline{x}^i=\overline{0}.
    But is this polynomial now in A/a[X]?

    Has somebody got answers or ideas?
    Banach
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  2. #2
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    Quote Originally Posted by Banach View Post
    Hi! Let A be a subring of a ring B which is integral over A. Then the following assertions hold:

    (i) B[X] is integral over A[X]
    you only need to show that for any f \in B[X], the ring A[X][f] is contained in a ring C \subseteq B[X] which is finitely generated A[X]-module. now let f(X)=\sum_{j=0}^nb_jX^j.

    since B is integral over A, the ring \overline{A}=A[b_0,b_1, \cdots, b_n] is finitely generated A-module. let \overline{A}=\sum_{j=0}^mAc_j. clearly \forall k : f^k \in \overline{A}[X]. thus: A[X][f] \subseteq \sum_{j=0}^mA[X]c_j. \ \ \ \Box

    (ii) Let b be an ideal of B and define a:=A \cap b, then B/b is integral over A/a.
    we have the ring monomorphism f: A/\bold{a} \longrightarrow B/\bold{b} defined by f(\alpha + \bold{a})=\alpha + \bold{b}. also note that we can define \forall \alpha \in A, \ \forall \beta \in B: \ (\alpha + \bold{a})(\beta + \bold{b})=\alpha \beta + \bold{b}. see that

    this is well-defined. now if \beta \in B, then \sum_{j=0}^n{\alpha_j}\beta^j=0, for some \alpha_j \in A. thus: \sum_{j=0}^n(\alpha_j + \bold{a})(\beta + \bold{b})^j=\sum_{j=0}^n(\alpha_j+\bold{a})(\beta^  j + \bold{b})=\sum_{j=0}^n\alpha_j \beta^j + \bold{b}=\bold{b}=0_{B/\bold{b}}. \ \ \ \Box
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  3. #3
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    Thank you for your helping answer and sorry for replying so late. Just one more question: What exactly do you mean by writing \sum_{j=0}^mAc_j.?
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  4. #4
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    I thought about it. Now everything is clear. Thank you very much.
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