In the previous post, we glue together affine spaces X_i = \text{Spec} A_i = \text{Spec} k\left[x_0, \ldots, x_n, \frac{1}{x_i} \right] to get \mathbb{P}^n(k).
To show that \mathbb{P}^n(k) is not affine, we first compute its global sections.
The global sections of \mathbb{P}^n are in bijective correspondence with tuples (s_0, \ldots, s_n) \in A_0 \times \cdots \times A_n that are compatible with respect to restrictions. However as restrictions A_i \to A_{ij} are simply inclusion maps, we have (s_i)_{X_{ij}} = (s_j)_{X_{ij}} iff s_i = s_j as elements of A_{ij}. So we have
s_0 = s_1 = \ldots = s_n \text{ as elements of } k(x_0, \ldots, x_n).
Thus the global sections \mathbb{P}^n is \bigcap_i A_i = k.
If \mathbb{P}^n(k) is affine, then it must then be \text{Spec} k. But if n >0, then \mathbb{P}^n(k) consists of more than one point. In fact, its closed points are in bijective correspondence with points of the classical projective Space.
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