Associated points and reduced scheme

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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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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.