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. 



Is every connected scheme path connected?
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](http://en.wikipedia.org/wiki/Topologist%27s_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](https://mathoverflow.net/questions/35793/prime-ideals-in-c0-1) 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 $g\in\mathcal{p}\setminus\mathcal{m}\_y$.


