13

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.

lee
  • 2,900
  • 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. – Loki Clock 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. – Cocopuffs 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? – rschwieb Jun 18 '13 at 17:03
  • 1
    @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) – Dhruv Ranganathan Jun 18 '13 at 17:18
  • @DhruvRanganathan You're overwhelming me a little bit :) 1) Are you confirming my intuition that spectra do not determine their rings up to isomorphism? If so, thanks. That's what I was trying to convey to lee 2) You're saying that when the scheme structure is taken into consideration, you get a lot more information (but does having that extra information isomorphic imply the rings are isomorphic?) – rschwieb Jun 18 '13 at 17:42
  • Dear @rschwieb, Correct, there are lots of schemes consisting of just one point (the spectrum of any Artin local ring is such a scheme), but why they all have homeomorphic underlying set, they are not all isomorphic. An isomorphism between the spectra of two fields is the same as an isomorphism between the two fields, for example. – Keenan Kidwell Jun 18 '13 at 18:07
  • Dear @lee, I don't think that A and M claim that $A$ is a product of two subrings, but instead that $A$ is just a product of two rings. In the product $R=\prod_{i=1}^n R_i$ of rings $R_i$, the individual factors $R_i$ are quotients of $R$, but they are generally not subrings. – Keenan Kidwell Jun 18 '13 at 18:36
  • @KeenanKidwell I think I am experiencing a misconception that you and Dhruv are not picking up on. As far as I know, an isomorphism of spectra is just a (topological) homeomorphism. But judging from the last two answers I think there must be more. I had thought that the "more" version was an isomorphism of schema, not spectra. What do you additionally need to claim two spectra are isomorphic besides the homeomorphism? – rschwieb Jun 18 '13 at 19:36
  • Dear @rschwieb, I think it is just a misunderstanding of language. When you say "isomorphism of spectra," I think "isomorphism of schemes." I would just say "homeomorphism of spectra" to mean a map on the underlying topological spaces that is a homeomorphism, but may not be the underlying map of topological spaces of a morphism of schemes. So if by isomorphism (of spectra) you meant homeomorphism, not necessarily a morphism of schemes, then yes, most definitely the spectra of all fields are all isomorphic. – Keenan Kidwell Jun 18 '13 at 19:45
  • Oh, and to answer your question, two affine schemes, $\mathrm{Spec}(A)$ and $\mathrm{Spec}(B)$ are isomorphic (as schemes) if and only if they are isomorphic as rings. Any ring map between them induces a morphism between the associated affine schemes, meaning a map of topological spaces and map of structure sheaves. The underlying map of topological spaces is determined entirely by the ring map. But a continuous map between the spectra may not come from a ring map. – Keenan Kidwell Jun 18 '13 at 19:49
  • @KeenanKidwell Great, thanks: now everything fits together. – rschwieb Jun 18 '13 at 19:52

1 Answers1

13

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

Community
  • 1
Keenan Kidwell
  • 26,770
  • 1
    But you haven't show that the intersection of I and J is null. Howeverm, I know your way is right, just with a slight change. – lee Jun 19 '13 at 13:32
  • 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. – Keenan Kidwell Jun 19 '13 at 13:57