Finitely generated torsion-free prime ideals

Posted: February 19, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

Problem. let  R  be a commuative ring and P  a finitely generated prime ideal of R. Suppose that \text{ann}_R(P) = 0. Prove that \text{ann}_R(P/P^2) = P.

Solution. Let P = \sum_{i = 1}^n Rx_i and suppose that a \in \text{Ann}(P/P^2). So aP \subseteq P^2 and thus ax_i = \sum_{j = 1}^nb_{ij}x_j, \ \ 1 \leq i \leq n, for some b_{ij} \in P. Write this system of equations as Ax = 0, where A is the n \times n matrix of coefficients and x = [x_1 \ x_2 \ \cdots x_n]^T. Then multiplying Ax = 0 from the left by \text{adj}(A) gives us: (\det A)x = 0 and hence \det A = 0, since \text{Ann}(P) = 0. But \det A = a^n + b, for some b \in P. So a^n \in P and hence a \in P. This proves that \text{Ann}(P/P^2) \subseteq P. The other side of the inclusion is trivial.

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Comments
  1. Abdulrasool Azizi says:

    Hi Yaghoub. Please answer to the e-mails I sent you, recently.

    According to Exercise 2.2 of the book {\bf Commutative ring
    theory, by H. Matsumura}, we have:

    If $M$ is a finitely generated $R$-module, then
    $\sqrt{ann(M/IM)}=\sqrt{ann(M)+I},$ where $I$ is an ideal of $R.$

    Your exercise is a particular case of the above result, where
    $M=I=P.$

    To prove Exercise 2.2, consider $a\in ann(M/IM)$ and define
    $\varphi:M\longrightarrow M,\; \varphi(m)=am.$ Now use Theorem 2.1
    of the same reference.

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