0
$\begingroup$

From hartshorne exercise, it is know that an affine scheme is disconnected iff the ring is a product of rings iff the ring has one idempotent.

So I want to consider two rings, $k[x]\times k[y]$ and $k[x,y]$ where $k$ can be assumed to be $C$ or some algebraically closed field.

So $k[x]\times k[y]$ is really $A^1\cup A^1$ or 2 copies of $A^1$ where $k[x,y]$ is really $A^2$ with one extra point.

Q1: So product of schemes should look like disjoint union. Then it is more or less glueing schemes without identification. Fiber product of schemes is really identifying something and then glue them. Is this interpretation correct?

Q2. Is there a scheme that is connected but locally disconnected? Or some affine scheme is disconnected but scheme as a whole is connected.

Q3. The connected component of an affine scheme is exactly the number of its idempotent?

$\endgroup$
  • $\begingroup$ I don't know what you mean by "$k[x,y]$ is really $A^2$ with one extra point". What "extra point" do you think there is? $\endgroup$ – Eric Wofsey Oct 21 '17 at 23:08
  • $\begingroup$ @EricWofsey I am thinking about $Spec k[x,y]$. However for this case, I do not intend to make significant amount of distinction between affine plane with $Spec k[x,y]-\{(0\}$. $\endgroup$ – user45765 Oct 21 '17 at 23:23
  • 2
    $\begingroup$ But $\operatorname{Spec} k[x,y]$ has a lot more points than just points corresponding to elements of $k^2$ and the generic point of the plane. It also has a generic point for every irreducible curve. $\endgroup$ – Eric Wofsey Oct 21 '17 at 23:45
3
$\begingroup$

(1) No, the fiber product of schemes doesn't really involve gluing or identification at all. Instead, it is like a cartesian product (but more complicated since schemes are weird). For instance, the fiber product of two copies of $\mathbb{A}^1$ over $\operatorname{Spec} k$ is $\mathbb{A}^2$, and (assuming $k$ is algebraically closed) the you can think of closed points of $\mathbb{A}^2$ as ordered pairs of two closed points of $\mathbb{A}^1$. So on closed points, the fiber product is just the cartesian product in this case (though on non-closed points things are more complicated).

More generally, if $X$ and $Y$ are schemes with morphisms $f:X\to Z$ and $g:Y\to Z$, then the fiber product $X\times_Z Y$ is like an algebro-geometric version of the set $\{(x,y)\in X\times Y: f(x)=g(y)\}$ (so if $Z$ has just one point, this is all of $X\times Y$). In general, there is a natural surjection from $X\times_Z Y$ to this set, and in nice cases this surjection will be injective on the closed points.

(2) It is much easier to find a counterexample to your second question here. For instance if $A=k[x,y]/(xy)$ and $X=\operatorname{Spec} k[x,y]/(xy)$, then $X$ is connected but the affine open subset $U=\operatorname{Spec} A_{x+y}$ is disconnected. Geometrically, $X$ is two lines which meet at a point and $U$ is the complement of the intersection point, which is disconnected since you can decompose it into the two punctured lines.

Finding a connected but not locally connected scheme is a lot harder; for instance, any Noetherian space is locally connected (given a point, the complement of all the irreducible components that don't contain it is a connected neighborhood). Here's one example. Let $K$ be your favorite compact Hausdorff space that is connected but not locally connected (e.g., a topologist's sine curve), and let $C(K)$ be the ring of all continuous functions $K\to\mathbb{C}$. Then the topology of $\operatorname{Spec} C(K)$ is closely related to the topology of $K$ (see this MO post, for instance) and in particular it will also be connected but not locally connected.

(3) No. If $A$ is a commutative ring, then the idempotent elements of $A$ are in bijection with the clopen subsets of $\operatorname{Spec} A$. If there are only finitely many idempotents in $A$, then the connected components of $\operatorname{Spec} A$ are exactly the minimal nonempty clopen subsets, and every clopen subset is a union of some collection of connected components. So if there are $n$ connected components, there will be $2^n$ idempotents, one for each union of connected components.

If $A$ has infinitely many idempotents, the relationship between idempotents and connected components is much more complicated (the connected components correspond to the ultrafilters on the Boolean algebra of idempotents of $A$).

| cite | improve this answer | |
$\endgroup$

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.