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Let $F\in k[x_0,x_1,..,x_n]$, and $X=V(F)=\{p\in k^n: F(p)=0\}$. Let $A=\frac{k[x_1,..,x_n]}{(F)}$.

Suppose $A$ has a non-trivial idempotent, then prove that $X=X_1\sqcup X_2$, that is $X$ is disconnected.

This is the problem I am trying to work out, from Miles Reid book. This is my attempt.

Let $f\in A$ be the idempotent. Then if $g=1-f$, then $A$ can be written as $A=Af\oplus Ag$. Is it true that $X=V(F)=V(f)\sqcup V(g)$? Thank you.

In the book, we have not yet identified, $Spec A$ and $V(F)$.

user349424
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1 Answers1

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In $\mathbb A_k^n$, we have the inclusion $X=V(F) \subset V(f) \cup V(g)$, because $fg=f-f^2 \in (F)$, i.e. any root of $F$ is a root of $fg$ and thus of $f$ or $g$.

Clearly $V(f) \cap V(g) = \emptyset$ by the virtue of $f+g=1$.

Thus, with $X_1 = X \cap V(f)$ and $X_2 = X \cap V(g)$, we have the desired $$X=X_1\sqcup X_2.$$

MooS
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