# If $\operatorname{Spec} A$ is not connected then there is a nontrivial idempotent

I'm solving a problem from Atiyah-Macdonald.

I have to show that if $$X=\operatorname{Spec} A$$ is not connected then $$A$$ contains idempotents $$e \neq 0,1$$.

The converse is easy. If $$e \in A$$ is an idempotent then $$(e)+(1-e)=(1)$$ and $$(e)\cdot(1-e)=0$$ so that $$V(e) \cup V(1-e) = V( (e) \cdot(1-e))=V(0) = X, \\ V(e) \cap V(1-e) = V( (e)+(1-e))=V(1)=\varnothing$$ then $$V(e)$$ and $$V(1-e)$$ are both closed and open and $$X$$ is not connected.

Now let $$\mathfrak{a}$$ and $$\mathfrak{b}$$ be ideals in $$A$$ such that $$V(\mathfrak{a}) \cup V(\mathfrak{b})=X$$, $$V(\mathfrak{a}) \cap V(\mathfrak{b}) = \varnothing$$. Then $$V(\mathfrak{a}) \cup V(\mathfrak{b}) = V( \mathfrak{a} \cap \mathfrak{b} ) = X,$$ i.e. $$\left\{ \mathfrak{p} - \text{prime} \mid \mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{p} \right\} = X$$, i.e. $$\mathfrak{a} \cap \mathfrak{b} \subseteq \cap \mathfrak{p} = \mathfrak{n}$$ (nilradical). On the other hand since $$V(\mathfrak{a}) \cap V(\mathfrak{b}) = V(\mathfrak{a}+\mathfrak{b})=\varnothing$$ we have $$\left\{ \mathfrak{p} - \text{prime} \mid \mathfrak{a}+\mathfrak{b} \subseteq \mathfrak{p} \right\} = \varnothing$$. Then $$\mathfrak{a}+\mathfrak{b}=(1)$$ because any ideal that is not equal to $$(1)$$ is contained in some maximal ideal. Then $$\mathfrak{a}$$ and $$\mathfrak{b}$$ are comprime and $$\mathfrak{a} \cdot \mathfrak{b} = \mathfrak{a} \cap \mathfrak{b}$$. So I have two ideals $$\mathfrak{a}$$ and $$\mathfrak{b}$$ with properties $$\mathfrak{a} + \mathfrak{b} = (1), \\ \mathfrak{a} \cdot \mathfrak{b} = \mathfrak{a} \cap \mathfrak{b} = \mathfrak{n}.$$ I don't see any way to obtain a nontrivial idempotent $$e \in A$$ here. Please help me.

• @BenjaLim: It is not true that $x$ is idempotent. A priori only $x^2-x$ is nilpotent. And then one has to do some computations in order produce some idempotent element. Anyway, no calculations are needed if one uses the structure sheaf (see my answer). Mar 10, 2013 at 14:25
• @MartinBrandenburg do you know which kind of computations is needed here? Is there some standard trick? Mar 10, 2013 at 14:43
• By the way, more connections between the topology of Spec(A) and the algebra of A can be found here: math.stackexchange.com/questions/299765 Mar 11, 2013 at 1:36

This is most easily solved using the structure sheaf. More generally, let $X$ be any locally ringed space. Then there is a bijection between the clopen subsets of $|X|$ (the underlying space) and the idempotent elements of $\Gamma(X,\mathcal{O}_X)$. Essentially this comes down to the fact that a local ring has only trivial idempotents. Then for idempotents $e$ we have that $D(e)=\{x \in X : e_x=1\}$ is clopen with complement $V(e)=\{x \in X : e_x=0\}$, and conversely if $U \subseteq X$ is clopen then there is a unique idempotent $e$ satisfying $e|_U=1$ and $e|_{U^c}=0$ (by definition of a sheaf).

This bijection implies immediately that $X$ is connected iff $0,1$ are the only idempotents in $\Gamma(X,\mathcal{O}_X)$. And this has really nothing to do with spectra, it also holds for example for the sheaf of smooth functions on a manifold.

EDIT: Since not everyone is familiar with the structure sheaf, here is a more down-to-earth proof. I hope that this motivates to get familiar with the structure sheaf, because it is quite useful and gives geometric intuition.

Lemma: Let $A$ be a commutative ring, then every idempotent of $A/\sqrt{0}$ lifts to some idempotent of $A$ (in fact uniquely, but we won't need that).

Once we have proven the lemma, we can solve the problem: With the notation as in the question, choose $x \in \mathfrak{a}$ and $y \in \mathfrak{b}$ with $x+y=1$. Then $x^2+xy=x$ shows that $x^2-x \in \mathfrak{a} \cap \mathfrak{b}$ and is therefore nilpotent, thus $x$ becomes idempotent in $A/\sqrt{0}$, and we can apply the Lemma.

For the proof of the lemma, assume that $x \in A$ and $x^2-x$ is nilpotent, so there is some $n \in \mathbb{N}$ with $0 = (x^2-x)^n=x^n (x-1)^n$. Since $x^n$ and $(x-1)^n$ are coprime, the Chinese Remainder Theorem gives us $A \cong A/x^n \times A/(x-1)^n$. The preimage of $(0,1)$ is an idempotent $e \in A$ such that $x-e$ is nilpotent (since this is the case in both factors), so that $e$ is the desired lift.

