# Exercise from Eisenbud & Harris's The Geometry of Schemes

I've just started learning about schemes, so maybe I'm missing something basic.

This is exercise I-24(a):

Take Z = Spec$\mathbb{C}[x]$, let $X$ be the result of identifying the two closed points (x) and (x-1) of |Z|, and let $\phi: Z \to X$ be the natural projection. Let $\mathcal{O}$ be $\phi_* \mathcal{O}_Z$, a sheaf of rings on $X$. Show that $(X, \mathcal{O})$ satisfies condition (i) above for all elements $f \in \mathcal{O}(X) = \mathbb{C}[x]$.

The condition (i) referred to: For any $f \in \mathbb{C}[x]$ define $U_f \subset X$ as the set of points $x \in X$ such that $f$ maps to a unit of the stalk $\mathcal{O}_x$. (i) means that $\mathcal{O}(U_f) = \mathbb{C}[x][f^{-1}]$ for all f.

But how can this be? Put f = x. Then

$U_f = X \setminus \{(x)\}$

$\phi^{-1}(U_f) = Z \setminus \{ (x), (x-1) \}$

$\mathcal{O}(U_f) = \mathcal{O}_Z(\phi^{-1}(U_f)) = \mathbb{C}[x][ ((x)(x-1))^{-1} ]$.

And that is not $\mathbb{C}[x][f^{-1}]$.

evgeniamerkulova's answer reassures me that I'm not out of my mind, but obviously Matt E and Mariano know what they're talking about, so I don't know what to think.

Both Mariano and Matt E imply that $\mathcal{O}(X)$ is not $\mathbb{C}[x]$, but that seems obviously wrong (and contradicts the book itself).

Here's my reasoning, spelled out. O(X) is C[x]. This is because $\mathcal{O}_Z(\phi^{-1}(X)) = \mathcal{O}_Z(Z) = \mathbb{C}[x]$. In order for the condition to be satisfied, we need $\mathcal{O}(U) = \mathbb{C}[x,x^{-1}]$ for some open U in X. So we need $\mathcal{O}_Z(\phi^{-1}(U)) = \mathbb{C}[x,x^{-1}]$. For that to happen we need $\phi^{-1}(U) = Z \setminus \{ (x) \}$. But all the inverse images of sets in X either include both (x) and (x-1) or neither of them, so this can never happen.

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What is $R$ here? Note that $x$ is not a well-defined function on $X$ (since it doesn't know what value to take at the node, where $x = 0$ and $x = 1$ have been identified). – Matt E Oct 30 '10 at 3:10
R = C[x], sorry. I'll edit. – Mike Benfield Oct 30 '10 at 3:20
I do realize x is not a function on X; I was taking inverse images of open sets back to Z to figure out what the rings are. – Mike Benfield Oct 30 '10 at 3:24
But $X$ is not equal to Spec $\mathbb C[x]$. The calculation you are making is a calculation on $Z$ (even though you are trying not to phrase it that way). To compute correctly on $X$, you need to figure out what $\mathcal O(X)$ is. This will give you the correct choice of $R$, and then the statement you are trying to prove will actually be true (for $f$ chosen from this ring $R$). – Matt E Oct 30 '10 at 4:21
That comment has me scratching my head... it says right in the problem statement that $\mathcal{O}(X)=\mathbb{C}[x]$, and isn't that obvious anyway? I'll think about this some more. – Mike Benfield Oct 30 '10 at 5:00

Everything you say is right and Eisenbud and Harris are false. I don't inderstand Matt E ' comments either because you did no mistake:

a) That "x" is not function on $X$ has nothing to do with problem, and you never said it was function.

b) He writes "To compute correctly on $X$, you need to figure out what $\mathcal O(X)$ is" : you have computed that it is $\mathbb C[x]$ and you are right.

For completeness stalk $\mathcal O_a$ of $\mathcal O$ at quotient point $a\in X$ corresponding to $0,1$ [you write x, but you may not because x is already polynomial] is ring $S \subset \mathbb C(x)$ of all fractions $f(x)/g(x)$ ($f(x), g(x) \in \mathbb C [x])$) such that $g(0)\neq 0$ and $g(1)\neq 0$

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If $\mathcal O(X)$ is a polynomial ring, then there is a globally defined regular function on $X$ which generates $\mathcal O(X)$: can you describe it? – Mariano Suárez-Alvarez Oct 30 '10 at 14:39
(By the way, if you agree that "x" is not a function on $X$, then you cannot say that $\mathcal O(x)$ is $\mathbb C[x]$...) – Mariano Suárez-Alvarez Oct 30 '10 at 14:40
I misread the question, and am posting a new comment above to reflect this. – Matt E Oct 30 '10 at 15:55