There exists $v_1,v_2\in A$ such that $v_1+v_2=1,v_1v_2=0$ and $v_i(x)=0$. Show that $A = A_1 \oplus A_2$ with $X_i= $ $^{a}\varphi(Spec(A_i))$.

Asked Sep 07 '22 at 15:42

Active Sep 07 '22 at 16:13

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Suppose that $X_1,X_2$ are closed subsets of $Spec A$. There exists $v_1,v_2\in A$ such that $v_1+v_2=1,v_1v_2=0$ and $v_i(x)=0$ for all $x\in X_i,i=1,2$. Prove that there exist $A_1,A_2$ such that $A = A_1 \oplus A_2$ with $X_i= $ $^{a}\varphi(Spec(A_i)), \ i=1,2$.

What I think is that let $A_i=A/(u_i)$ and in this way we have $Spec(A/(u_i))=V(u_i)$ which is closed in $SpecA$. Since $A=(u_1)+(u_2)$, then we need to show that $(u_1)\cap(u_2)=\emptyset$ and then $A=A/(u_1)\oplus A/(u_2)$. But I don't know how to show $(u_1)\cap(u_2)=\emptyset$. Please help. Thanks!

asked Sep 07 '22 at 15:42

dddxdddy

In fact, you cannot show that the intersection of the ideals is the empty set, because $0$ is always in that intersection. See the linked duplicate for an explanation. – KReiser Sep 07 '22 at 15:56
@KReiser Thanks. So how can we show $A_1,A_2$ exist to make $A$ a direct sum of them? – dddxdddy Sep 07 '22 at 16:00
$A_i=Av_i$..... – KReiser Sep 07 '22 at 16:00
@KReiser Sorry I don't quite understand why in this way $v_i(x)=0$ for all $x\in$ $^a\varphi (Spec(A_i))$. – dddxdddy Sep 07 '22 at 16:06
$v_2$ is sent to $0$ in $Av_1$ and vice versa. – KReiser Sep 07 '22 at 16:14

There exists $v_1,v_2\in A$ such that $v_1+v_2=1,v_1v_2=0$ and $v_i(x)=0$. Show that $A = A_1 \oplus A_2$ with $X_i= $ $^{a}\varphi(Spec(A_i))$.

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