TITLE: Connected components of a scheme are irreducible
QUESTION [13 upvotes]: 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.

REPLY [7 votes]: 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 $Y = Z \cap U$. $\blacksquare$