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Definition: An affine scheme X is connected if: $X\ne \emptyset$ If $X_1$, $X_2$ are open subschemes of $X$ such that $X_1 ∪X_2= X$ and $X_1\cap X_2=\emptyset$, then $X_1$ or $X_2$ is empty.

I aim to prove that An affine scheme is connected $\Leftrightarrow$ $\mathcal O(X)\ne \{0\}\land \mathcal O(X)$ only have trivial nilpotents, that is, no more nilpotent than $0,1$.

I have proved $X$ is connected $\Rightarrow \mathcal O(X)$ is nonzero and the only idempotent in it are $0$ and $1$. And stuck on the other direction.

Could someone please help? Thanks.

Y.X.
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It is basically trivial.

Assume that $X=specA$ is not connected, then there is open sets $U, V$, such that $X=U\cup V$ and $U\cap V=\emptyset$. Let $f\in \mathcal{O}_X(X)=A$, such that $f|_U=1$ and $f|_V=0$. Then $f$ is idempotent, which is a contradiction.

Feng Hao
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  • Thanks for answering. Could you please explain your notation? Is $f|_U$ meaning that the polynomial $f$ restricted to $U$? If so, I haven't seen such a thing... How does it defined and how does it work? – Y.X. Sep 30 '17 at 14:10
  • @PropositionX I define $f$ to be $f|_U=1$ and $f|_V=0$, which is of course a polynomial and lives in $A$. – Feng Hao Sep 30 '17 at 14:14