There is a connection between these two proofs: The Chinese Remainder Theorem is just the sheaf property of $\mathcal{O}_{\mathrm{Spec}(A)}$ applied to disjoint open subsets.

• Oh, unfortunately I'm not familiar with structure sheafs, my problem is from the first chapter of the book Mar 10, 2013 at 14:33
• I don't get this. The last sentence of paragraph one is the result OP is trying to prove. I don't see how the language of structure sheaves helps.
– user29743
Mar 10, 2013 at 15:03
• @countinghaus: This is just the definition of a sheaf, sections can be glued. Here we just have two disjoint open subsets. Mar 10, 2013 at 15:08
• I like the use of the Chinese Remainder Theorem (proposition 1.10 in Atyah-Macdonald), I think it is the fastest way to solve this exercise if we know that existence of nontrivial idempotents is equivalent to nontrivial representability $A \simeq A_1 \times A_2$ (exercise 22 (II)-(III)). Mar 10, 2013 at 15:38
• Dear Martin Brandenburg, could you please explain what you mean by "a local ring has only trivial idempotents"? I first thought that you meant that all idempotents in a local ring are in fact 1, but obviously this is not the case since any root of unity is an idempotent in $\mathbb C$. Nov 2, 2013 at 15:24

This is similar to BenjaLim's answer. I know this is an old question, but I am answering only because I was trying to come up with a much simpler solution than the ones I have found. It seemed like even using something as elementary as the Chinese Remainder Theorem was unnecessary. Eisenbud breaks down the problem much more clearly than Atiyah and MacDonald in exercise 2.25 in his commutative algebra text, which helped me a lot.

Since $$\mathfrak{a} + \mathfrak{b} = (1)$$, we have $$a \in \mathfrak{a}$$ and $$b \in \mathfrak{b}$$ such that $$a + b = 1$$. Now let's look at $$1 = (a + b)^{2n}$$ where $$n$$ is such that $$(ab)^n=0$$. Since commutative rings have the binomial theorem, we can expand:

$$(a+b)^{2n} = a^{2n} + \ldots + b^{2n}$$

Let $$e_1 = a^{2n} + \binom{2n}{1}a^{2n-1}b + \ldots + \binom{2n}{n-1}a^{n+1}b^{n-1}$$

and $$e_2 = b^{2n} + \binom{2n}{1}b^{2n-1}a + \ldots + \binom{2n}{n-1}b^{n+1}a^{n-1}$$

Notice that I left out $$\binom{2n}{n}a^nb^n$$ since this is just zero. Now $$e_1 + e_2 = 1$$ and $$e_1e_2 = 0$$ (every term in $$e_1$$ has $$a^{n}$$ and every term in $$e_2$$ has $$b^{n}$$). So $$e_2 = 1 - e_1$$ and $$e_1e_2=e_1(1-e_1) = e_{1} - e_{1}^2=0 \Rightarrow e_1=e_{1}^2$$

Similarly for $$e_2$$, we see that $$e_1$$ and $$e_2$$ are idempotents. Well, how do we know that they are nontrivial? We only need to see that $$e_1$$ is an element in $$\mathfrak{a} \neq (1)$$ and $$e_2 \in \mathfrak{b} \neq (1)$$, so neither of them are $$1$$, and thus by the identity $$e_1 + e_2 = 1$$, neither of them can be zero as well.

I was wrong previously and indeed we need to do some work. If $\mathfrak{a} + \mathfrak{b} = 1$, there exists $x \in \mathfrak{a}$ and $y \in \mathfrak{b}$ such that $x +y =1$. Now by your observation above we have $(xy)^n = 0$ for some $n$ since $xy \in \mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{n}$. Now we have

$$1 = (x+y)^n = x^n + y^n +xy(\operatorname{some terms})$$

and so $x^n + y^n = 1 - xyz$ where $z = (\operatorname{some terms})$. Now $xy$ is nilpotent and $1$ is a unit so by Exercise 1.1 we have there exists $v \in A$ such that $v(x^n +y^n)= 1$. Then

$$(vx^n) = (vx^n)(v(x^n + y^n)) = v^2x^{2n}$$

and similarly for $vy^n$. Now show that one of these is not equal to $1$ or $0$.

If $$X={\rm Spec} A$$ is the disjoint union of (cl)open subsets $$X_1$$ and $$X_2$$, then since $${\scr O}_X$$ is a sheaf, the ring homomorphism $$\varphi:A={\scr O}_X(X)\to{\scr O}_X(X_1)\times{\scr O}_X(X_2)$$ is an isomorphism, since the images of any elements of $${\scr O}_X(X_1)$$ or $${\scr O}_X(X_2)$$ in $${\scr O}_X(X_1 \cap X_2)={\scr O}_X(\varnothing)=0$$ agree. If moreover $$X_i$$ is nonempty, we have $$0\neq1$$ in $${\scr O}_X(X_i)$$, since for $$x_i\in X_i$$, $${\scr O}_X(X)\to{\scr O}_{X,x_i}$$ factors through $${\scr O}_X(X_i)$$ and $$0\neq1$$ in $${\scr O}_{X,x_i}$$ (it's a local ring, i.e. has a unique maximal ideal, while the trivial ring has no maximal ideal). Therefore if $$X_1$$ and $$X_2$$ are both nonempty, then $$\varphi^{-1}((1,0))$$ and $$\varphi^{-1}((0,1))$$ are nontrivial idempotents in $$A$$.