## Finitely generated torsion-free prime ideals

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

1. Abdulrasool Azizi says:

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