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.


marked as duplicate by Martin Brandenburg, Martin, Myself, Quixotic, Julian Kuelshammer Jun 19 '13 at 8:46

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  • $\begingroup$ 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. $\endgroup$ – Loki Clock Jun 18 '13 at 13:55
  • $\begingroup$ 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. $\endgroup$ – Cocopuffs Jun 18 '13 at 14:04
  • $\begingroup$ Is the equivalent used to show that if two rings have isomorphic spectrum, then they are isomorphic? $\endgroup$ – lee Jun 18 '13 at 15:02
  • $\begingroup$ @lee That doesn't sound right. Shouldn't the spectrum of any two fields be isomorphic? $\endgroup$ – rschwieb Jun 18 '13 at 17:03
  • 1
    $\begingroup$ @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) $\endgroup$ – Dhruv Ranganathan 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

  • $\begingroup$ 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. $\endgroup$ – lee Jun 19 '13 at 13:32
  • $\begingroup$ 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. $\endgroup$ – Keenan Kidwell Jun 19 '13 at 13:57

Every connected component $V$ of $\operatorname{Spec}(A)$ is a closed set. Therefore $\operatorname{Spec}(A/I(V))$ is just that component. Therefore one can partition $\operatorname{Spec}(A)$ by the $\operatorname{Spec}(A/I(V))$. Every ring $B$ that has morphisms to each $A/I(V)$ has a universal morphism to $\prod A/I(V)$. Is this a quotient, when $B=A$? What's the kernel? In the other category, the morphism from $A$ induces a natural map from the spectrum of the product to $A$, whose topology certainly matches that of $A$.


So, what I was hoping would make the statement elementary was that if $\operatorname{Spec}(A)$ had subspaces that matched the product's spectrum, the disjointness of the subspaces would imply the only topological difference there could be was whether all the subspaces of $\operatorname{Spec}(A)$ were closed like they are in the product spectrum. Since I wasn't doing the math, I forgot that if $A$ wasn't reduced it could have the same topology.

The surjectivity condition on the universal map reinforces the connection. That the kernel is the nilradical can checked several ways, but see Kidwell's answer for proving the nilradical is $\{0\}$.

W.l.o.g., let $f, g$ be quotient maps from $A$. Then the universal map $<f,g>:A\rightarrow A/\ker f \prod A/\ker g$ is surjective iff

$$\not \exists <[y]_f,[z]_g>: f^{-1}(y) \bigcap g^{-1}(z)=\emptyset$$ $$\iff \forall y, z \in A, \exists i \in \ker f, j \in ker g : y+i = z+j$$ $$\iff \forall y, z, y-z \in \ker f+ \ker g.$$

Thus surjectivity of the universal map holds iff the kernels are comaximal. Furthermore, comaximality implies disjoint varieties, and the radical of the sum of the kernels must be $A$ - which is just the sum of the kernels, if one is initially defining the kernels as $I(V)$ for $V$ a connected component.

  • $\begingroup$ Is I(V) mean the intersection of all the prime ideals belonging to V? $\endgroup$ – lee Jun 18 '13 at 14:57
  • $\begingroup$ $I(V),$ or $Z(V)$ in some people's notation, is the ideal of elements of $A$ that vanish on every point of $V$. $\endgroup$ – Loki Clock Jun 18 '13 at 15:00
  • $\begingroup$ The kernel is the intersection of I(V)'s, however I only know that it is contained in the nilradical of A. And I don't know how to show that the universal morphism you constructed above is surjective. Sorry that I'm new to this subject, so can you explain more concisely to me about it? Thanks a lot. $\endgroup$ – lee Jun 18 '13 at 15:05
  • $\begingroup$ The more I started to think about it, the less convincing it was. See Kidwell's answer instead. I'm working on a proof of the surjectivity of the product, but I might fall asleep before I give it to you, so just note that the universal map can be described as $x \mapsto (f(x),g(x))$ for $f,g$ the quotient maps. $\endgroup$ – Loki Clock Jun 18 '13 at 18:23

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