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Let $R$ be a commutative ring. How can I prove that $\operatorname{Spec}(R)$ is not connected if and only if $R$ is isomorphic to the product of nonzero ring $A_1$ and $A_2$?

If we consider $R=\mathbb{C}[x,y]/(xy)$ we have that $\operatorname{Spec}(R)$ is connected, but I think that I can write $R$ as $\mathbb{C}[x,y]/(y) \times \mathbb{C}[x,y]/(x)$...

ArthurStuart
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    The proof you ask for is standard and googleable. Your example is wrong. You can't split since $(x)$ and $(y)$ aren't comaximal (they generate the ideal of polys with zero constant coefficient). It is definitely true that $\text{Spec}(A)$ is connected. Note that if $V(I)$ contains both $(x)$ and $(y)$ it's everything. Indeed, note then that $V(I)\supseteq V((x))$ and $V(I)\supseteq V((y))$ so $V(I)\supseteq V((x))\cup V((y))=V((xy))=\text{Spec}(A)$. So, we can assume that $\text{Spec}(A)=V(I)\cup V(J)$ where $(x)\in V(I)$ and $(y)\in V(J)$. Note then that $(x,y)\in V(I)\cap V(J)$ so they can't – Alex Youcis Sep 22 '13 at 09:13
  • be disjoint. Geometrically, any non-stupid closed subset of the cross $\times$ cut out by $xy$ will contain $(0,0)$ and so it can't be a disjoint union. – Alex Youcis Sep 22 '13 at 09:14
  • @AlexYoucis I'm sorry... I was wrong. But I can write $\mathbb{C}[x,y]/(xy-1)$ as product... – ArthurStuart Sep 22 '13 at 09:40
  • @YACP I think that this is the product of $\mathbb{C}[x]$ and $\mathbb{C}[y]$ – ArthurStuart Sep 22 '13 at 09:51
  • @YACP In the quotient we have that $xy=1$ so we have only $p(x)+q(y)$. Now I can send $(p(x)+q(y))$ in $(p(x),q(y))$... – ArthurStuart Sep 22 '13 at 10:01
  • @ArthurStuart Is this a ring homomorphism? –  Sep 22 '13 at 10:04
  • @YACP Not... I have a problem with costants – ArthurStuart Sep 22 '13 at 10:06
  • @YACP You suggest me to think that this ring is an integral domain, so there aren't nilpotents – ArthurStuart Sep 22 '13 at 10:19
  • @YACP But... if the quotient is an integer domain, then $Spec(\mathbb{C}[x,y]/(xy-1))$ is irreducible, so is also connected. – ArthurStuart Sep 22 '13 at 10:30
  • And so... what is a significant example of non connected $spec(A)$? – ArthurStuart Sep 22 '13 at 10:51
  • @ArthurStuart The CRT is a theorem about non-connected spectra :) – Alex Youcis Sep 22 '13 at 21:42

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Let me show you why $R=\mathbb{C}[X,Y]/(XY)$ can't be isomorphic with a direct product of two (or more) rings.

Assume the contrary and look at the idempotents of both rings. A direct product of two (or more) rings has nontrivial (that is, different from $0,1$) idempotents (for instance, $(1,0,\dots)$) while $R$ don't. Write $R=\mathbb C[x,y]$ with $xy=0$ and let $f(x,y)\in R$ an idempotent. We get $f^2(X,Y)-f(X,Y)\in(XY)$, so $XY\mid f(X,Y)[f(X,Y)-1].$ Since $X$ and $Y$ are prime elements we have $X\mid f$ or $X\mid f-1$, and analogously for $Y$. I claim that $XY\mid f$ or $XY\mid f-1$. Consider the following situation: $X\mid f$ and $Y\mid f-1$, that is, $f=Xf_1$ and $f=1+Yf_2$. Then $Xf_1=1+Yf_2$ and now send $X$ and $Y$ to $0$ in order to obtain a contradiction. (The other case, $X\mid f-1$ and $Y\mid f$ can be discarded by the same argument.) If $XY\mid f(X,Y)$ we have $f(x,y)=0$ while $XY\mid f(X,Y)-1$ we get $f(x,y)=1$, and we are done.