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I am trying to solve the following question, which appears as the part of an exercise in Atiyah-Macdonald Chapter I.22:

For a ring $A$, if $\mathrm{Spec}(A)$ is disconnected, then we have $A\cong A_1\times A_2$ for some rings $A_1,A_2$.

Actually, the original question is to show three equivalent conditions and we can avoid the above. But I would like to see if there is a possibility to prove this.

Here is my attempt:

Let $\mathrm{Spec}(A)=V(I_1)\cup V(I_2)$, where $V(I_i)$ are proper closed sets and $V(I_1)\cap V(I_2)=\emptyset$. Then we have $$I_1+I_2=(1)$$

So if we have $I_1\cap I_2=\{0\}$, we can use the Chinese remainder theorem to show $$A\cong A/\{0\}\cong (A/I_1)\times (A/I_2)$$ then we are done. Unfortunately, from $V(I_1)\cup V(I_2)=V(I_1\cap I_2)=\mathrm{Spec}(A)$, we can only obtain $$I_1\cap I_2\subset \mathfrak{N}$$ where $\mathfrak{N}$ is the nilradical of $A$.

So I was wondering if we can shrink $I_1$ and $I_2$ so that complete the above argument. The obvious try is $I_1$ and the ideal generated by $I_2\setminus I_1$, but it is apparent to see $I_1\cap \langle I_2\setminus I_1 \rangle\neq \{0\}$...

I really appreciate any kind of help!

user26857
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sysMirror
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    Note that you can write $1=x+y$ based on the fact $I_1+I_2=A$. Do you see that we would be done if $x^2=x$? – Steve D Jun 13 '18 at 05:00
  • @SteveD I see. Just take $I_1'=(x)$ and $I_2'=(1-x)$. Now I know how to prove this. Thanks! – sysMirror Jun 13 '18 at 05:06
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    No, you need to be slightly more careful. You want $x(1-x)=0$ for this to be a true split. All you know is that $x(1-x)$ is in the nilradical. Proposition 1.16 is helpful here.... – Steve D Jun 13 '18 at 20:22
  • @SteveD Yes thanks for your information. I know how to prove this. I saw is in the proof from some online documents which is to show (ii) implies (iii). Since they are equivalent conditions so sometimes I can find some useful ideas there. – sysMirror Jun 13 '18 at 21:48
  • @SteveD how does proposition 16 helps us to get around the fact that $x(1-x)$ is in the nilradical? – rmdmc89 Mar 19 '20 at 11:51

1 Answers1

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We have $X=specA$ is disconnected, then $X=U_1\cup U_2; U_1\cap U_2=\phi$. and $A=O_X(X)= O_X(U_1\cup U_2)=O_X(U_1)\times O_X(U_2)=A_1\times A_2 $, where $O_X(X)$ is a sheaf of ring

Mustafa
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