What are basic examples for irreducible schemes which are not affine? What happens if I also demand the scheme to be Noetherian and/or locally Noetherian?
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7Projective spaces?! – Martin Brandenburg Jun 30 '13 at 14:57
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@MartinBrandenburg why the question mark, the irreducibility of projective spaces over a field is not particularly surprising, is it? Or do you refer to their non-affineness? – May 13 '19 at 15:55
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I was just saying that this is the most obvious and well-documented answer. – Martin Brandenburg Dec 22 '19 at 14:34
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I think the plane minus the origin is an example, and it's Noetherian if you take it over a field:
It is covered by the open sets $V(x)^c$ ad $V(y)^c$ where $V(x) = \text{Spec} k[x,y]/(x)$ and similarly for $V(y)$. Taking their complements gives open sets in $\text{Spec}k[x,y] = \mathbb{A}_k^2$ and so about any point in the plane we have an open set which is an affine scheme (exercise II.2.1 in Hartshorne shows that $D(f) \simeq \text{Spec} A_f$ as schemes, so in this case, $V(x)^c = D(x) \simeq \text{Spec} k[x,y]_{x}$ and similary for $D(y)$).
This scheme is irreducible by example I.1.1.3 in Hartshorne since the plane minus a point is a nonempty open (complement of the origin which is closed as it is a variety cut out by $(x,y)$) subset of an irreducible space, $\mathbb{A}^2_k$.
Derek Allums
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Thanks. And I guess an easy way to see that it is not affine is to see that the sections of the structure sheaf are just $k[x,y]_x \cap k[x,y]_y = k[x,y]$, but the underlying space lacks a point, right? – Guest Jun 30 '13 at 14:42
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@Guest Exactly: if it were some $\text{Spec}(A)$ then we could recover the ring $A$ by taking global sections, but as you point out, the global sections are elements of $k[x,y]$, but the spectrum of this ring is $\mathbb{A}^2_k$, not what we have here. See also: http://math.stackexchange.com/questions/122821/mathbba2-not-isomorphic-to-affine-space-minus-the-origin – Derek Allums Jun 30 '13 at 15:18