13
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Hint for density: Being irreducible is equivalent to "any two non-empty open subsets intersect non-trivially" (by taking complements in the usual definition). $\endgroup$ – Tobias Kildetoft Aug 5 '13 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 '13 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. $\endgroup$ – Tobias Kildetoft Aug 5 '13 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 '13 at 8:35
  • $\begingroup$ No, closed subsets of open subset are not necessarily closed in the larger space. $\endgroup$ – Tobias Kildetoft Aug 5 '13 at 8:38
19
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $U \neq \overline{U}$. $\endgroup$ – G. Chiusole Nov 2 '18 at 16:13
  • $\begingroup$ @G.Chiusole But $U \subsetneq \overline{U}$ $\endgroup$ – Ping Wan Mar 4 at 2:17
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.