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In Vakil's algebraic geometry notes, a problem asks us to find an example of a ring $A$ where $\operatorname{Spec}A$ is reducible but connected. The hint he gives is to consider the symbol "$\times$".

The hint has led me to consider the ring $\mathbb{C}[x,y]/((y-x)(y+x))$ which (if I'm correct) is just the union of the lines cut out by $y=x$ and $y=-x$. We then have that $\mathbb{C}[x,y]/((y-x)(y+x))=V(y-x)\cup V(y+x)$. Since neither $V(y-x)$ nor $V(y+x)$ equals the whole space (and both are nonempty), it follows that the space is reducible.

Intuitively, it seems like the space is connected. However, I'm having trouble actually proving this. In an answer to similar question asked here, an answer given uses the theory of idempotents. However, this theory hasn't been developed yet in the notes I'm reading. Is there a way to prove that the space is connected from first principles?

Bernard
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ponchan
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    You have two lines (lines are connected, right?) meeting in a point. – Angina Seng Jul 31 '20 at 14:30
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    @AnginaSeng Yes, I want to know how to rigorously prove this given open sets in the Zariski topology. – ponchan Jul 31 '20 at 14:32
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    ?? Aren't we talking about closed subsets in the Zariski topology? And, surely lines are connected?? – Angina Seng Jul 31 '20 at 14:34
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    Isn't the quotient you're suggesting just $\mathbb C$? Wouldn't you need to use the product of $(x-y)$ and $(x+y)$ to get the union of lines as the zero set? – rschwieb Jul 31 '20 at 14:44
  • @rschwieb Why is the quotient I suggested $\mathbb{C}$? – ponchan Jul 31 '20 at 15:08
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    $\mathbb C[x,y]/(x-y)\cong \mathbb C[x]$ using the substitution homomorphism $y\mapsto x$. Then the second quantity $x+y=2x$, so quotienting further by $(2x)$ gets you just $\mathbb C$. But on the other hand the zero set of $(x-y)(x+y)$ is like you are describing. – rschwieb Jul 31 '20 at 15:09
  • @rschwieb Thanks. Edit made. Question remains the same regarding connectedness. – ponchan Jul 31 '20 at 15:13
  • @ponchan show that if $X$ is a topological space, $X = A \cup B$, and $A$ and $B$ are connected and intersect each other non-trivially, then $X$ is connected. – hunter Jul 31 '20 at 15:16
  • @hunter Ok, but how to actually show, say, that $V(y-x)$ is connected? Sure, you can look at it, but I mean rigorously, if we were to assume $V(y-x)=(\operatorname{Spec}A-V(I))\cup(\operatorname{Spec}A-V(J))$? – ponchan Jul 31 '20 at 15:23
  • @ponchan show that it gets the co-finite topology. thus any two non-empty opens in it intersect. – hunter Jul 31 '20 at 15:30

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Vakil's text states in the introduction "[t]he reader should be familiar with some basic notions in commutative ring theory", and I would argue that the definition of what an idempotent is would be one of those basic notions. Especially since the concept completely answers the question about when an affine scheme is connected!

Anyways, if you're looking for a solution which doesn't use idempotents, you're most of the way there already. As you've observed, $V((x+y)(x-y))=V(x+y)\cup V(x-y)$ as sets, and $V(x+y)\cong V(x-y)\cong \Bbb A^1$, which is irreducible and thus connected. As a set which can be written as $A\cup B$ with $A,B$ connected and $A\cap B\neq\emptyset$ is connected, we have that $V((x+y)(x-y))$ is connected.

KReiser
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  • It's not a matter of "covering [my] ears", it's a matter of wanting to know a particular solution from first principles. Also, since Vakil defines "idempotent" a page earlier, and the theorem used in the link answer isn't discussed up to this point, I think it's safe to say that the intended a solution from first principles, whether or not you happen to think it's the best one. – ponchan Jul 31 '20 at 18:28
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    @ponchan I did not intend that phrase to come off as seriously as you've taken it - if I've offended you that deeply, my apologies, and I've removed it. The rest of the post explains how to solve your problem in the manner you're looking for. – KReiser Jul 31 '20 at 18:38
  • I'm not sure what part of what I just wrote indicates that I was "offended deeply". I sure hope that wasn't your intention! – ponchan Jul 31 '20 at 19:30
  • It's not just what you've written, it's also that it seems like you downvoted a correct answer (your reputation dropped by 1 around the same time the score of this post went from 0 to -1). Anyways, I hope you find the solution to your problem in the manner you asked useful - do you have any questions about that? – KReiser Jul 31 '20 at 19:44