In Hartshorne chapter 2 proposition 2.6,Hartshorne shows that there is a fully faithful functor $t:\mathcal{Var}\rightarrow \mathcal{Sch}(k)$ from the category of varieties over $k$ to the category of schemes over $k$.He proceed as follows:

First,for any topological space $X$,define $t(X)$ to be the set of irreducible closed subsets of $X$.We can define a topology on $t(X)$ by taking as closed sets the subsets of the form $t(Y)$,where $Y$ is closed subset of $X$.If $f:X_1\rightarrow X_2$ is continuous ,then define a map $$t(f):t(X_1)\rightarrow t(X_2)\\ Y\mapsto \overline{f(Y)},$$where $Y$ is a irreducible closed subset of $X_1$. And for any topological space $X$,define $\alpha :X\rightarrow t(X)$ by $\alpha(p)=\overline{\{p\}}$.

Question 1:why is $\overline{f(Y)}$ a irreducible closed subset of $X_2$?

Next,let $V$ be an affine variety over $k$ with coordinate ring $A$,and $O_V$ its sheaf of regular functions,he then shows that $(t(V),\alpha_*(O_V))$ is isomorphic to the affine scheme $(X,O_X)$,where $X=\operatorname{Spec}A$.

Question 2:why is $(t(V),\alpha_*(O_V))$ a locally ringed spaces,I mean why is $(\alpha_*(O_V))_Y$ a local ring for any $Y\in t(V)$?

Now define a morphism of locally ringed spaces $\beta:(V,O_V)\rightarrow X=\operatorname{Spec}A,$by $\beta(p)=m_p$.And for any open set $U\subset X$,define a homomorphism of rings $O_X(U)\rightarrow \beta_*(O_V)(U)$:given $s\in O_X(U)$,$p\in \beta^{-1}(U)$,we get a regular function $g$ on $\beta^{-1}(U)$ by $g(p):=\overline{s_{\beta(p)}}\in A_{m_p}/m_p=k$,where $s_{\beta(p)}\in O_{X,\beta(p)}$ and we identify the stalk $O_{X,\beta(p)}$ with the local ring $A_{m_p}$.

Question 3:why is $g$ a regular function?It seems to me that $g$ is locally constant,is this ture?

Then he claim that the above homomorphism $O_X(U)\rightarrow \beta_*(O_V)(U)$ is a isomorphism and use the fact that there is a 1-1 correspondence $t(V)\leftrightarrow \operatorname{Spec}A=X$,then $(X,O_X)\cong (t(V),\alpha_*(O_V))$ as locally ringed spaces.I can't follow this part of proof.

Question 4:what is this $\beta$ used for?Shouldn't we construct a morphism $(t(V),\alpha_*(O_V))\rightarrow (X,O_X)$?And why $O_X(U)\rightarrow \beta_*(O_V)(U)$ is a isomorphism?

  • $\begingroup$ question 1: because continuous maps preserve irreducibility and closure also preserves irreducibility. $\endgroup$ – Seth Sep 7 '14 at 16:00
  • $\begingroup$ @Seth:You are right.This is easier than I thought.Thanks! $\endgroup$ – Wei Xia Sep 8 '14 at 15:08

Question 1 is derived exactly as Seth's comment.

Question 2. We know that $$(\alpha_*(O_V))_Y=\varinjlim_{Y\in U}[\alpha_*(O_V)](U)=\varinjlim_{Y\in U}O_V(\alpha^{-1}(U)),$$ now,each open subset of $t(V)$ is of the form $t(V)-t(W)$ for some irreducible closed subset $W$ of $V$,let $U=t(V)-t(W)$ for some $W$ ,then $Y\in U \Leftrightarrow Y\notin t(W)\Leftrightarrow \exists p\in Y,s.t. p\notin W$$\Leftrightarrow Y\cap (V-W)\neq \emptyset$,but $V-W=\alpha^{-1}(U)$,hence $Y\in U \Leftrightarrow Y\cap \alpha^{-1}(U) \neq \emptyset$.And $\alpha$ induce a 1-1 correspondence between open subsets of $V$ and $t(V)$,we have $$\varinjlim_{Y\in U}O_V(\alpha^{-1}(U))=\varinjlim_{Y\cap U\neq \emptyset}O_V(U),$$ where $U$ runs over all open subsets of $V$ s.t. $Y\cap U\neq \emptyset.$ Note that $\varinjlim_{Y\cap U\neq \emptyset}O_V(U)$ is just the local ring of a subvariety which has been shown to be a local ring(excercize 3.13 chapter 1).

