# Show that a ring with disconnected spectrum is a product of two subrings. [duplicate]

It's an exercise from the book introduction to commutative algebra by Atiyah and Macdonald. If $\operatorname{Spec}(A)$ is disconnected, I'm asked to show that $A$ is a product of two subrings.

I know that $A/R$ is a product of two rings, where $R$ is the nilradical of $A$. What should I do? Also, my friend told me that it can be solved using the fact(I don't know it) that the category of affine schemes is isomorphic to the opposite category of rings, I don't understand how would this fact help. I want an elementary solution to this exercise. Thanks for any help.

• Quotients of rings induce inclusions of spectra. So he's reminding you to think of those disconnected pieces as subschemes, as spectra of quotients of A. Jun 18 '13 at 13:55
• The fact that the categories are equivalent is not too difficult. In one direction you send a ring $A$ to its spectrum $\mathrm{Spec}\, A$. In the other direction you take global sections. It's like how an affine variety is described by its regular functions. Jun 18 '13 at 14:04
• Is the equivalent used to show that if two rings have isomorphic spectrum, then they are isomorphic?
– lee
Jun 18 '13 at 15:02
• @lee That doesn't sound right. Shouldn't the spectrum of any two fields be isomorphic? Jun 18 '13 at 17:03
• @rschwieb. Certainly topologically the spectrum of any two fields is just a point. But there is an interesting ring of functions on the point! The topological information is far from determining the scheme structure (just as a smooth manifold is much more than a topological space) Jun 18 '13 at 17:18

Suppose the spectrum $X=\mathrm{Spec}(A)$ is disconnected. This means that it is the disjoint union of two closed sets (which are also open). Closed sets of the spectrum are, by definition, the subsets of the form $V(I)$ for $I$ an ideal of $A$. So the assumption is that $X=V(I)\cup V(J)$ for some ideals $I,J$ with $V(I)\cap V(J)=\emptyset$. Now, in general, from the definition, one has $V(I)\cap V(J)=V(I+J)$, because a prime ideal of $A$ contains both $I$ and $J$ if and only if it contains $I$ and $J$, if and only if it contains $I+J$. So $V(I+J)=\emptyset$. This means that $I+J$ does not lie in any prime ideal of $A$, which means it cannot be a proper ideal: $I+J=A$. On the other hand, again from the definition, one has $V(I)\cup V(J)=V(IJ)$. Because $I+J=A$, one has $IJ=I\cap J$ (this is proved early on in A and M). In fact, for such ideals $I$ and $J$ (sometimes called comaximal ideals), A and M prove that the natural ring map $A\rightarrow(A/I)\times(A/J)$ is an isomorphism (this is an instance of the Chinese remainder theorem). So indeed $A$ decomposes as a product.
EDIT: As is pointed out by the OP in the comments, my argument is not complete. It only shows what the OP already apparently knew, that $A$ modulo its nilradical decomposes as a product. One needs to argue that idempotents lift modulo the nilradical. This is proved in both the answers to If $\mathop{\mathrm{Spec}}A$ is not connected then there is a nontrivial idempotent
• Dear @lee, You are absolutely right. As a result I haven't really answered your question, particularly since you already knew that $A$ modulo its nilradical decomposed into a product, which is all that my argument shows ($V(IJ)=\mathrm{Spec}(A)$ shows that $IJ$ is contained in the nilradical, and by replacing $I$ and $J$ with their radicals, one can assume $IJ$ is the nilradical). You still need to know why you can lift idempotents modulo nil ideals. This is shown nicely in the previous question mentioned at the top of your post, so I guess I'll link to that in an edit to my answer. Jun 19 '13 at 13:57