# Eisenbud Exercise on schemes

I was trying to solve exercise 1.24 part b) of Eisenbud's book on Schemes and to deduce from it an example of locally ringed space which is not an affine scheme. However I am having many problems in one computation

Consider the affine scheme (Spec($\mathbb C[x,y],O_{Spec\mathbb C[x,y]})$ and set $U=D(x)\cup D(y)$ and say $O$ is the restriction sheaf to $U$. I want to prove that $(U,O)$ is not an affine scheme. To do so, I would like to compute $O(U)$ which should be $\mathbb C[x,y]$ according to the text. How can I prove this? How can I deduce from this that the pair $(U,O)$ is not an affine scheme?

I mean, an element of $O(U)$ is a function from $D(x)\cup D(y)$ into the disjoint union of $\mathbb C[x,y]_{p}$ where $p$ is a prime ideal not containing $x,y$. But from this I really don't see an isomorphism to $\mathbb C[x,y]$. I tried to use the sheaf axioms but I failed.

Sections of $D(x)$ are elements of the localized rings: $\frac{p(x,y)}{x^n}$ for some $p \in \mathbb{C}[x,y]$ and $n \in \mathbb{N}$. Your sections of $U$ are rational functions that are sections of $D(x)$ and $D(y)$ at the same time... this is exactly $\mathbb{C}[x,y]$.
Now you use the correspondence between affine schemes and rings. The ring $\mathbb{C}[x,y]$ is already taken by $\mathbb{A}^2$, so there's no room for $D(x) \cup D(y) = \mathbb{A}^2 \backslash \{0\}$ to be affine.
• Ok but how can I prove that if $f\in \mathbb C[x,y,1/x]$ and $g\in \mathbb C[x,y,1/y]$ such that they coincides in $\mathbb C[x,y,1/(xy)]$ then they must be equal and belong to $\mathbb C[x,y]$ ? – Blacksmith Dec 2 '17 at 23:46
• @Blacksmith If they coincide then they're equal... there's no need to overthink this. The second question: sections of $D(x)$ are polynomial in $y$ for fixed $x$, and sections of $D(y)$ are polynomial in $x$ for fixed $y$; to be both at once they must be in $\mathbb{C}[x,y]$. – user16394 Dec 2 '17 at 23:56
• I'm a bit confused about the end of your answer: I mean we found out that $O(U)$ is isomorphic to $\mathbb C[x,y]$, not equal. So how exactly can we use the correspondence to conclude? – Blacksmith Dec 3 '17 at 0:57
• I mean, if for absurd we assume that $(U,O)$ is an affine scheme, then it means there is a pair $(Spec(R),O_{Spec(R)})$, for some commutative ring $R$, isomorphic to it as locally ringed space. By the correspondence, this induces an isomorphism to $(O_{Spec(R)}(Spec(R))\to O(U)$. Ho do we conclude by this? – Blacksmith Dec 3 '17 at 1:02
• @Blacksmith The correspondence is more precise; given a fixed morphism of rings (here it's the identity $\mathbb{C}[x,y] \rightarrow \mathbb{C}[x,y]$) in particular it tells you what the function on underlying sets is; here it has to be the inclusion. That's not a bijection since $D(x) \cup D(y)$ doesn't contain the origin. – user16394 Dec 3 '17 at 1:09