# Question of Hartshorne book's Proposion II.(2.6)

In Proposition II. (2.6) of Hartshorne book "algebraic geometry". I can't understand proof of part. In proof, let $V$ be an affine variety over field $k$ with the sheaf of regular function $\mathcal{O}_V$ and affine coordinate ring $A$. Let $X={\rm Spec}A$. Now, define a morphism of locally ringed spaces

$$\beta : (V, \mathcal{O}_V) \rightarrow X$$

as follows: For each point $p \in V$, $\beta(p)=\mathcal{m}_p$, the maximal ideal corresponding at $p$. And for any open set $U \subseteq X$, define a ring homomorphism $\mathcal{O}_X(U) \rightarrow \beta_*(\mathcal{O}_V(U)=\mathcal{O}_V(\beta^{-1}(U))$. Givena section $s\in \mathcal{O}_X(U)$, and given a point $p\in \beta^{-1}(U)$, we define $s(p)$ by taking the image of $s$ in the stalk $\mathcal{O}_{X,\beta(p)}$, which is isomorphic to the local ring $A_{\mathcal{m}_p}$, and then passing to the quotient ring $A_{\mathcal{m}_p}/\mathcal{m}_p$ which is isomorphic to the field $k$. Thus, we regard $s$ as a function from $\beta^{-1}(U)$ to $k$.

In book, this homomorphism gives an isomorphism $\mathcal{O}_X(U) \cong \mathcal{O}_V(\beta^{-1}(U))$.

But, I don't understand this part... why isomorphism??

I want to see Detailed description or reference of this part.

• Could you please elaborate which part exactly you do not understand? In your quote there are a lot of different things happening. Surely you don't want us to explain what a local ring is. Commented Nov 19, 2012 at 12:23

Let $V$ be an affine variety over an algebraically closed field $k$. We don't assume that $V$ is irreducible. Let $A$ be the affine coordinate ring of $V$. Let $X = Spec(A)$. Let $U$ be an open subset of $X$. Let $s \in \mathcal{O}_X(U)$. Let $p \in \beta^{-1}(U)$. We denote by $s(p)$ the image of $s_{m_p}$ in $A_{m_p}/m_pA_{m_p} = k$, where $m_p$ is the maximal ideal corresponding to $p$. We denote by $\bar s$ the function $\bar s(p) = s(p)$. If $U = X$, then $A = \mathcal{O}_X(U)$. Since $A_{m_p}/m_pA_{m_p}$ is canonically isomorphic to $A/m_p$, $s(p)$ is the value of $s$ at $p$. Hence $\bar s = s$ is regular on $V = \beta^{-1}(X)$. Similarly if $U = D(f)$ for some $f \in A$, then $A_f = \mathcal{O}_X(U)$. Hence $s = g/f^n$ for some $g \in A$ and an integer $n \ge 0$. Since $s(p) = g(p)/f(p)^n$, $\bar s$ is regular on $\beta^{-1}(D(f))$. Hence, $\bar s$ is regular on $\beta^{-1}(U)$ for any open subset $U$. We define a homomorphism $\psi_U\colon \mathcal{O}_X(U) \rightarrow \mathcal{O}_V(\beta^{-1}(U))$ by $\psi_U(s) = \bar s$.

We claim that $\psi_U$ is an isomorphism. Suppose $\psi_U(s) = 0$. Then $\bar s|D(f) = \bar {(s|D(f))} = 0$ for every open subset $D(f) \subset U$. Suppose $s|D(f) = g/f^n$ for $g \in A$ and an integer $n \ge 0$. Since $\bar {(s|D(f))} = 0$, $g(p) = 0$ for every $p \in \beta^{-1}(D(f))$. Hence $f(p)g(p) = 0$ for every $p \in V$. Hence $fg = 0$. Hence $s|D(f) = g/f^n = 0$ in $A_f$. Since $D(f) \subset U$ is arbitrary, $s = 0$. Hence $\psi_U$ is injective.

It remains to prove that $\psi_U$ is surjective. We first note that if $U = D(f)$, $\psi_U$ is surjective, hence an isomorphism. This is clear from the fact that $\mathcal{O}_V(\beta^{-1}(D(f)))$ is canonically isomorphic to $A_f$ (for the proof, see this question).

Let $t \in \mathcal{O}_V(\beta^{-1}(U))$. Let $(D(f_i))_{i\in I}$ be a cover of $U$. Let $t_i = t|\beta^{-1}(D(f_i))$. Since $\psi_{D(f_i)} \colon \mathcal{O}_X(D(f_i)) \rightarrow \mathcal{O}_V(\beta^{-1}(D(f_i)))$ is an isomorphism, there exists $s_i \in \mathcal{O}_X(D(f_i))$ such that $\bar s_i = t_i$. Since $t_i = t_j$ on $\beta^{-1}(D(f_i)) \cap \beta^{-1}(D(f_j))$, $s_i = s_j$ on $D(f_i) \cap D(f_j)$. Hence there exists a unique $s \in \mathcal{O}_X(U)$ such that $s|D(f_i) = s_i$. Since $\bar s |\beta^{-1}(D(f_i)) = \bar s_i = t_i$, $\bar s = t$. Hence $\psi_U$ is surjective.

• Followup question: But what if $\beta^{-1}(U)$ is empty? In this case the claim is $\mathcal O_X(U) = 0$ but I don't see this immediately. Commented Dec 10, 2019 at 7:52
• Can $\beta^{-1}U$ be empty? Suppose, we have a basic open set in $U$ which we call $D(f) = \{\mathfrak{p}\mid f\not\in\mathfrak{p}\}$ for some $f\in A$. Let us suppose, $D(f)$ is non-empty, then there exists a $\mathfrak{p}$ not containing $f$. We know every ideal is contained in a maximal ideal. Let $\mathfrak{m}$ be that ideal which may or may not be $\mathfrak{p}$. Thus, $f\not\in \mathfrak{m}$ which shows that $\mathfrak{m}\in D(f)$. The image of $\beta$ is all maximal ideals and we have shown that $U$ must contain a maximal ideal. Thus, $\beta^{-1}U$ is non-empty. Commented Jun 8, 2021 at 11:16