Errata: On the Relationship Between Connectedness of Spectrum and Idempotents

In Commutative Algebra and Algebraic Geometry (4): Zariski topology on affine scheme we discuss a lots of topology property of $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$ and the relationship with the algebraic property of $A$$A$.

Errata: This part is wrong, we need the condition that $A$$A$ is reduced, or we need to say that $A/\sqrt{(0)}\cong A_1\times A_2$$A/\sqrt{(0)}\cong A_1\times A_2$.

Proposition.Let $A$$A$ be a commutative ring, then $\mathrm{Spec}(A)$$\mathrm{Spec}(A)$​ is disconnected if and only if there exists some nontrivial idempotent elements in $A/\sqrt{(0)}$$A/\sqrt{(0)}$.

Lemma.

Let $x\in \mathrm{Spec}(A)$$x\in\mathrm{Spec}(A)$, and for $f\in A$$f\in A$ such that $f^2=f$$f^2=f$, $f(x)$$f(x)$ is either equal to $1$$1$ or $0$$0$.

Proof.

Since $f^2(x)=f(x)\in A/{\mathfrak{p}}_x$$f^2(x)=f(x)\in A/\mathfrak p_x$, whcih is an integral domain. Hence

$$\begin{array}{}\text{(1)}& f^2(x)=f(x)⟺f^2(x)-f(x)=0⟺f(x)(f(x)-1)=0⟹f(x)=0\vee f(x)-1=0◻\end{array}$$

Lemma.

Let $f\in A$$f\in A$ satisfies that $f^2=f$$f^2=f$, then $(1-f{)}^2=1+f-2f=1-f$$(1-f)^2=1+f-2f=1-f$.

Lemma. $f(x)=0⟺(1-f)(x)=1$$f(x)=0\iff (1-f)(x)=1$.

Proof of the proposition.

Let $f^2=f,f\ne 0,f\ne 1$$f^2=f,f\ne 0,f\ne 1$

By $f(x)=0⟺(1-f)(x)=1$$f(x)=0\iff (1-f)(x)=1$ we see that $V(f)=D(1-f)$$V(f)=D(1-f)$, hence $D(f)=V(1-f)$$D(f)=V(1-f)$.

Hence $D(1-f)=\mathrm{Spec}(A)-D(f)$$D(1-f)=\mathrm{Spec}(A)-D(f)$. Hence $\mathrm{Spec}(A)=D(f)\cup D(1-f)$$\mathrm{Spec}(A)=D(f)\cup D(1-f)$, it is disconnected.

Conversely, let $\mathrm{Spec}(A)=V(I)\cup V(J)$$\mathrm{Spec}(A)=V(I)\cup V(J)$ such that $V(I)\cap V(J)=V(I+J)=\mathrm{\varnothing }$$V(I)\cap V(J)=V(I+J)=\empty$,

then $V(I+J)=V(1),I(V(I+J))=\sqrt{I+J}=A⟹I+J=A$$V(I+J)=V(1),I(V(I+J))=\sqrt{I+J}=A\implies I+J=A$. Hence $I$$I$ and $J$$J$ are coprime.

Also, $I(\mathrm{Spec}(A))=I(V(I)\cup V(J))=IV(I)\cap IV(J)=\sqrt{I}\cap \sqrt{(J)}=\sqrt{I\cap J}=\sqrt{(0)}$$I(\mathrm{Spec}(A))=I(V(I)\cup V(J))=IV(I)\cap IV(J)=\sqrt I\cap\sqrt{(J)}=\sqrt{I\cap J}=\sqrt{(0)}$.

WLOG, assume that $I,J$$I,J$ are radical ideals, then

$$\begin{array}{}\text{(2)}& A/\sqrt{(0)}\cong A/I\times A/J\end{array}$$

Hence $(1,0)$$(1,0)$ is idempotent in $A/\sqrt{(0)}$$A/\sqrt{(0)}$. $◻$$\square$