Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{C}[x][f^{-1}]$.

Edit: Regarding the answer and comments.

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.

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.