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For a scheme $X$ to be reduced there is a criterion that says that $X$ is reduced if there is a covering $(U_i)$ by affines open subsets such that for all $i$, $\mathcal{O}_X(U_i)$ is reduced. I guess that it doesn't works anymore when the $U_i$ are not affines. Does someone has a example.

I tried with the classical $\operatorname{Spec}(k[X]/(X^2))$ but all its open sets are affine... something like $\operatorname{Spec}k[x,y]/(x^2)$ maybe?

Thanks!

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    $\begingroup$ Since $X$ is an open cover of itself, it suffices to find a nonreduced scheme $X$ with reduced ring of global sections - this will imply that reducedness cannot be checked on global sections of an arbitrary cover. The answer here should be an example of this. $\endgroup$
    – Stahl
    Nov 25, 2019 at 8:09

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As Stahl says it suffices to have some non reduced scheme $X$ with global sections ring $\mathcal{O}_X(X)$ non reduce. The example given here is perfect: $X=\operatorname{Proj}k[s,x_0,x_1]/(s^2)$ is not reduce because with $$U=D_+(x_0)=\operatorname{Spec}\left(k\left(\frac{s}{x_0},\frac{x_1}{x_0}\right)/\left(\left(\frac{s}{x_0}\right)^2\right)\right)$$ one has $\mathcal{O}_X(U)$ not reduced but the global sections ring $\mathcal{O}_X(X)$ is reduced because $$\mathcal{O}_X(X)=k$$ se the link above.

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