Lemma 26.13.1. Let $X$ be a scheme. Let $R$ be a local ring. The construction above gives a bijective correspondence between morphisms $\mathop{\mathrm{Spec}}(R) \to X$ and pairs $(x, \varphi )$ consisting of a point $x \in X$ and a local homomorphism of local rings $\varphi : \mathcal{O}_{X, x} \to R$.

Proof. Let $A$ be a ring. For any ring homomorphism $\psi : A \to R$ there exists a unique prime ideal $\mathfrak p \subset A$ and a factorization $A \to A_{\mathfrak p} \to R$ where the last map is a local homomorphism of local rings. Namely, $\mathfrak p = \psi ^{-1}(\mathfrak m)$. Via Lemma 26.6.4 this proves that the lemma holds if $X$ is an affine scheme.

Let $X$ be a general scheme. Any $x \in X$ is contained in an open affine $U \subset X$. By the affine case we conclude that every pair $(x, \varphi )$ occurs as the end product of the construction above the lemma.

To finish the proof it suffices to show that any morphism $f : \mathop{\mathrm{Spec}}(R) \to X$ has image contained in any affine open containing the image $x$ of the closed point of $\mathop{\mathrm{Spec}}(R)$. In fact, let $x \in V \subset X$ be any open neighbourhood containing $x$. Then $f^{-1}(V) \subset \mathop{\mathrm{Spec}}(R)$ is an open containing the unique closed point and hence equal to $\mathop{\mathrm{Spec}}(R)$. $\square$

Comment #5879 by Zhenhua Wu on

I don't see how bijection is proved in the general scheme case. It's not that the proof is wrong, but morelike it's not enough. The rest of the proof still needs some non-trivial logic. $\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Spec}{Spec}$ Define as above. We first show it's surjective. Given any $(x,\varphi)$, there exists an open affine $U=\Spec A$ containing $x$, and a map $A \to \mathcal{O}\_{X, x}$, by composing with $\varphi:\mathcal{O}\_{X, x}\to R$, we have $f:\Spec R\to \Spec A\to X$. It's easy to see $c(f)=(x,\varphi)$. Next we show it's injective. Given any two maps $g,h:\Spec R\to X$, s.t. $c(g)=c(h)=(x,\varphi)$. We know $g(\mathfrak{m})=h(\mathfrak{m})=x$. Pick an open affine $U=\Spec A$ containing $x$ we know both $g$ and $h$ factor through $U$ using the orginal argument. So we have $g_0,h_0:\Spec R\to U$. It suffices to show $g_0=h_0$. Clearly $c$ is functorial w.r.t. the open immersion $U\to X$, and the canonical map is an identity map. We have shown $c_U$ is a bijection (defined as $c$ with $U$ replacing $X$). So $c\_U (g\_0)=c\_U(h\_0)$ implies that $g\_0=h\_0$, which in turn implies $g=h$.

Comment #5880 by Zhenhua Wu on

The typo above is an example of this issue \ref{https://stackoverflow.com/questions/29014573/mathjax-how-to-deal-with-this-strange-behavior/29040570#29040570}. Note that adding underscore after right curly bracket instantly turned into italc form (you can see the italic button on the tool row becomes highlighted). I added backslash sign before underscore sign to make it looks right in preview. But now in real output it screws up. Let me test again. One backslash: $\mathcal{O}\_{X,x}, \mathcal{O}\_{X,x}$. Correct in preview, probably wrong in real output. No backslash: $\mathcal{O}_{X,x}, \mathcal{O}_{X,x}$. Wrong in preview, probably correct in real output.

Comment #5882 by on

@#5879: In stead of giving your own proof, could you comment on what sentence(s) of the current proof should be clarified? I'm a bit surprised you are complaining about the deduction of the general case from the affine case, as that seems to me is explained quite well.

@#5880. Yes. I'm sorry about the mismatch between the editor and what readers get to see. As you say, this has been a problem for a while. We don't quite know how to fix this.

Comment #5888 by Zhenhua Wu on

@#5882, denote the functor as $c$, for me the problem of confusion is that the relation between $c$ restricted to open affine $U$ and $c$ on $X$ is not clear. For example, before we prove that $U\hookrightarrow X$ is a monomorphism in the category of schemes in tag 01L7 (so for this tag not proved yet), we cannot use the fact $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ injects into $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$. So in this case I prefer we emphasis the difference between the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ and the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$. Actually I would drow a square for it if I know how to draw it in this editor.

And we don't really need the injection in this proof. For the shortest suggestion on the proof, I would add something like:

We proved every element of $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$ can be lifted to an element in $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ for some open affine $U$. Providing $c$ is functorial. Then the injectivity of $c$ on $X$ reduces to the injectivity of $c$ on $U$, which is already proved.

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