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An paragraph from Vakil’s book. Here $U$ is plane minus the origin.

I know this question has appeared on this site. But I really get stuck by an step which is essentially the same as the above one from Vakil’s book. The only thing I can’t see is the reason why the prime ideal $(x,y)$ of $k[x,y]$ should cut out a point of $U$ if $U \cong \mathbb{A}^2_k$, which is used to derive an contra diction.

Could you help me? Thanks in advance.

A little edit: It seems that in scheme $U$ we could talk about points cut by ideals, as in scheme $\mathbb{A}^2_k$, but why these two ways of cutting should cut the same number of points? I couldn’t see why...

Mugenen
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  • Well, by assumption $U$ is affine, so $U = \operatorname{Spec}(A)$ for a ring $A$. But this precisely means that a prime ideal of $A$ corresponds to a point of $U$. In other words: $V(x,y) \subseteq U$ is non-empty. – Claudius Dec 16 '19 at 17:55
  • @Claudius But we assume $U$ is just isomorphic to $\mathbb{A}^2$ rather than an real spectrum, why this bijection is still available in $U$? – Mugenen Dec 17 '19 at 02:11
  • $X=\Bbb{A}^2_k- (0,0) = Spec(k[x,y,x^{-1}])\cup Spec(k[x,y,y^{-1}])$ is an integral scheme so if it was affine it would be the $Spec$ of its ring of regular functions $\bigcap_U O_X(U)$ which is $k[x,y]$. Also removing $(0,0) $ from $\Bbb{A}^2_k$ adds a new morphism $(x,y)\to [x:y]$ to $\Bbb{P}^1_k$ – reuns Dec 17 '19 at 03:28
  • @reuns Yes I could see that as a set of prime ideals $U$ can’t be $Spec(k[x,y])$, but why can’t they be isomorphic as schemes? – Mugenen Dec 17 '19 at 03:45
  • Here are several proofs. – Georges Elencwajg Dec 17 '19 at 07:43

1 Answers1

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Like he says: there is bijection between prime ideals of $k[x,y]$ and points of $U$. Since $(x,y)$ is a maximal ideal, it corresponds to a closed point—although you don't need to know that the point is closed.

Since $V(x,y)$ is empty, by the constructive nature that he mentions, we should be able to recover $(x,y)$ as $I(V(x,y)) = I(\varnothing) = k[x,y]$, which is a contradiction.

Trevor Gunn
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  • But we assume $U$ is just isomorphic to $\mathbb{A}^2$ rather than an real spectrum, why this bijection is still available in $U$? – Mugenen Dec 17 '19 at 02:10
  • @likemath I probably should have wrote $U$ there but the bijection is the same. – Trevor Gunn Dec 17 '19 at 03:27
  • Sorry to bother again, I’m still confused. Suppose $B$ is any affine scheme, then we could talk about the bijection between $B$ and the ring of its global section, but through the isomorphism between $B$ and some ring spectrum $Spec(A)$, right? But here it seems that we talk about the bijection directly rather than through an isomorphism $U \cong \mathbb{A}^2$, then why these two ways of bijections coincide? – Mugenen Dec 17 '19 at 03:54
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    @likemath Yes, but the global sections of $U$ is $k[x,y]$, not isomorphic to, equal to. Look at 4.4.1.1 again. – Trevor Gunn Dec 17 '19 at 04:03
  • Actually I see that, but I’m still confused at some point...So given an ideal of $k[x,y]$, we could use it to cut out points in $U$ directly, we could also cut out points in $\mathbb{A}^2$ firstly, and then by the isomorphism $U \cong \mathbb{A}^2$ get the points in $U$. I can’t see why finally we get the same point.. – Mugenen Dec 17 '19 at 04:26
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    If $U$ is affine, then $U$ is the Spec of its ring of global sections. So $U = \operatorname{Spec}(k[x,y])$ where $\pmb x$ and $\pmb y$ are the coordinate functions on $\pmb U$. Then, $V(x) \cap V(y)$ is empty because there is no point of $U$ whose first and second coordinates are simultaneously $0$. Therefore the ideal $(x,y)$ does not correspond to a point of $U$. @likemath The bold part is what's covered in 4.4.1.1 – Trevor Gunn Dec 17 '19 at 04:29
  • Oh...Finally I see that...It seems that I didn’t think valuation of a function should behave equally under isomorphism....Anyway thanks a lot. – Mugenen Dec 17 '19 at 04:47
  • @TrevorGunn Hi, Trevor. Thanks for the answer; I think that I was confused for a similar reason as the author of the post. To check I understand this well enough, could I confirm that, intuitively, the only "difficult" part of the proof is determining the global sections of $U$, and given that fact, for someone who understands the reasoning, the conclusion that $U$ is not affine should be immediately obvious? – legionwhale Dec 11 '23 at 03:01