Every irreducible closed has a generic point

Question

If $X$ is a scheme and $Z$ an irreducible closed subset, then $\exists\,\xi\in Z$ with $Z=\overline{\{\xi\}}$

Here is my attempt:

Let $X=\bigcup_i U_i$ be an open affine cover. Since $Z$ is irreducible, then $Z\cap U_i$ is a closed irreducible subset of the affine scheme $U_i$. This means there is a prime ideal $\mathfrak{p}_i\in U_i$ such that $Z\cap U_i=V_i(\mathfrak{p}_i)$ (where $V_i(I):=\{\mathfrak{p}\in U_i\mid \mathfrak{p}\supset I\}$)

On the other hand, $V_i(\mathfrak{p}_i)=\overline{\{\mathfrak{p}_i\}}^{U_i}$ (closure in $U_i$) or, equivalently, $V_i(\mathfrak{p}_i)=\overline{\{\mathfrak{p}_i\}}\cap U_i$. Therefore: $$Z=\bigcup_iZ\cap U_i=\bigcup_i\overline{\{\mathfrak{p}_i\}}\cap U_i\subset\bigcup_i\overline{\{\mathfrak{p}_i\}}$$

Conversely, since $\mathfrak{p}_i\in Z$, then $\overline{\{\mathfrak{p}_i\}}\subset Z$ for all $i$, so $\bigcup_i\overline{\{\mathfrak{p}_i\}}\subset Z$ and: $$Z=\bigcup_i\overline{\{\mathfrak{p}_i\}}$$

By irreducibility of $Z$, we have $Z=\overline{\{\mathfrak{p}_i\}}$ for some $i$. $_\blacksquare$

Is this correct?

I'm not sure about the conclusion since the union may be infinite, which made me suspicious.

Answer 1

It's not quite correct that $Z\cap U_i$ is $V$ of a prime ideal (if you mean $V$ to give a subscheme, not just a subset, which you should): think about the case $V(x^2)$. This is irreducible but not reduced. This is an error, but it's a fixable one, since you can just take the reduction everywhere and end up with the same topological space and then apply your argument.

The final part of your argument is indeed problematic: just because an irreducible topological space is a union of infinitely many closed subsets, one cannot conclude that one subset is necessarily the whole thing. Consider the underlying topological space of $\Bbb A^1_\Bbb C$ without the generic point, for instance - it's irreducible and also the union of infinitely many closed points.

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A way to fix this is to show that the generic point is in every affine open and generic points of irreducible affine schemes are unique. This shouldn't be too difficult, but if you have trouble, drop me a comment.

Answer 2

I'll write down how I fixed the final argument using KReiser's suggestion.

In case $X$ is affine, $X$ is an affine open cover of itself, so $Z=V(\mathfrak{p})=\overline{\{\mathfrak{p}\}}$ for some $\mathfrak{p}\in X$ (notice that here $V(\mathfrak{p})$ is merely a set, not a scheme). Such $\mathfrak{p}$ is unique because $V(\mathfrak{p})=V(\mathfrak{p}')\Leftrightarrow \mathfrak{p}=\mathfrak{p}'$, since $\mathfrak{p},\mathfrak{p}'$ are prime. So we have proven:

Lemma: every irreducible closed subset of an affine scheme has a unique generic point.

Now suppose $X$ is any scheme. For simplicity, assume $Z\cap U_i\neq\emptyset$ for every affine open $U_i$ in the cover.

We already know that $Z\cap U_i=\overline{\{\mathfrak{p}_i\}}\cap U_i$ for some $\mathfrak{p}_i\in U_i$.

Claim: $\mathfrak{p}_i\in U_i\cap U_j$ for all $i,j$.

Proof: Suppose $\mathfrak{p}_i\notin U_j$. Then $\mathfrak{p}_i$ is in $U_i\setminus U_j$, which is closed in $U_i$, therefore $\overline{\{\mathfrak{p}_i\}}\cap U_i\subset U_i\setminus U_j$. Consequently $Z\cap U_i\cap U_j=\overline{\{\mathfrak{p}_i\}}\cap U_i\cap U_j=\emptyset$. On the other hand, $Z$ is irreducible and $Z\cap U_i$ and $Z\cap U_j$ are nonempty open sets of $Z$, so their intersection $Z\cap U_i\cap U_j$ is nonempty (contradiction). $\blacksquare$

Now take an affine open set $V$ such that $\mathfrak{p}_i\in V\subset U_i\cap U_j$. Notice that: $$\overline{\{\mathfrak{p}_i\}}\cap V=(Z\cap U_i\cap U_j)\cap V=\overline{\{\mathfrak{p}_j\}}\cap V$$

Since $V$ is affine, $\mathfrak{p}_i=\mathfrak{p}_j$ by the lemma. We might as well write $\mathfrak{p}=\mathfrak{p}_i$ for all $i$, so

$$Z=\bigcup_i\overline{\{\mathfrak{p}_i\}}=\overline{\{\mathfrak{p}\}}$$

Finally, if $\exists\,\mathfrak{p}'$ with $Z=\overline{\{\mathfrak{p}'\}}$, then $\overline{\{\mathfrak{p}'\}}\cap U_i=\overline{\{\mathfrak{p}_i\}}\cap U_i$ for any $i$. Since $U_i$ is affine, by the lemma, $\mathfrak{p}'=\mathfrak{p}_i=\mathfrak{p}\,\,_\blacksquare$