Hartshorne Algebraic Geometry Proposition 1.5 (page5) In a notherian topological space $X$, every nomempty closed subset $Y$ can be expressed as a finite union $Y=Y_1 \cup \cdots \cup Y_r$ of irreducible closed subsets $Y_i$. If we require that $Y_i \nsupseteq Y_j$ for $i \neq j$, then the $Y_i$ are uniquely determined. They are called the irreducible components of $Y$.

I understood the existence of such a representation. But I can't understand a sentence in the uniqueness part, which I highlighted: Let $Y=Y_1 \cup \cdots \cup Y_r$ and $Y=Y'_1 \cup \cdots \cup Y'_s$ be two such representation. Then we can deduce that $Y_1=Y'_1$ (the details are in the textbook). Now let $Z=(Y-Y_1)^-$, then $Z=Y_2 \cup \cdots \cup Y_r$ and also $Z=Y'_2 \cup \cdots \cup Y'_s$.

What is the operator $^-$? I understood it as a closure, is it right? But then how can I deduce that $Z$ is represented as $Z=Y_2 \cup \cdots \cup Y_r=Y'_2 \cup \cdots \cup Y'_s$?

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    $\begingroup$ Yes, it does mean closure, and the deduction is because $Y_1 = Y'_1$ is irreducible and the union of the remainder is a closed set. $\endgroup$ – Zhen Lin Jun 22 '11 at 8:47
  • $\begingroup$ Dear Zhen, you should post this as an answer so that it can be (deservedly) upvoted and accepted. $\endgroup$ – Amitesh Datta Jun 22 '11 at 8:55

The bar must mean closure. It's clear that $Y - Y_1$ is a subset of $Y_2 \cup \ldots \cup Y_r$ and as the right hand side is a finite union of closed sets, it is closed and so the closure of this set is also a subset of it. So $\overline{Y - Y_1} \subset Y_2 \cup \ldots \cup Y_r$.

We do not miss any points of $\overline{Y - Y_1}$: $Y - Y_1$ is open (as $Y_1$ is closed, being irreducible) and intersects $Y_i$ for $i \ge 2$ (otherwise $Y_i \subset Y_1$, contrary to assumption) and as $Y_i$ is irredicible, every open subset of it is dense in it, which means that the closure of $Y - Y_1$ in $Y_i$ equals $Y_i$, or equivalently $\overline{Y - Y_1} \cap Y_i= Y_i$, for $i=2 \ldots r$, which implies the other inclusion, and hence equality.

We can hold the exact same argument for the $Y'_i$, $i = 2 \ldots s$.

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    $\begingroup$ +1 for filling in the details I was too lazy to write up. $\endgroup$ – Zhen Lin Jun 22 '11 at 9:45
  • $\begingroup$ Thanks, but I have a question. Since $(Y-Y_1) \cap Y_i$ is non-empty open in $Y_i$, it's closure in $Y_i$ equals $Y_i$ so that $\overline{(Y-Y_1)\cap Y_i} \cap Y_i = Y_i$. But how can you deduce that $\overline{Y - Y_1} \cap Y_i= Y_i$? $\endgroup$ – Gobi Jun 23 '11 at 1:00
  • $\begingroup$ Oh, I found the answer while I'm commenting. $Y_i = \overline{(Y-Y_1)\cap Y_i} \cap Y_i \subseteq \overline{Y-Y_1} \cap \overline{Y_i} \cap Y_i = \overline{Y-Y_1} \cap Y_i$. And $Y_i \supseteq \overline{Y-Y_1} \cap Y_i$ is obvious so that $ Y_i = \overline{Y - Y_1} \cap Y_i$. $\endgroup$ – Gobi Jun 23 '11 at 1:06

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