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Suppose $X$ is a non-reduced finite type seperated projective scheme over a field $k$, can it happen that $\Gamma(X,O_X)=k$?

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Let $X=\textrm{Proj}(S)$ where $S=k[s,x_0,x_1]/(s^2)$ equipped with the standard grading: $s,x_0,x_1$ have degree one and $k=S_0$. The open sets $U_i=D_+(x_i)$ for $i=0,1$ cover $X$. Now let $f \in \Gamma(X,\mathcal{O}_X)$. We can write $f|_{U_0}=x_0^{-d}(s\cdot p_1+p_2)$ and $f|_{U_1}=x_1^{-e}(s\cdot q_1+q_2)$ where $p_1,p_2,q_1$ and $q_2$ are homogeneous polynomials in $x_0,x_1$ of degree $d-1,d,e-1$ and $e$. Thus we have $f|_{U_0 \cap U_1}=x_0^{-d}(s\cdot p_1+p_2)=x_1^{-e}(s\cdot q_1+q_2)$. This implies $p_1=q_1=0$ and $p_2=\lambda x_0^d$ and $q_2= \lambda x_0^e$ where $\lambda \in k$, thus $f=\lambda \in k$.

Therefore $\Gamma(X,\mathcal{O}_X)=k$.

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  • $\begingroup$ Details: the $U_i$ cover $X$ because as usual $D_+(s)$, $D_+(x_0)$ and $D_+(x_1)$ cover $X$ (because if $x_0,x_1,s\in\mathfrak{p}$ then $\mathfrak{p}$ contains the irrelevant ideal which is forbidden) but here $D(s)=\emptyset$ because $s$ is nilpotent hence all prime $\mathfrak{p}$ contains $s$. $\endgroup$ Commented Nov 25, 2019 at 8:44

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