Affine local property of reducedness

Question

Suppose $X$ is a scheme, and $\operatorname{Spec}A$ is an affine open sunset of $X$ with $(f_1\dots f_n)=A$. Then $\bigcup_{i=1}^n \operatorname{Spec}A_{f_i}=\operatorname{Spec}A$. Further assume that each $\operatorname{Spec}A_{f_i}$ is reduced, where we define reduced to mean that $\mathcal{O}_{\operatorname{Spec}A}(U)$ has no nonzero nilpotents for every open subset $U$ of $\operatorname{Spec}A$. I'd like to understand why this means that $\operatorname{Spec}A$ is also reduced.

So let $U$ be an open subset of $\operatorname{Spec}A$. Then $U$ is the union of some of the $\operatorname{Spec}A_{f_i}$. Then perhaps we could look at the restriction map $\mathcal{O}_{\operatorname{Spec}A}(U)\rightarrow \mathcal{O}_{\operatorname{Spec}A}(\operatorname{Spec}A_{f_i})$. However, I am not sure where to go from here. If we have some nilpotent $f\in\mathcal{O}_{\operatorname{Spec}A}(U)$, then this restricts to an element in a reduced ring.

Answer 1

Exercise: $X$ being reduced (in the sense that $\mathcal{O}_X(U)$ is a reduced ring for all open $U\subset X$) is equivalent to $\mathcal{O}_{X,x}$ being reduced for all points $x\in X$.

Hint/Solution outline:

For $\Rightarrow$, consider what would happen if you had a nonzero nilpotent in the stalk and derive a contradiction. For $\Leftarrow$, embed $\mathcal{O}_X(U)$ in the product of it's stalks.

With this in hand, since the stalk at $\mathfrak{p}\in\operatorname{Spec} A$ is the same as the stalk at $\mathfrak{p}_f \in\operatorname{Spec} A_f$, we have the result immediately.