(Intuition: any two points in a connected space may be connected by a path. I would like to know if something like this holds in certain category of `connected algebraic spaces'. I formulate the precise question in terms of commutative algebra.)

Let ${\cal A}$ be the category of finitely generated (associative unital) commutative ${\mathbb{C}}$-algebras with exactly two idempotents.

Let ${f, g : A \rightarrow \mathbb{C}}$ be two maps in ${\cal A}$.

Does it exist a map ${c : A \rightarrow C}$ (always in ${\cal A}$) with $C$ a curve such that both $f$ and $g$ factor through $c$? (I.e. such that there are maps ${f', g' : C \rightarrow \mathbb{C}}$ such that ${f' c = f}$ and ${g' c = g}$.)

Here by "curve" I mean a connected finitely generated $\mathbb{C}$-algebra of Krull dimension 1.

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    $\begingroup$ How do you map an algebra to a curve? $\endgroup$ – abx Jan 5 at 16:52
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    $\begingroup$ Curve := finitely generated algebra of Krull dimension $1$ ? Also, are your algebras automatically commutative? $\endgroup$ – darij grinberg Jan 5 at 17:00
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    $\begingroup$ Related: rationally connected varieties: en.wikipedia.org/wiki/… $\endgroup$ – YCor Jan 5 at 17:01
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    $\begingroup$ To clarify: "exactly two idempotents" means that 0 and 1 are the only idempotents (and they are distinct), and this condition is the same as "connected", right? $\endgroup$ – Tim Campion Jan 5 at 18:03
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    $\begingroup$ Any two points of an irreducible variety can be connected by an irreducible curve. A beautiful proof was given by C. P. Ramanujam and can be found in his collected works. If dimension of $X$ s one, there is nothing to prove. If dimension is larger, blow up the two points and by Bertini, a general hyperplane intersects it in an irreducible variety. Since the exceptional divisors are codimension one, this variety intersects both and thus its image in $X$ is an irreducible variety passing through both points and one less dimension. $\endgroup$ – Mohan Jan 6 at 23:21

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