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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 proceeds 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 true?

Then he claims that the above homomorphism $O_X(U)\rightarrow \beta_*(O_V)(U)$ is an isomorphism and uses 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 is $O_X(U)\rightarrow \beta_*(O_V)(U)$ an isomorphism?

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  • $\begingroup$ question 1: because continuous maps preserve irreducibility and closure also preserves irreducibility. $\endgroup$
    – Seth
    Commented Sep 7, 2014 at 16:00
  • $\begingroup$ @Seth:You are right.This is easier than I thought.Thanks! $\endgroup$
    – Wei Xia
    Commented Sep 8, 2014 at 15:08

1 Answer 1

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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 point 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 point 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}

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