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Let $\operatorname{Spec}(A_1), \operatorname{Spec}(A_2)$ be two affine scheme, assume we glue then together along the distinguished open set $D(f_1)\subset \operatorname{Spec}(A_1),D(f_2) \subset\operatorname{Spec}(A_2)$. The isomorphism between $D(f_1)\cong D(f_2)$ is given as follows, we define an isomorphism $$\phi:({A_1})_{f_1}\to (A_2)_{f_2} $$ then it will induce isomorphism of $D(f_1)$ and $D(f_2)$ therefore we can glue them together.To form a scheme $X$. Assume it happens to be an affine scheme with $X= \operatorname{Spec}(A)$

The question is what's the relation between coordinate ring $A$ and $A_1,A_2$ ?

I see in this post that the coordinate ring $A$ is limit of $A_1,A_2$ under identification $\phi$ , I don't know how to see it

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Question: "The question is what's the relation between coordinate ring A and A1,A2 ?"

Answer: @yi li - If an affine scheme $X:=Spec(A)$ has an open cover $D(f_1),D(f_2)$ of two basic open sets you may write down the Cech-complex corresponding to this covering. Take a global section $s\in A$ and restrict it to $U_i:=D(f_i)$ to get $s_i$. Restricting further to $V:=U_i \cap U_j$ you get $(s_i)_V=s_V=(s_j)_V$. Given two sections $s_i \in A_i:=A_{f_i}$ agreeing on $V$ there is by the sheaf property an element $s\in A$ with $s_{U_i}=s_i$. Hence $A$ is isomorphic to the set of pairs $(s_1,s_2) \in A_1\oplus A_2$ with $(s_1)_V=(s_2)_V$.

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  • $\begingroup$ Thank you this idea looks very concrete for example the line with double origin (if it's affine scheme) then since it's cover by two pieces of $\Bbb{A}^1_k$, therefore the coordinate ring is of the form $(s_1,s_2)\in k[x]\oplus k[y]$ which agree on the intersection $V$ $\endgroup$
    – yi li
    Jan 6 at 9:43
  • $\begingroup$ @yi li - the line with "double origin" is non-separated hence cannot be affine (over the base field $k$). Any affine scheme over a field $k$ is separated (see Hartshorne, Prop.II.4.1) $\endgroup$
    – hm2020
    Jan 6 at 9:47
  • $\begingroup$ Yeah I know it's not affine, I just use it to illustrate the idea above. $\endgroup$
    – yi li
    Jan 6 at 9:48

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