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.

2 suppressed an "n" in "questionn"

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

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# Is every connected scheme path connected?

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