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To prove $X$ is not affine, let us first prove that the global sections $\Gamma(X,\mathcal{O}_X)$ is isomorphic as a $k$-algebra to the polynomial ring in one variable $k[x]$. By definition of gluing, $\Gamma(X,\mathcal{O}_X)$ consists of pairs $\langle s_1,s_2\rangle$ where $s_1\in\Gamma(X_1,\mathcal{O}_{X_1})\simeq k[x]$ and $s_2\in\Gamma(X_2,\mathcal{O}_{... 0 (1) Note that the closure of a point$x \in X$is a closed subset of$X$. So if you show that every closed subset of$X$contains a closed point, then in particular the closure of$x$contains a closed point. So strictly speaking you're actually proving something stronger than what the exercise wants, but phrasing it this way makes it less confusing, I think.... 3 In an integral scheme, you can check that all restriction maps are injective. Thus, it follows that if$U \subset X$is an open subset covered by the affine open subsets$V_i$,$\mathcal{O}(U) = \bigcap_i\mathcal{O}(V_i)$(as subrings of$K$) by the sheaf property. So it is enough to show the statement for affine open subsets$U\$. In other words, we just ...