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All rings here are commutative rings with units, and $k$ is an algebraically closed field.

Definition: a ring $A$ is connected if it has only the trivial idempotent elements, i.e., if $e\in A$ satisfies $e^2 =e$, then $e=0$ or $e=1$.

Problem: Assumue $X$ is a nonempty affine algebraic set over $k$, then $X$ is connected (as topological subspace with Zariski topology) $\Leftrightarrow$ the coordinate ring $A(X):=k[x_1,...,x_n]/I(X)$ is connected (as a ring in the above definition).

My thoughts: "$\Rightarrow$": We can show that that an integral domain is connected. So if $X$ is an irreducible algebraic set, then $I(X)$ is a prime ideal and thus $A(X)$ is an integral domain, as a result $A(X)$ is connected. But I'm not sure about the case when $X$ is not irreducible, and how to apply the connectedness of $X$.

Background: I am a beginner at algebraic geometry. I've covered Hilbert Nullstellensatz and Noether normalization lemma so far.

cxh007
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  • If a topological space is not connected (and not empty) then it is the disjoint union of two non-empty closed subspaces. Use the Nullstellensatz to see what this says in terms of ideals. – Zhen Lin Mar 12 '22 at 11:30
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    @ZhenLin Well, if $X$ is not connected, then $X=Y\cup Z$, where $Y,Z$ are nonempty algebraic sets, and $Y\cap Z= \varnothing$. Write $Y=V(I),Z=V(J)$ where $I,J$ are ideals in $k[x_1,...,x_n]$. So $X=V(I) \cup V(J)=V(I\cap J)$, and $V(I)\cap V(J)= V(I+J)=\varnothing$. But by Nullstellensatz, $V(I+J)=\varnothing \Rightarrow I+J=k[x_1,...,x_n].$ – cxh007 Mar 12 '22 at 12:05
  • Yes, good. So now you should find some non-trivial idempotents modulo $I \cap J$. Do you know the Chinese remainder theorem? – Zhen Lin Mar 12 '22 at 12:38
  • @ZhenLin Yes, I do. It was used in my linear algebra class to prove the Jordan Chevalley decomposition. But I am not quite familiar with it. – cxh007 Mar 12 '22 at 12:54
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    Well, then you might want to take the opportunity to familiarise yourself with the CRT. It is essential when you want to do anything with non-connected rings. – Zhen Lin Mar 12 '22 at 13:04
  • @ZhenLin Thank you for this advice. I'm looking at https://math.stackexchange.com/questions/1102037/the-chinese-remainder-theorem-for-rings now. – cxh007 Mar 12 '22 at 13:21
  • @ZhenLin So I know that $I+J/ (I\cap J) \cong (I+J/I) \times (I+J/J)$, and the later has non-trivial idempotent $(0+I,1+J)$ right? – cxh007 Mar 12 '22 at 13:30
  • Not quite... $I + J$ is not necessarily a ring, in the first place. What are the coordinate rings of $X$, $Y$, and $Z$? – Zhen Lin Mar 12 '22 at 13:38
  • @ZhenLin The coordinate rings of $X,Y,Z$ are $k[x_1,...,x_n]/I(X),k[x_1,...,x_n]/I(Y),k[x_1,...,x_n]/I(Z)$ or $k[x_1,...,x_n]/I \cap J,k[x_1,...,x_n]/I,k[x_1,...,x_n]/J$, respectively. ($Y=V(I),Z=V(J),X=V(I\cap J)$ and we can take $I,J$ to be radical ideals for simplicity, so $I(Z)=I(V(J))=\sqrt{J} = J$ and $I(X)=I\cap J,I(Y)=I$). – cxh007 Mar 12 '22 at 13:59
  • OK. So can you use the CRT to relate these rings and find that the coordinate ring of $X$ has a non-trivial idempotent? – Zhen Lin Mar 12 '22 at 22:16
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    @ZhenLin Yes. In this case $I+J=k[x_1,...,x_n]=:R$, so by the CRT, we know that $A(X)=R/I\cap J \cong R/I \times R/J \cong A(Y) \times A(Z)$. While in $A(Y) \times A(Z)$, we already have a non trivial idempotent $(0+I, 1+J)$. So in $A(X)$ we also have a non trivial idempotent $x+I\cap J$, where $x$ is a solution of $x \equiv 0 \pmod {I}$ and $x \equiv 1 \pmod{J}$. And the solution's existense is guaranteed by the CRT. – cxh007 Mar 13 '22 at 01:25

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