Are there enough curves (to connect 'points' of f.g. algebras)?

(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.

• How do you map an algebra to a curve? – abx Jan 5 at 16:52
• Curve := finitely generated algebra of Krull dimension $1$ ? Also, are your algebras automatically commutative? – darij grinberg Jan 5 at 17:00
• Related: rationally connected varieties: en.wikipedia.org/wiki/… – YCor Jan 5 at 17:01
• 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? – Tim Campion Jan 5 at 18:03
• 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. – Mohan Jan 6 at 23:21