Perhaps this is obvious, but it really confuses me. It seems that I have to accept the fact:
Let $X$ be a noetherian topological space, and let $Y$ be an irreducible closed subset in $X$. Then the only irreducible component of $Y$ is $Y$ itself.
I follows the definition from Hartshorne:
In a noetherian topological space $X$, every nonempty closed subset $Y$ can be expressed as a finite union $Y=Y_{1}\cup\cdots\cup Y_{r}$ of irreducible closed subsets $Y_{i}$ in $Y$. WLOG we can require that $Y_{i}$ does not contain $Y_{j}$ for $i\neq j$, then the $Y_{i}$ are uniquely determined.
These $Y_{i}$ are called the irreducible component of $Y$. I understand that there is another way to define $Y_{i}$, as the maximal irreducible subset in $Y$, but I prefer to staying with Hartshorne, since I am reading it.
I have some attempt to prove the fact I need to accept:
Let $X$ be a noetherian topological space, and let $Y$ be an irreducible closed susbset in $X$. Suppose that $Y$ has irreducible components $Y_{1},\cdots, Y_{n}$, we need to show that $Y_{i}=Y$ for all $i=1,\cdots, n$.
We write $Y=Y_{1}\cup\cdots\cup Y_{n}$ as a finite union of closed irreducible subsets in $Y$, then $$Y=Y\cap \Bigg(\bigcup_{k=1}^{n}Y_{k}\Bigg)=Y\cap Y_{1}\cup \Bigg(\bigcup_{k=2}^{n}Y\cap Y_{k}\Bigg).$$ As every one is closed in $Y$, $Y\cap Y_{1}$ is closed in $Y$ and the big union is also closed $Y$, since $Y$ is irreducible, at least one of them is $Y$. If $Y\cap Y_{1}=Y$, then $Y\subseteq Y_{1}$. If the big union is not proper, then $$Y=\bigcup_{k=2}^{n}(Y\cap Y_{k})=Y\cap Y_{2}\cup \Bigg(\bigcup_{k=3}^{n}Y\cap Y_{k}\Bigg),$$ by the same argument, $Y\cap Y_{2}=Y$ or the big union equals to $Y$. Continue like this, and we will stop at some point because the union is finite, and we can conclude that $Y\subseteq Y_{j}$ for some $j=1,\cdots, n$.
Say $Y\subseteq Y_{1}$, since $Y_{1}$ is a subset of $Y$, we actually have $Y=Y_{1}$.
OK but then what I should do? I can certainly write $$Y=Y_{1}\cup Y_{2}\cup\cdots\cup Y_{n}=Y\cup Y_{2}\cup\cdots \cup Y_{n},$$ but this only implies that $Y_{2}\cup\cdots\cup Y_{n}\subseteq Y$ which is not helpful..
Thank you in advance!
Best Answer
Note that Hartshorne requires $Y_i$ not contained in $Y_j$ for $i\neq j$. Once you have that some $Y_j=Y$, then since $Y_i$ is contained in $Y$ for all $i$, we conclude that there is no $i\neq j$. That is, there is only the one irreducible component.