In Chapter 1 of Robin Hartshorne's "Algebraic Geometry," exercise 1.10 is:

If $Y$ is any subset of a topological space $X$, then $\dim Y \le \dim X$.

Here, $\dim X$ is the supremum of integers $n$ such that there exists a chain of distinct irreducible closed subsets $X_0\subsetneq X_1\subsetneq \cdots \subsetneq X_n$ of $X$.

I figured that if I can create a chain of irreducible closed subsets in $X$ from a chain in $Y$, $Y\cap X_0\subsetneq Y\cap X_1\subsetneq \cdots \subsetneq Y\cap X_n$, where the $X_i$ are closed subsets of $X$, since then, $n\le \dim X$.

However, the only chain in $X$ I can create from this chain in $Y$ is $X_0\subsetneq X_0\cup X_1\subsetneq \cdots \subsetneq \bigcup_{i=0}^n X_i,$ which I cannot prove consists of irreducible sets.

I feel like this exercise should be quite trivial, but I can't figure it out, so I would appreciate any help you can give me.

  • $\begingroup$ If you start with a chain in $Y$, say $Y_i := Y \cap X_i$, then as stated $X_i$ is not well-defined. There is, however a canonical choice of $X_i$ starting from $Y_i$, and you don't need to further manipulation (i.e. you get $X_0 \subset X_1 \subset \cdots \subset X_n$ for free, although you will need to justify why they are distinct). Can you see what $X_i$ to choose? $\endgroup$ – Pig Jul 2 '19 at 3:56
  • $\begingroup$ $X_i=\bar{Y_i}$? $\endgroup$ – Kenta S Jul 2 '19 at 3:59
  • $\begingroup$ :) The question is then 1. the sets are distinct and 2. why they are irreducible $\endgroup$ – Pig Jul 2 '19 at 5:17
  • $\begingroup$ Thanks, it was very helpful. $\endgroup$ – Kenta S Jul 2 '19 at 5:40

Let $Y_0\subsetneq Y_1 \subsetneq\cdots\subsetneq Y_n$ be an arbitrary chain of irreducible, closed subsets of $Y$. Let $\overline{Y_i}$ denote the closure of $Y_i$ in $X$. $\overline{ Y_{i-1}}\subsetneq \overline{ Y_i}$, since, otherwise, $Y_{i-1}=Y\cap \overline{Y_{i-1}}=Y\cap \overline{Y_i}=Y_i,$ contrary to the hypothesis.

If $\overline{Y_i}$ is irreducible in $X$ by Proposition 1.6 of Robin Hartshorne's "Algebraic Geometry":

Any nonempty open subset of an irreducible topological space is dense and irreducible, If $Y$ is a subset of a topological space $X$, which is irreducible in its induced topology, then the closure $\overline Y$ is also irreducible.

Hence, we have created a chain $\overline{Y_0}\subsetneq \overline{Y_1} \subsetneq\cdots\subsetneq \overline{Y_n}$ in $X$. By the definition of $X$'s dimension, $n\le \dim X$. Since $Y_0\subsetneq Y_1 \subsetneq\cdots\subsetneq Y_n$ was arbitrary, $\dim Y\le \dim X.$


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