Atiyah Macdonald define presheaf (chapter 3, exercise 23) on the base of $Spec(A)$, where $A$ is commutative ring with $1$, as follows $$ \mathfrak{F}(X_f) = A_f, $$ where $X_f$ is a basic open set associated with $f$ and $A_f$ is localization of $A$ at $\{1,f,f^2,...\}$. I proved that this definition is correct (i.e. independent on the choice of $f$) and constructed restriction mappings. However I am in trouble with showing that this is also a sheaf, i.e. if we take a cover of $Spec(A)$ by basic open sets $\{U_i\}$ and take $s_i \in \mathfrak{F}(U_i)$ such that they coincide on intersections $U_i\cap U_j$, then one may find some section $s\in\mathfrak{F}(Spec(A))=A_1=A$ such that restriction of $s$ on each $U_i$ is $s_i$.

I understand that I probably have to use compactness of $Spec(A)$, however I am not able to finish the work even in case of cover by two basic open sets. Suppose $Spec(A)=U_1\cup U_2$ and $\mathfrak{F}(U_1)=A_f$ and $\mathfrak{F}(U_2)=A_g$, let $s_1 = a/f^m$ and $s_2 = b/g^n$. Note that $U_1\cap U_2$ in this case is just basic open set corresponding to the element $fg$, hence restricting $s_1$ to $U_1\cap U_2$ we obtain $ag^m/(fg)^m$, similarly, restricting $s_2$ to $U_1\cap U_2$ we obtain $bf^n/(fg)^n$ (by construction of restriction mappings). Conditions which are necessary to glue $s_1$ and $s_2$ in global section $s$ are

$$ \frac{ag^m}{(fg)^m} = \frac{bf^n}{(fg)^n}, \text{in $A_{fg}$} $$ $$ U_1\cup U_2 = Spec(A), $$ the last condition is equivalent to the following, there exists $c,d \in A$ such that $cf + dg = 1$. Following these awful amount of letters, to show that $s_1$ and $s_2$ may be glued is exactly to show that $a=b$ and $m=0,n=0$, i.e. that we just took $s_1 = a/1\in A_f$ and $s_2=a/1\in A_g$. However I cannot conclude it from above conditions. Maybe there is a better way, please, help me.

I looked through MOF and MSE and did not find anything on this question. I am sorry if it has been asked before.


1 Answer 1


Using your notation: we may assume from the outset that $m = n$ (if $n > m$, $\dfrac{a}{f^m} = \dfrac{af^{n-m}}{f^n}$). The condition $\dfrac{ag^n}{(fg)^n} = \dfrac{bf^n}{(fg)^n}$ in $A_{fg}$ means that some power of $fg$ annihilates $ag^n - bf^n$ in $A$, say $(fg)^N(ag^n - bf^n) = 0$, i.e. $g^{N+n}af^N = f^{N+n}bg^N$. Choose $c, d$ such that $1 = cf^{N+n} + dg^{N+n}$.

Then, for $s := caf^N + dbg^N$, we have

$$f^{N+n}s = af^{N}cf^{N+n} + d(f^{N+n}bg^N) = acf^{N+n} + d(g^{N+n}af^N) = af^N$$

$$g^{N+n}s = c(g^{N+n}af^N) + bg^Ndg^{N+n} = c(f^{N+n}bg^N) + bg^Ndg^{N+n} = bg^N$$

and thus $\dfrac{s}{1} = \dfrac{af^{N}}{f^{N+n}} = \dfrac{a}{f^n}$ in $A_f$, and $\dfrac{s}{1} = \dfrac{bg^{N}}{g^{N+n}} = \dfrac{b}{g^n}$ in $A_g$.

If $s'$ is another section which restricts to $s_1, s_2$ respectively, then $s - s'$ restricts to $0$ on $A_f$ and $A_g$, hence is annihilated by suitable powers of both $f$ and $g$, and hence also by $1$, which gives uniqueness. The argument also extends straightforwardly to the general (finite) case.

  • $\begingroup$ It is a coincidence. Look as King Liu(page 42) or Hartshorne(page?), they also reasoned as in your answer! $\endgroup$
    – Hamou
    Aug 19, 2014 at 2:37
  • $\begingroup$ @Hamou: Yes, but to avoid duplication of content, if there is already an answer with that reasoning, there should not be another $\endgroup$
    – zcn
    Aug 19, 2014 at 2:38
  • $\begingroup$ I have a connection failure , and that's why I did not see your post. But what I wrote was not a copy of your answer. $\endgroup$
    – Hamou
    Aug 19, 2014 at 2:52
  • $\begingroup$ @Hamou: Very well. $\endgroup$
    – zcn
    Aug 19, 2014 at 2:54

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