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Let $x\in X$ be a general point of an irreducible set s.t. {$\bar x$} is (locally) the support of some section $s$ of a sheaf of $\mathcal {O}_X$-modules $\mathscr F$. Then why does it follow that $s$ annihilates every element in $\mathfrak m_x$ , i.e. that $x$ is an associated point of $\mathscr F$? (cf. Prop 1.2.)

I have a feeling that these concepts of associated and general point are related but I don't understand it quite yet. Is an associated point always a general point of some irreducible set?

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What is the definition of a general point and a general point of an irreducible set? Without the condition "general", this may not be true. Let $R = \mathbb{C}[[X]]$ and $X = \hbox{Spec} (R)$. Let $x \in X$ be the point corresponding to the maximal ideal $(X)$ of $R$. Then $\{ \bar{x} \} = \{ x \}$, and this is the support of $Y_n = \hbox{Spec} (R / (X^n))$ for $n \ge 1$. When $n > 1$, $(X)^n$ annihilates $R / (X^n)$, but not $(X^{n-1})$. Here $1 + (X^n)$ can be taken as the section $s$.

I believe that your question has a positive answer if you replace $\mathscr{F}$ by $s\mathscr{O}_X$ (I'm not sure if this is the right notation) and $\mathfrak{m}$ by some power of $\mathfrak{m}$ provided that things are Noetherian. Since $x$ is minimal in the support, $\dim_{\mathscr{O}_{X,x}} (s\mathscr{O}_x) = 0$. Therefore $\sqrt{\hbox{ann}_{\mathscr{O}_{X,x}} (s\mathscr{O}_x)} = \mathfrak{m}_x$.

However, I have no intuition about the word "general".

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  • $\begingroup$ Okay, but what do you mean by {$\bar x$} being the support of $Y_n = \hbox{Spec} (R / (X^n))$? So far I have only heard of a subset of a scheme being the support of a section, or of a sheaf (of modules, function, etc.), not the support of a subscheme. Furthermore, if I am not wrong, $Y_n$ has many more points than just {x}. In fact, all it contains all irreducible polynomials of degree up to n-1, i.e. all linear polynomials, since $\mathbb{C}$ is algebraically closed. And 1 + ($X^n$) vanishes only on the irreducible factors of 1 + ($X^n$), i.e. its support is much larger than only {x}. $\endgroup$
    – Rodrigo
    Jun 28, 2013 at 6:59
  • $\begingroup$ There are only two point in $X$. The generic point which corresponds to the ideal $(0)$ and a closed point which corresponds to the maximal ideal $(X)$. The ring $\mathbb{C}[[X]]$ is a DVR. Maybe it will help me to understand your comment if you can provide me an irreducible polynomials of degree two when $n = 3$. $\endgroup$
    – Youngsu
    Jun 28, 2013 at 14:52

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