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I am (slowly) reading Eisenbud and Harris and trying to get my head around (affine) schemes.

Let $R$ be a ring, and $X=\text{spec}(R)$, as usual with the Zariski topology. We have a basis of open sets; for $f \in R$, $X_f=\{ p \subset R | f \notin p \}$ where $p$ is a prime ideal. Then the structure sheaf is $\mathcal{O}(X_f) = R_f$, the localization of the ring $R$ with respect to the multiplicative subset $\{ 1,f,f^2,\ldots \}$.

Exercise I-20 in E-H is to calculate the points and sheaf of functions for some schemes


1) $X_1 = \text{Spec } \mathbb{C}[x]/(x^2)$

This shouldn't be to hard - there is exactly one (closed) point corresponding to the maximal ideal $(x)$ in which case $X_1 = \{(x)\}$. Thus the only open sets are $\emptyset \subset X_1$ and then $\mathcal{O}(\emptyset) = 0$ and $\mathcal{O}(X_1) = \mathbb C[x]/(x^2)$

Is this correct?

2) $X_2 = \text{Spec } \mathbb{C}[x](x^2-x)$ Here we should have exactly two (closed) points: $(x),(x-1)$. Call these $\{ a,b \}$. The topology should then be $\{\emptyset,\{a \}, \{ b \}, \{a, b\} \}$ (the discrete topology). Again we have $\mathcal{O}(\emptyset) =0 $ and $\mathcal{O}(\{ a,b \}) = \mathbb C[x]/(x^2-x)$.

Now $$ \begin{align} \mathcal{O}(\{ a \}) &= [\mathbb C[x]/(x^2-x)]_{(x)} \\ &\simeq [\mathbb C[x]/(x(x-1)]_{(x)} \end{align} $$

Am I now localizaing with respect to the multiplicative set $R - \mathfrak{p}$ where $\mathfrak{p}=(x)$ and $R = \mathbb C[x]/(x^2-x)$? And then is this just: $$ \begin{align} \mathcal{O}(\{ a \}) &\simeq [\mathbb C[x]/(x(x-1)]_{(x)} \\ &\simeq [\mathbb C[x]/(x-1)]_{(x)} \\ &\simeq \mathbb{C} \end{align} $$ ?

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Just note that in the definition of the open set $X_f$, you definitely want $f\notin p$, and not $f\neq p$. –  M Turgeon Jun 1 '12 at 14:48
    
@M-Turegon: thank you for noticing this typo! –  Juan S Jun 3 '12 at 23:48
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Keep in mind that, in general, if $\mathfrak{p}$ is a principal prime ideal in a ring $R$ with generator $x$, then $R_x$ and $R_\mathfrak{p}$ are different. The ring $\mathbf{C}[x]/(x-1)$ localized at the prime ideal $(x)$ is zero. Localized at the element $x$ (or the image of $x$), it is $\mathbf{C}$. Your second and third isomorphisms (in the second set of equations) is incorrect. The first localization is $\mathbf{C}$ but the second is zero. In this case, you have $R_x=R_{(x)}$, but that's not true in general. –  Keenan Kidwell Jun 4 '12 at 0:15
    
@Keenan: Thank you for this - I think this kind of goes to the root my misunderstanding. Localizing at a prime ideal $\mathfrak{p}=(x)$ in a ring $R$ is localizing w.r.t $R - \mathfrak{p}$ while localizing at the element $x$ is localizing w.r.t the multiplicative subset $\{1,x,x^2,\ldots \}$? –  Juan S Jun 4 '12 at 0:35
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@Juan Yes, that's right. The terminology is kind of confusing. –  Keenan Kidwell Jun 4 '12 at 1:03
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up vote 2 down vote accepted

Your arguement are correct. I am not sure what are your question? But what you write is correct. In fact, as you already know, as for the first example, the underlying topology is easy which is clearly irreducible, but note that the ring of global sections is not reduced; as for the second example, please note that the underlying toplogy is not irreducible, and the ring of global sections is isomorphic to $\mathbb{C}\times \mathbb{C}$, which is reduced.

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Thank you for this. I was just after confirmation I was on the right track (self-studying, so it's hard to know sometimes) –  Juan S Jun 3 '12 at 23:49
    
You are welcome. I also learnt it by myself. So go ahead bravely! –  Joy-Joy Jun 4 '12 at 3:31
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