Question 3.In fact,the function $g$ is defined by the following map:let $p\in \beta^{-1}(U)$, $$O_X(U)\rightarrow O_{X,\beta(p)}\rightarrow A_{m_p}\rightarrow k\\s\mapsto s(\beta(p))\mapsto \frac{a}{b}\mapsto \frac{a(p)}{b(p)}=g(p),$$ where $O_{X,\beta(p)}\cong A_{m_p}$ and $a,b\in A,b\notin m_p$.Note that $a,b$ are polynomial functions on $V$ since $A$ is the coordinate ring of $V$ ,and by the definition of $s\in O_X(U)$,for any $p\in \beta^{-1}(U)$,$s(\beta(p))$ is locally quotient of elements in $A$,so $g$ is a regular function on $\beta^{-1}(U)$.Clearly,$g$ is a regular function in the usual sense and need not to be locally constant.

Question 4.Our original plan is to show $(t(V),\alpha_*(O_V))\cong(X,O_X)$.There is a natural homeomorphism $\gamma:t(V)\rightarrow X$,so it's left to show $\gamma':O_X\rightarrow \gamma_*(\alpha_*(O_V))$ is an isomorphism.But $\beta=\gamma\circ\alpha$,so it's enough to show $O_X\cong \beta_*(O_V)$.

We next show that for any open set $U\subset X$,$\varphi:O_X(U)\rightarrow \beta_*(O_V)(U)$ is an isomorphism.

$\varphi$ is injective: suppose $g(p)=0$ for all $p\in\beta^{-1}(U)$.For any $p\in\beta^{-1}(U)$,there exist an open set $p\in W\subset U$ and $a,b\in A,b\notin m_p$ s.t. for any $q\in W$ we have $s(q)=\frac{a}{b}$ and $b\notin q$. For any $q\in\beta^{-1}(W),\beta(q)\in W$,we have $s(\beta(q))=\frac{a}{b}$ and $g(q)=\frac{a(q)}{b(q)}=0$. It follows that $a(q)=0$ for all $q\in\beta^{-1}(W)$,since $\beta^{-1}(W)$ is open and dense in $V$,we see that $a=0$ in $A$.So $s(\beta(q))=0$ for all $\beta(q)\in W$.Since $p$ is a arbitrary piont in $U$,these W can cover U,we see that $s=0$ in $O_X(U)$.

$\varphi$ is surjective:let $g\in O_V(\beta^{-1}(U))$,then for any $p\in\beta^{-1}(U)$,there exist $W\subset \beta^{-1}(U)$,$p\in W$ and $a,b\in A,b\notin \beta(p)=m_p$ such that $g=\frac{a}{b}$ on $W$.Let $s(\beta(q)):=\frac{a}{b}$ for all $q\in W\subset\beta^{-1}(U)$,since $p$ is a arbitrary piont in $\beta^{-1}(U)$ ,these open sets $\beta(W)$ cover $U$,in this way we get $s$ as an element of $O_X(U)$.Then one can check that $s$ is well- defined and $\varphi(s)=g$.

Alternatively,one can show $O_X\cong \beta_*(O_V)$ by the following diagram: \begin{matrix} O_X(U) &\longrightarrow & \beta_*(O_V)(U)=O_V(\beta^{-1}(U)) \\\ \downarrow & & \downarrow \\\ O_{X,p}& \longrightarrow& [\beta_*(O_V)]_p=O_{V,\beta^{-1}(p)} \\\ \downarrow & & \downarrow \\\ A_p & \longrightarrow & A_p \\\ \end{matrix}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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