I'm trying to understand this example in Hartshorne's algebraic geometry book

In order to prove the irreducible part, suppose $Y$ is an irreducible space and $Y'$ a open subset of $Y$ with $Y'=Y'_1\cup Y'_2$ with $Y'_1,Y'_2$ proper closed subsets. Then $Y=(Y-Y')\cup (Y'_1\cup Y'_2)$ contradiction because $Y$ is irreducible.

Am I right? I need help also in the density part, I'm really stuck I don't know even how to begin.

Thanks a lot.

  • 6
    $\begingroup$ Hint for density: Being irreducible is equivalent to "any two non-empty open subsets intersect non-trivially" (by taking complements in the usual definition). $\endgroup$ Aug 5, 2013 at 7:17
  • $\begingroup$ @TobiasKildetoft Is my solution in the irreducible part right? In the density part, do I need to use the definition of Zariski topology or this is true in general? $\endgroup$
    – user42912
    Aug 5, 2013 at 7:42
  • 1
    $\begingroup$ In your argument, why should $Y_1\cup Y_2$ be closed in $Y$? (Note that this is $Y'$ which is assumed open in $Y$. It would probably be easier also for that part to rewrite the definition in the form I mentioned above. And no, this is all for arbitrary irreducible spaces, not just those coming from the Zariski topology. $\endgroup$ Aug 5, 2013 at 7:48
  • $\begingroup$ @TobiasKildetoft because I suppose by contradiction that $Y'$ is irreducible, then by definition of irreducible $Y'$ can be written as a union of closed proper subsets, say $Y'=Y'_1\cup Y'_2$, $Y'_1, Y'_2$ closed subsets in $Y'$, so they are closed in the whole space "Y" also. $\endgroup$
    – user42912
    Aug 5, 2013 at 8:35
  • 2
    $\begingroup$ No, closed subsets of open subset are not necessarily closed in the larger space. $\endgroup$ Aug 5, 2013 at 8:38

2 Answers 2


Your proof for the irreducible part is wrong. For example what do you mean by $Y' = Y_1 \cup Y_2$ with $Y_1,Y_2$ closed? The key question is closed in which space? First I would prove that an open subset of an irreducible space is dense: If $U$ open and non-empty is such that $\overline{U} \neq Y$ then immediately we can write $Y= U^c \cup \overline{U}$ contradicting $Y$ being irreducible.

Then now you can use this result to prove that any open subset of an irreducible space is irreducible: If $U = A \cup B$ with $A,B$ closed in $U$ then taking closures in $Y$ we get that $Y = \overline{U} = \overline{A} \cup \overline{B}$ and so this forces $\overline{A} = Y$ say. But now the closure of $A$ in $U$ which is equal to $\overline{A} \cap U$ is also equal to $A$ because $A$ was closed in $U$ by assumption. Thus $A = \overline{A} \cap U = Y \cap U = U$. It follows $U$ is irreducible in the subspace topology on it.

  • $\begingroup$ I don't understand how $Y$ is equal to $ U^{\complement} \cup \overline{U}$ and not merely a subset, since $U$ is open and thus $U \neq \overline{U}$. $\endgroup$ Nov 2, 2018 at 16:13
  • $\begingroup$ @G.Chiusole But $U \subsetneq \overline{U}$ $\endgroup$
    – Peanica
    Mar 4, 2019 at 2:17
  • $\begingroup$ Not necessarily a strict inclusion, for example, when $U=Y,$ but $U\subset \bar U$ is correct. $\endgroup$
    – Kenta S
    Jun 27, 2019 at 8:23

For a slightly different proof not using the density, consider that if an open subset $U \subset Y$ is reducible we can write $U = (U \setminus V) \cup (U \setminus W)$ for $V, W$ open non-empty subsets of $U$. Since $U$ is itself open, $V$ and $W$ will also be open subsets of $Y$. Hence we can write $Y = (Y \setminus V) \cup (Y \setminus W)$, so $Y$ is reducible.

  • $\begingroup$ I think that we also need to mention $V\cap W=\varnothing$ to make $Y=(Y\setminus V)\cup(Y\setminus W)$ work. Otherwise the union missing the intersection part. But this follows from $U=(U\setminus V)\cup (U\setminus W)$, if $V\cap W\neq \varnothing$, then $$(U\setminus V)\cup (U\setminus W)=U\setminus(V\cap W)\subsetneq U,$$ contradiction. $\endgroup$ Jan 4, 2022 at 14:49

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