# Connected vs irreducible Variety

I am asking if there is any particular criterion for a connected component of given variety to be irreducible (you can assume suitable conditions on the variety)

thanks

(See the book "Algebraic Geometry I" by Görtz and Wedhorn, Exercise 3.16.)

If $X$ is a scheme such that its set of irreducible components is locally finite (i.e. every point of $X$ has an open neighborhood which is disjoint to all but finitely many irreducible components of $X$), then the following are equivalent:

1. Every connected component of $X$ is irreducible.
2. $\forall x \in X :$ the nilradical of $\mathcal{O}_{X,x}$ is a prime ideal (where $\mathcal{O}_{X,x}$ denotes the local ring of $X$ in $x$)

Note that, for example, any locally noetherian scheme satisfies the requirement that the set of its irreducible components is locally finite.

Feel free to ask for a proof, the above result is not difficult.

Edit (Proof): We shall use the following facts:

Fact 1: For any scheme $X$ and $x \in X$, there is a bijection between

• the set of those irreducible components of $X$ that contain $x$; and
• the set of minimal prime ideals of the local ring $\mathcal{O}_{X,x}$.

(Sketch of proof: Let $U = \operatorname{Spec} A$ be an affine open subscheme containing $x$. First, use basic topology to relate the irreducible components of $U$ to those of $X$. Then, translate this to a statement involving the minimal prime ideals of $A$. Finally, use the well known 1-1 correspondence between the prime ideals of $A_\mathfrak{p}$ and the prime ideals of $A$ which are contained in $\mathfrak{p}$ (where $\mathfrak{p}$ denotes a prime ideal of $A$ and $A_\mathfrak{p}$ the localization of $A$ by $A \setminus \mathfrak{p}$.)

Fact 2: Let $A$ be any ring (commutative, with unit). Then $A$ contains exactly one minimal prime ideal if and only if its nilradical is a prime ideal. (This follows easily from $\operatorname{nil}(A) = \bigcap_{\mathfrak{p} \in \operatorname{Spec} A} \mathfrak{p}$.)

Fact 3: Every irreducible topological space is connected.

Fact 4: Let $X$ be a topological space and $(X_i)_{i \in I}$ a family ob subspaces of $X$. Assume that all $X_i$ are connected and that $\bigcap_{i \in I} X_i \neq \emptyset$. Then $\bigcup_{i \in I} X_i$ is connected.

Fact 5: Let $X$ be a topological space satisfying the following two assumptions:

• The set of the irreducible components of $X$ is locally finite; and
• $X$ is the disjoint union of its irreducible components.

Then every irreducible component of $X$ is open and closed in $X$. (Sketch of proof: Let $Z$ be an irreducible component of $X$; we only have to show that $Z$ is open. Let $(U_i)_{i \in I}$ be an open cover of $X$ such that every $U_i$ meets only finitely many irreducible components of $X$. It suffices to show that $Z \cap U_i$ is open in $U_i$ for all $i$. This follows from the assumptions by writing $U_i = \bigcup (Y \cap U_i)$ where $Y$ ranges over all irreducible components of $X$.)

The proof of the original statement will now be easy:

1. implies 2.: (Note that we won't need the assumption that the set of the irreducible components of $X$ is locally finite for this implication.) Let $x \in X$. By Fact 1 and Fact 2, it suffices to show that there is exactly one irreducible component of $X$ that contains $x$. So let $(X_i)_{i \in I}$ be those irreducible components of $X$ that contain $x$. By Fact 3, all $X_i$ are connected; by construction, $\bigcap_{i \in I} X_i \neq \emptyset$ (because the intersection contains at least $x$). Thus, by Fact 4, $\bigcup_{i \in I} X_i$ is connected and thereforce contained in a connected component, say $Y$, of $X$. That is, for all $j \in I$, we have inclusions $$X_j \subset \bigcup_{i \in I} X_i \subset Y ,$$ where $X_j$ is an irreducible component and $Y$ a connected component of $X$. But by assumption, $Y$ is even irreducible! Therefore, $X_j = Y$ because $X_j$ is an irreducible component. We have seen, that there is exactly one ireducible component of $X$ that contains $x$, namely $Y$.

2. implies 1.: By Fact 1, Fact 2 and the assumption about the local rings, $X$ is the disjoint union of its irreducible components. (Clearly, $X$ is the union of its irreducible components; this union is not disjoint if and only if there is an $x \in X$ which is contained in more than one irreducible component of $X$. But by Fact 1 and Fact 2, the local rings $\mathcal{O}_{X,x}$ for such $x$ would violate our assumption.) We may therefore use Fact 5 in the following.

From now on, the proof proceeds by purely topological arguments. Let $Y$ be a connected component of $X$ and let $Z'$ be an irreducible component of $Y$. Let $Z$ be an ireducible component of $X$ which contains $Z'$. By (the purely topological) Fact 5, $Z$ is open and closed in $X$. (This is the only place where the requirement that the set of the irreducible components of $X$ is locally finite is needed.) Thus, $Z \cap Y$ is open and closed in $Y$. Since $Y$ is connected, we must have either $Z \cap Y = Y$ or $Z \cap Y = \emptyset$; however, the latter is impossible since, by construction, $Z \cap Y \supset Z'$ (and $Z'$, being an irreducible component of some space, can't be empty). Therefore, $Z \cap Y = Y$; equivalently, $Z \supset Y$. But $Y$ is a connected component and $Z$ is connected (by Fact 3, because $Z$ is irreducible by definition). Thus, $Y = Z$; in particular, $Y$ is irreducible.

• thanks for your answer, yes I want to see a prove please? – Z.A.Z.Z Apr 3 '15 at 7:28
• @AllyMath Sure, I've added a proof to my answer. – c_c_chaos Apr 3 '15 at 16:55
• thank very mutch @c_c_chaos – Z.A.Z.Z Apr 3 '15 at 18:04