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Every (?) algebraic geometer knows that concepts like homotopy groups or singular homology groups are irrelevant for schemes in their Zariski topology. Yet, I am curious about the following.

Let's start small. Consider a local ring $A$ with maximal ideal $M$; is the affine scheme $X=Spec(A)$ connected? Sure, because every open subset of $X$ containing $M$ is equal to $X$ itself. Or because the only idempotents of $A$ are $0$ and $1$. But is it path connected? Yes, because if you take any point $P$ in $X$ the following path $\gamma$ joins it to $M$ (reminds you of the hare and the tortoise...):

$ \gamma(t)=P \quad for \quad 0\leq t < 1\quad , \quad \gamma (1)=M $.

The same trick shows that the spectrum of an integral domain is path connected: join the generic point to any prime by a path like above. More generally, in the spectrum of an arbitrary ring $R$ you can join a prime $P$ to any bigger prime $Q$ $(P \subset Q)$ by adapting the formula above:

$ \gamma(t)=P \quad for \quad 0\leq t < 1\quad , \quad \gamma (1)=Q $.

[Continuity at $t=1$ follows from the fact that every neighbourhood of $Q$ contains $P$ and so its inverse image under $\gamma$ is all of $[0,1]$ ]

The question in the title just asks more generally: Is a connected scheme path connected ?

Edit (after reading the comments) If an arbitrary topological space is connected and if every point has at least one path connected open neighbourhood, then the space is path connected. But I don't see why the local condition holds in a scheme, affine or not, even after taking into account what I proved about local rings.

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Yes if $X$ is noetherian, because your argument shows that $X$ is locally path-connected. –  Martin Brandenburg Oct 1 '10 at 11:22
Local path-connectedness implies local connectedness, and the latter can fail for Spec(R) when R is not noetherian. I believe that in Spec of an infinite product of fields every point is closed, only the obvious points are isolated, and there enough idempotents in the ring to show that the only connected sets are the points. (So each non-obvious point fails to have a connected nbhd.) –  Tom Goodwillie Oct 1 '10 at 12:56
I wonder if there can be a (non-noetherian) ring R having exactly two maximal ideals, such that no prime ideal belongs to both of them, but such that Spec(R) is connected. –  Tom Goodwillie Oct 1 '10 at 12:58
There's an old paper of Hochster, "Prime ideal structure of commutative rings" which answers exactly what topological spaces come from Spec $R$ (for arbitrary commutative rings $R$). This sounds like it may be relevant here. –  Karl Schwede Oct 1 '10 at 16:02
@Ying Zhang & Karl: One of the most interesting applications in the referenced paper of Hochster (his thesis, actually) is that, for any (commutative) ring R, there is another ring R' such that the prime ideals of R and R' are in bijection, but with containment reversed. A nice example: R = k[x], for an algebraically closed field k. Then Spec R' has infinitely many irreducible components, but just one closed point (so it's connected). –  Dave Anderson Oct 3 '10 at 2:49
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1 Answer

up vote 14 down vote accepted

There exist connected affine schemes which are not path connected. Let E be a compact connected metric space* which is not path connected (e.g., the closed topologist's sine curve) and consider the following.

$X={\rm Spec}(A)$ where $A$ is the ring of continuous functions $f\colon E\to\mathbb{R}$.

Then X is connected, since any idempotent f satisfies $f(x)\in\{0,1\}$ and, by connectedness of E, $f=0$ or $f=1$. The maximal ideals of A are $$ \mathcal{m}_x=\left\{f\in A\colon f(x)=0\right\} $$ for $x\in E$. There will also non-maximal primes (see this question for example) but, every prime ideal will be contained in one and only one of the maximal ideals**. So, we can define $\pi\colon X\to E$ by $\pi(\mathcal{p})=x$ for prime ideals $\mathcal{p}\subseteq\mathcal{m}_x$.

