Lemma 10.66.2. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\mathfrak p$ be a prime of $R$. The following are equivalent:
$\mathfrak p$ is weakly associated to $M$,
$\mathfrak pR_{\mathfrak p}$ is weakly associated to $M_{\mathfrak p}$, and
$M_{\mathfrak p}$ contains an element whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$.
Proof. Assume (1). Then there exists an element $m \in M$ such that $\mathfrak p$ is minimal among the primes containing the annihilator $I = \{ x \in R \mid xm = 0\} $ of $m$. As localization is exact, the annihilator of $m$ in $M_{\mathfrak p}$ is $I_{\mathfrak p}$. Hence $\mathfrak pR_{\mathfrak p}$ is a minimal prime of $R_{\mathfrak p}$ containing the annihilator $I_{\mathfrak p}$ of $m$ in $M_{\mathfrak p}$. This implies (2) holds, and also (3) as it implies that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$.
Applying the implication (1) $\Rightarrow $ (3) to $M_{\mathfrak p}$ over $R_{\mathfrak p}$ we see that (2) $\Rightarrow $ (3).
Finally, assume (3). This means there exists an element $m/f \in M_{\mathfrak p}$ whose annihilator has radical equal to $\mathfrak pR_{\mathfrak p}$. Then the annihilator $I = \{ x \in R \mid xm = 0\} $ of $m$ in $M$ is such that $\sqrt{I_{\mathfrak p}} = \mathfrak pR_{\mathfrak p}$. Clearly this means that $\mathfrak p$ contains $I$ and is minimal among the primes containing $I$, i.e., (1) holds. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: