# Connected components of a scheme are irreducible

Update 2: I posted an answer to this question.

Update 1: Problem is now solved because of the excellent hint by Qil. So, if someone wants to post an answer just for the sake of closing this question you are more than welcome. I will post an answer myself after I am back from a short vacation, if someone hasn't already left an answer by then.

The following question is problem in Goertz and Wedhorn's Schemes With Examples:

Let $X$ be a scheme. Then consider the following assertions:

(i) Every connected component of $X$ is irreducible.

(ii) $X$ is the disjoint union of its irreducible components.

(iii) For all $x \in X$, the nilradical of $O_{X,x}$ is a prime ideal.

Show that $(i) \Rightarrow (ii) \Rightarrow (iii)$. Show that all assertions are equivalent if the set of irreducible components of $X$ is locally finite (i.e. for all $x \in X$ there exists an open neighborhood of $x$ such that only finitely many irreducible components of $X$ pass through that open neighborhood).

So, I have managed to prove $(i) \Rightarrow (ii)$ which is a topological property, $(ii) \Rightarrow (iii)$ which follows from the fact the the irreducible components of $Spec(O_{X,x})$ are in bijection with the irreducible components of $X$ pasing through $x$. In fact, you can even use this fact to prove $(iii) \Rightarrow (ii)$ (I have not used the fact that the irreducible components of $X$ are locally finite so far).

However, I have no idea how to prove $(iii) \Rightarrow (i)$ or say $(ii) \Rightarrow (i)$ using the hypothesis that the irreducible components of $X$ are locally finite.

I think I am missing something obvious, and any hint would be appreciated. I have been thinking about this problem for a while, and I would not like complete solutions, just hints to get me started in the right direction. If I am unsuccessful, then I will later edit this question and ask for a solution. Also, although this might sound like a homework problem, it is not.

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Show that each irreducible component is open (here the locally finiteness is needed). – user18119 Mar 8 '12 at 22:23
Okay thanks. Will think about this and get back to you. – Rankeya Mar 8 '12 at 22:24
Thanks that was an excellent hint. – Rankeya Mar 8 '12 at 23:47
Dear Rankeya, you write "So, if someone wants to post an answer just for the sake of closing this question you are more than welcome ". I strongly advise you to write an answer yourself, using QiL's hint. This is (maybe surprisingly) explicitly encouraged on this site and will be an excellent exercise for you and profitable to users here. – Georges Elencwajg Mar 9 '12 at 9:03
Dear Georges, I was planning on doing it once I come back from a short vacation to Chicago. I don't have my laptop with me. But, I was in no way indicating that someone should post the answer and make my work easier. I have posted answers to my own questions in the past. Thanks for the heads up! – Rankeya Mar 9 '12 at 11:55

## 2 Answers

Okay here is answer based on Qil's hint:

Qil said, "Show that each irreducible component is open (here the locally finiteness is needed)."

Step 1 (Showing the sufficiency of Qil's claim): Assume $(ii)$. We will show that $(ii) \Rightarrow (i)$. Let $Z \subset X$ be a connected component of $X$, and let $x \in Z$. By $(ii)$ let $Y \subset X$ be the unique irreducible component of $X$ that contains $x$ (Note we don't really need $(ii)$ here because every point of a topological space is contained in an irreducible component. But, we do need $(ii)$ to prove Qil's claim). Since $Y$ is irreducible it is connected, and $Y \cap Z \neq \emptyset$ because $x \in Y \cap Z$. Thus, $Y \cup Z$ is connected. Also, $Z$ is a connected component, and so it follows that $Y \cup Z = Z$. Thus, $Y \subset Z$. Now $Y$ is an irreducible component, so closed $X$. By Qil's claim $Y$ is open in $X$. Thus, $Y$ is clopen in $Z$, and so, $Z - Y$ is open in $Z$. We have $Z = Y \cup (Z-Y)$, and since $Z$ is connected and $Y \neq \emptyset$, we get $Z = Y$. This shows that $Z$ is irreducible.

Step 2 (Proving Qil's Claim. We will assume $(ii)$ in this proof): Let $Y \subset X$ be an irreducible component and $y \in Y$. Now, the irreducible components of $X$ are locally finite. Hence, $\exists$ open $U_x \subset X$, such that $x \in U_x$ and only finitely many irreducible components of $X$ pass through $U_x$. Since the affine opens form a basis of $X$, we can assume w.l.o.g. that $U_x$ is an affine open, say $U_x = Spec(A)$.

Let $x = [p]$ be the corresponding point of $Spec(A)$. We know that the irreducible components of $U_x$ are in bijection with the irreducible component of $X$ passing through $U_x$. Thus, $U_x$ has only finitely many irreducible components, which implies that $Spec(A)$ has only finitely many irreducible components. So, $A$ has finitely many minimal primes (because the irreducible components of $Spec(A)$ are in bijection with the minimal primes of $A$).

Let $q_1,...,q_n$ be the distinct minimal primes of $A$, numbered in such a way that $[p]$ is a specialization of $[q_1]$, or in other words $V(q_1)= U_x \cap Y$, where by $(ii)$, $Y$ is the unique irreducible component of $X$ passing through $x$. If $q_1$ is the unique minimal prime of $A$, then $Spec(A) = V(q_1) = U_x \cap Z$, so $U_x \subset Z$, and we are done.

If not, then $n >1$. Note that $q_2...q_n \neq (0)$, as otherwise we would have $q_1 \supset q_2...q_n$, which would imply that $q_1 = q_i$ for some $i \neq 1$, contradicting the fact that all the $q_i$ are distinct. Since $Y$ is the unique irreducible component of $X$ passing through $x$, hence, $V(q_1)$ is the unique irreducible component of $Spec(A)$ passing through $[p]$. Thus, $[p]$ is not a specialization of $[q_2],...,[q_n]$,i.e., $q_2,..., q_n \not\subset p$, which implies that $q_2...q_n \not\subset p$. Thus, $\exists f\in q_2...q_n$ such that $f \neq 0$ and $f \notin p$. Then, $[p] \in D(f)$, and $[q_2],...,[q_n] \notin D(f)$. Note that as $[p]$ is a specialization of $[q_1]$, $[q_1] \in D(f)$.

Now, $D(f)$ is an open subset of $Spec(A)$. So, the irreducible components of $D(f)$ are in bijection with the irreducible components of $Spec(A)$ passing through $D(f)$. But, there is only one irreducible component of $Spec(A)$ passing through $D(f)$, namely $V(q_1) = U_x \cap Y$. Thus, $D(f) = V(q_1) \cap D(f)$ (since the irreducible components of a topological space cover that space). So, $D(f) \subset V(q_1) \subset Y$. But, $D(f)$ contains $x = [p]$. This shows that for all $x \in Y$, there is an open neighborhood of $x$ contained in $Y$. So, $Y$ is open.

(I tried to be very detailed in the proof. If someone has a shorter proof, please post it. Also, please give me some indication that the proof is correct. I have spent a few nights thinking about this problem, and II hope this is the last time I have to think about it.)

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I stumbled upon this while working on the same exercise. For others who may stumble upon this, I think this is a more concise (purely topological) solution for Step 2:

Assume (ii). Let $Z$ be an irreducible component of $X$ and let $x \in Z$ be given. Choose a neighborhood $U$ of $x$ that intersects only finitely many irreducible components, $Z_1 = Z,\, Z_2,\, \ldots, Z_n$. If we let $U_i = Z_i \cap U$, then the $U_i$ are the irreducible components of $U$ (*), and are pairwise disjoint by assumption. Since there are only finitely many and they are each closed in $U$, we see that for each $i$ $$U_i = U \setminus \left(\bigcup_{j \neq i} U_j\right)$$ is open in $U$. By definition of the subspace topology, there exists an open subset $V$ of $X$ such that $$U_1 = Z \cap U = V \cap U$$ is open in $X$. Since $U_1$ contains $x$, there exists an open neighborhood $U^\prime \subseteq V \cap U \subseteq Z$ of $x$. Therefore $Z$ is open. $\blacksquare$

In case it's not clear, here is a proof of the claim (*) above.

If $U$ is an open subset of $X$, then the irreducible components of $U$ are precisely the non-empty sets $Z \cap U$ as $Z$ ranges over the irreducible components of $X$.

Proof: Let $Z$ be an irreducible component of $X$ that meets $U$. Since $U \cap Z$ is open in $Z$, it is irreducible and dense in $Z$. If $Y$ is an irreducible subset of $U$ such that $U \cap Z \subseteq Y$, then we have $$Z \subseteq \overline{U \cap Z} \subseteq \overline{Y},$$ and hence $\overline{Y} = Z$ by maximality of $Z$. Now, since $Y$ is closed in $U$ we have $$Y = \overline{Y} \cap U = Z \cap U.$$ Hence $U \cap Z$ is an irreducible component. Every irreducible component of $U$ arises in this way, for if $Y$ is an irreducible component of $U$, then it is irreducible and hence lies in an irreducible component $Z$ of $X$, giving $Y = Z \cap U$. $\blacksquare$

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