Lemma 26.12.2. A scheme $X$ is reduced if and only if $\mathcal{O}_ X(U)$ is a reduced ring for all $U \subset X$ open.

**Proof.**
Assume that $X$ is reduced. Let $f \in \mathcal{O}_ X(U)$ be a section such that $f^ n = 0$. Then the image of $f$ in $\mathcal{O}_{U, u}$ is zero for all $u \in U$. Hence $f$ is zero, see Sheaves, Lemma 6.11.1. Conversely, assume that $\mathcal{O}_ X(U)$ is reduced for all opens $U$. Pick any nonzero element $f \in \mathcal{O}_{X, x}$. Any representative $(U, f \in \mathcal{O}(U))$ of $f$ is nonzero and hence not nilpotent. Hence $f$ is not nilpotent in $\mathcal{O}_{X, x}$.
$\square$

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