In fact, $\pi$ is continuous, using the following argument. For any open ball $B_r(x)$ in E, choose $f\in A$ to be positive on $B_r(x)$ and zero elsewhere. Then $D_f=\left\{\mathcal{p}\in X\colon f\not\in \mathcal{p}\right\}$ is open and $\pi^{-1}(B_r(x))\subseteq D_f\subseteq \pi^{-1}(\bar B_r(x))$. Writing $B_r(x)=\cup_{s < r}B_s(x)=\cup_{s < r}\bar B_s(x)$, this shows that there are open sets $U_s$ lying between $\pi^{-1}(B_s(x))$ and $\pi^{-1}(\bar B_s(x))$. So, $\pi^{-1}(B_r(x))=\bigcup_{s < r} U_s$ is open, and $\pi$ is continuous.

So, $\pi\colon X\to E$ is continuous and onto. If X was path connected then E would be too.

It may be worth noting that ${\rm Specm}(A)$ is also connected but not path connected, being homeomorphic to E.

(*) I assume that E is a metric space in this argument so that the open balls give a basis for the topology, and there are continuous $f\colon E\to\mathbb{R}$ which are nonzero precisely on any given open ball. Actually, it is enough for the topology to be generated by the continuous real-valued functions. So the argument generalizes to any compact Hausdorff space (+ connected and not path connected, of course).

(**) Maybe I should give a proof of the fact that every prime $\mathcal{p}$ is contained in precisely one of the maximal ideals $\mathcal{m}_x$. Let $V(f)=\{x\in E\colon f(x)=0\}$ be the zero set of f. Then, $V(\mathcal{p})\equiv\bigcap\{V(f)\colon f\in\mathcal{p}\}$ will be non-empty. Otherwise, by compactness, there will be $f_1,f_2,\ldots,f_n\in\mathcal{p}$ with $V(f_1)\cap V(f_2)\cap\cdots\cap V(f_n)=\emptyset$. Then, $f=f_1^2+f_2^2+\cdots+f_n^2\in\mathcal{p}$ would be nonzero everywhere, so a unit, contradicting the condition that $\mathcal{p}$ is a proper ideal. Choosing $x\in V(\mathcal{p})$ gives $\mathcal{p}\subseteq\mathcal{m}_x$.

On the other hand, we cannot have $\mathcal{p}\subseteq\mathcal{m}_x\cap\mathcal{m}_y$ for $x\not=y$. Letting $f,g\in X$ have disjoint supports with $f(x)\not=0, g(y)\not=0$ gives $fg=0\in\mathcal{p}$ and, as $\mathcal{p}$ is prime, $f\in\mathcal{p}\setminus\mathcal{m}_x$ or $g\in\mathcal{p}\setminus\mathcal{m}_y$.

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@Victor: the ideal corresponding to $F$ is not a prime unless $F$ is a point: the quotient is $\mathcal{C}(F,\mathbb{R})$. –  Laurent Moret-Bailly Nov 10 '10 at 7:41
Dear Anonymous, there is just one point that I don't understand: why do we have $D_f\subseteq \pi^{-1}(\bar B_r(x))$ ? Imagine a prime ideal $p \subsetneq m_z$ (strict inclusion) with $z\notin \bar B_r(x)$. Then $f\in m_z$ of course, but could it not happen that $f\notin p$, in which case we would have $p\in D_f$ but $\pi (p)=z \notin \bar B_r(x)$ ? –  Georges Elencwajg Nov 10 '10 at 19:47
Choose a $g\in A$ with $g(z)\not=0$ and support disjoint from $B_r(x)$. Then $fg=0\in p$ but $g\not\in m_z$. So $f\in p$. –  Anonymous Nov 10 '10 at 21:31
Right, thank you. Dear Anonymous, I can't tell you how impressed I am by this brilliant answer : it is a real gem. I am sorry that I can't do more to show my gratitude and admiration than upvoting you and accepting your answer. Also, I wish you Godspeed in your projected change of career: you obviously more than deserve a position in Academia. –  Georges Elencwajg Nov 10 '10 at 22:23
Thanks for those kind words and encouragement. Regarding this answer, it is interesting to note that $X={\rm Spec}(A)$ deformation retracts onto ${\rm Specm}(A)$, which is homeomorphic to E. So X and E are actually homotopy equivalent. –  Anonymous Nov 11 '10 at 0:24
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