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If $K$ is a field, $\mathbb A^2_K=\textrm{Spec}K[X,Y]$ and $U=\mathbb A^2_K\setminus\{(X,Y)\}$, I want to prove that $(U,\mathscr O_{\mathbb A^2_K|U})$ is not an affine scheme. I know that this topic has been handled many times, but I want to point a specific passage in the language of schemes.

I have proved that $\mathscr O_{\mathbb A^2_K|U}(U)=\mathscr O_{\mathbb A^2_K}(\mathbb A^2_K)=K[X,Y]$. Now suppose that $(U,\mathscr O_{\mathbb A^2_K|U})$ is an affine scheme, there is a ring $R$ such that $(U,\mathscr O_{\mathbb A^2_K|U})\cong (\textrm{Spec }R,\mathscr O_{\textrm{Spec}R})$; by the equivalence between the opposite category of rings and the category of affine schemes we have that $R\cong \mathscr O_{\mathbb A^2_K|U}(U)=K[X,Y]$. The conclusion is that $(U,\mathscr O_{\mathbb A^2_K|U})\cong (\mathbb A^2_K,\mathscr O_{\mathbb A^2_K})$ but I don't understand where is the contraddiction. Why $\mathbb A^2_K$ and $U$ shouldn't be homeomorphic (look here for a similar question)? Even if they are homeomorphic (I don't think so) why there shouldn't exist an isomorphism as locally ringed spaces?

The fact that the open embedding is not an isomorphism doesn't imply that there is no isomorphisms between the two schemes.

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You've misunderstood the argument. Let's recall the fundamental theorem of affine schemes:

Theorem.

  1. For all commutative rings $A$, there is a natural isomorphism $A \cong \Gamma (\operatorname{Spec} A, \mathscr{O}_{\operatorname{Spec} A})$.
  2. For all locally ringed spaces $(X, \mathscr{O}_X)$ and all commutative rings $A$, there is a natural bijection between morphisms $(X, \mathscr{O}_X) \to \operatorname{Spec} A$ and ring homomorphisms $A \to \Gamma (X, \mathscr{O}_X)$.

Corollary. The functor $\operatorname{Spec} : \mathbf{CRing}^{\mathrm{op}} \to \mathbf{Sch}$ is fully faithful and has a left adjoint.

Now, let's apply this to the problem at hand.

Lemma. If $X$ and $Y$ are affine schemes, then a morphism $f : X \to Y$ is an isomorphism if and only if the ring homomorphism $f^* : \Gamma (Y, \mathscr{O}_Y) \to \Gamma(X, \mathscr{O}_X)$ is an isomorphism.

Proof. If $f : X \to Y$ is an isomorphism, then $f^* : \Gamma (Y, \mathscr{O}_Y) \to \Gamma(X, \mathscr{O}_X)$ must also be an isomorphism, by functoriality. Conversely, if $f^*$ is an isomorphism, say with inverse $g^* : \Gamma (X, \mathscr{O}_X) \to \Gamma(Y, \mathscr{O}_Y)$, then the fundamental theorem implies there is a unique scheme morphism $g : Y \to X$ inducing $g^*$, and moreover $g$ must be a two-sided inverse for $f$. ◼

Corollary. If $X$ is a scheme, $f : X \to \operatorname{Spec} A$ is a morphism but not an isomorphism, and $f^* : A \to \Gamma (X, \mathscr{O}_X)$ is an isomorphism, then $X$ is not affine.

Proof. This is the contrapositive of the above lemma. ◼

Of course, in the case of interest, $X = U$, $A = K[X, Y]$, and $f : X \to \operatorname{Spec} A$ is the inclusion. This is certainly not an isomorphism: it is not even a bijection on points!

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  • $\begingroup$ Are we considering the restriction homomorphisms as the inclusions in Frac$(K[X,Y])$? If this is true then the embedding of $(U,\mathscr O_{\mathbb A^2_k|U})$ in $(A^2_k,\mathscr O_{\mathbb A^2_k})$ induces the identity on global sections as you say. $\endgroup$ – Dubious May 19 '13 at 13:46
  • $\begingroup$ You can do that if you like, but that is not necessary. $\endgroup$ – Zhen Lin May 19 '13 at 14:44
  • $\begingroup$ You have cited the functions $f$ and $f^{\ast}$. If in our case $f^{\ast}$ is the identity on $K[X,Y]$, we should be sure that $f$ is really the embedding. In other words does the embedding of schemes induces the identity on the common ring of global sections? I think the answer is yes because the restrictions are the inclusions! $\endgroup$ – Dubious May 19 '13 at 15:12
  • $\begingroup$ No, you start with $f$ and then calculate $f^*$. $\endgroup$ – Zhen Lin May 19 '13 at 17:17
  • $\begingroup$ Ok, $f$ is the embedding, but is $f^*$ the identity? I'm sorry if I'm not clear. $\endgroup$ – Dubious May 19 '13 at 17:40

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