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  1. Let $X$ be a locally Noetherian scheme without embedded point, show that $X$ is reduced if and only if it is reduced at the generic points.

  2. Let $X$ be a locally Noetherian scheme (maybe has some embedded points), do we have $X$ is reduced if and only if it is reduced at the associated points?

Question (1) is from Liu Qing's book "Algebraic Geometry and Arithmetic Curves" exercise 7.1.2. If possible, I want to see a global proof, do not reduced to the affine scheme please. I guess there are some geometric meanings, maybe you can help me to point it.

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Yes to both questions, but I'm not sure what you mean by wanting a global proof. Reducedness is a local property! The proof will consist of picking a point $x \in X$ and an affine chart $U = \text{Spec}(R)$ containing $x$, then checking reducedness on that chart.

In particular, in the affine case, if $R$ is reduced, then all its localizations are reduced. On the other hand, if $R$ is nonreduced, let $f \in R$ be nilpotent. The annihilator $\text{Ann}(f)$ is contained in some associated prime ideal $P$ (this is a defining property of associated prime ideals), hence $f/1 \in R_P$ is nonzero, so $R_P$ is also nonreduced.

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    $\begingroup$ Thanks. I means that maybe we can do not use local method directly, set a global lemma first, then use the lemma instead of reducing to affine scheme. In fact, Liu's book has a hint to Lemma 1.9, which says O$_X$→i$_*$O$_U$ is injective iff AssO$_X$$\in$U, but I do not find a way to use it. $\endgroup$ – Strongart Jul 14 '14 at 13:56
  • $\begingroup$ Well, checking injectivity is still probably easiest using (distinguished) affine charts, so that the map $O_X \to i_*O_U$ becomes the ring map $R \to R[1/f]$, for some ring element $f$. I'm not sure how to phrase anything in terms of associated points without at some point referring to rings and ring elements, though. $\endgroup$ – Jake Levinson Jul 14 '14 at 14:05
  • $\begingroup$ OK, let us consider the affine situation. For the qusetions 1), we can use that lemma 2.4.11 which says generic point correspondings the minimal prime ideal. But how to do the question 2), what does the associated points (or embedded points) corresponding? $\endgroup$ – Strongart Jul 16 '14 at 6:31
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    $\begingroup$ Associated points correspond to annihilators $Ann(f)$ for $f \in R$ - namely, they are maximal among all such ideals. This is the reason I could say that $Ann(f)$ is contained in some associated prime ideal in my answer to 2) above. $\endgroup$ – Jake Levinson Jul 16 '14 at 23:02
  • $\begingroup$ Very helpful, thank you. $\endgroup$ – Strongart Jul 18 '14 at 6:05
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An addendum: we can actually use the lemma in Liu and refrain from making too many algebraical observations. Suppose that $X$ is reduced at all its associated points. Take an open $U$ with Ass$(\mathcal{O}_X) \subset U$ (this is possible because the locus where a locally Noetherian scheme is reduced is open; see Liu exercise 2.4.9 or here, in the comments). Now the map

$$\mathcal{O}_X \longrightarrow i_*\mathcal{O}_U$$

is injective by lemma 7.1.9 in Liu. Let $V \subset X$ be open. Then the map

$$\mathcal{O}_X(V) \longrightarrow \mathcal{O}_X(U \cap V)$$

is injective and therefore maps nilpotents to nilpotents, but $\mathcal{O}_X(U \cap V)$ has no nilpotents, so $\mathcal{O}_X(V)$ is reduced. It follows that $X$ is reduced.

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