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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