# A possible proof that the affine line with double origin is not affine

I'm trying to prove that the affine line with double origin is not an affine scheme.

Consider $$X_1:=\text{Spec}(k[x])$$ and the open subset $$U_1:=X_1\setminus\{(x)\}$$. In particular $$\mathcal{O}_{X_1}(U_1)=k[x]_x$$.

Similarly, $$X_2:=\text{Spec}(k[y])$$, $$U_2:=X_2\setminus\{(y)\}$$, so that $$\mathcal{O}_{X_2}(U_2)=k[y]_y$$.

The ring isomorphism $$k[x]_x\to k[y]_y$$ induces an isomorphism $$\varphi:U_1\stackrel{\sim}{\to} U_2$$. We define the line with double origin as the scheme $$X=X_1\sqcup_\varphi X_2,$$ i.e. the gluing of $$X_1$$ and $$X_2$$ along $$U_1\simeq U_2$$ via $$\varphi$$.

Here is my strategy: if we remove one of the origins from $$X$$, say $$(x)$$, we get the usual line $$\text{Spec}(k[y])$$, whose set of global sections is $$k[y]$$. On the other hand, if $$X$$ is affine, i.e. $$X\simeq \text{Spec}(A)$$ for some ring $$A$$, then the closed point $$(x)$$ corresponds to a maximal ideal $$\mathfrak{m}\subset A$$. Hence the removal of $$(x)$$ gives the scheme $$\text{Spec}(A)\setminus\{\mathfrak{m}\}$$, which should look like a line without an origin.

I hope that I can prove that $$\mathcal{O}_{\text{Spec}(A)}(\text{Spec}(A)\setminus\{\mathfrak{m}\})\not\simeq k[y]$$, but I don't know how to do it.

• It seems to me like you've skipped the rather important step of identifying $A$. This would be the thing to do before you go any further. Apr 17, 2020 at 20:34

A correct argument is by exploiting the fact that the restriction morphism $$r:\mathcal O(X)\to \mathcal O(X_1)$$ is a ring isomorphism, in which $$\mathcal O(X)=A=k[x]$$.
If $$X$$ were affine the dual inclusion map $$j=r^*:X_1\to X$$ would be an isomorphism of affine schemes,.
This is of course false since $$j$$ is not surjective. Hence $$X$$ is not affine.

• If $X$ were affine, why would the inclusion $X_1 \to X$ be an isomorphism? Jan 16, 2021 at 16:40
• How do we know $j$ is not surjective? Jan 16, 2021 at 19:52
• The category of affine schemes and commutative rings are equivalent. We know that $X_1$ is an affine scheme. If $X$ is an affine scheme, $j$, would be the category of affine schemes $X$ and since $j$ is not isomorphism (it is not surjective), the corresponding morphism in commutative rings category, $r$, should not either be isomorphism. Jan 28, 2022 at 12:21
• You can see the answer of this question to see relation of commutative rings and schemes categories: math.stackexchange.com/questions/56854/… Jan 28, 2022 at 12:25

I denote the line with double origin by $$X$$. It is obtained by gluing $$Spec\, k[u]$$ and $$Spec\, k[t]$$ along the isomorphism $$D(u) = k[u,1/u]\to k[t,1/t] = D(t)$$ which sends $$u$$ to $$t$$. The first step is to compute $$\Gamma(X,\mathcal O_X)$$, since if $$X$$ is isomorphic to the spectrum od some ring, then it is the ring $$\Gamma(X,\mathcal O_X)$$. By definition (this depends on how you do the gluing) $$\Gamma(X,\mathcal O_X)$$ is the limit of \begin{align*} k[u] \rightarrow k[u,1/u] \cong k[t,1/t] \leftarrow k[t] \end{align*}
in the category of rings. Hence giving an element of $$\Gamma(X,\mathcal O_X)$$ is the same as giving two polynomials $$\sum_{n}f_nu^n$$ and $$\sum_m g_mt^m$$ such that $$\sum_{n}f_nu^n = \sum_m g_mu^m$$ in $$k[u,1/u]$$. Note that this just means that $$f_n=g_n$$ for all $$n$$. Hence $$\Gamma(X,\mathcal O_X)$$ is isomorphic to $$k[u]$$. If $$X$$ is affine, then we have isomorphisms \begin{align*} (X,\mathcal O_X) \xrightarrow{(f,f^\#)} Spec \,\Gamma(X,\mathcal O_X) \xrightarrow{(g,g^\#)} Spec\, k[u] \end{align*} of locally ringed spaces. Now consider the vanishing set $$V(u)$$ of $$X$$. Here $$u$$ denotes the global section $$u =v$$ of $$\Gamma(X,\mathcal O_X)$$. $$V(u)$$ consists of all those points $$p \in X$$ such that $$u_p = 0$$ modulo $$\mathfrak m_p$$. Note that it contains at least two points, the two origins of $$X$$. Now any isomorphism of locally ringed spaces must send this vanishing set to a vanishing set of the same cardinality. But $$V(u)$$ in $$Spec \, k[u]$$ consists of only one point. (Note that $$u$$ in $$k[u]$$ corresponds to $$u$$ in $$\Gamma(X,\mathcal O_X)$$ after applying $$f^\#$$ and $$g^\#$$). This a contradiction.

Maybe it is helpful to abstract the main point of the argument. Given a locally ringed space $$X$$ define the vaishing set $$V(f)$$ of a global section in the usual way. $$p\in V(f)$$ if and only if $$f=0$$ in $$\kappa(p) = \mathcal O_{X,p}/\mathfrak m_p$$. If $$\pi:X\to Y$$ is an isomorphism of locally ringed spaces, then the following statement holds for every global section $$f$$ of $$Y$$: $$\pi (x)\in V(f)$$ if and only if $$x\in V(\pi^\#f)$$. In other words $$\pi^{-1}V(f) = V(\pi^\#f).$$

• Hi @Nico , How can I prove $\Gamma(X,\mathcal O_X)$ is limit of \begin{align*} k[u] \rightarrow k[u,1/u] \cong k[t,1/t] \leftarrow k[t] \end{align*} ? Jan 6 at 9:03
• @yili There is an adjunction $Hom_{Ring}(A,\mathcal O(X))= Hom_{Sch}(X, Spec(A))$ which implies that $\mathcal O$ sends colimit diagrams of schemes to limit diagrams of rings.
– Nico
Jan 6 at 9:19
• Ah, thank you @Nico , I have rough idea now, I will try to make it precise. Jan 6 at 9:25
• @yili Alternatively you can use that $\mathcal O$ is isomorphic to the representable functor $Hom(-,\mathbb A):Sch^{op}\to Ring$ and work from there. :) Have fun
– Nico
Jan 6 at 9:30
• Do you know where can I found this ? for example reference? Jan 6 at 9:35