Connected components of infinite type schemes I can't quite figure this out. Perhaps it's simple. 
If $S$ is an affine scheme over $\mathbb{C}$ and $s$ and $t$ are $\mathbb{C}$ points in the same connected component, is there a finite type connected $\mathbb{C}$ scheme mapping to $S$ such that $s$ and $t$ are contained in the image of this map? 
 A: That is false, and the suggestions by Ariyan Javanpeykar and Joe Berner lead to a counterexample.  Let $R_0$ be $\mathbb{C}[t]$.  Let $\Sigma$ be the multiplicative closed set of all elements $u(t)\in \mathbb{C}[t]$ such that $u(0)u(1)\neq 0$.  Let $R$ be the fraction ring $\Sigma^{-1}R_0$.  This is $\mathbb{C}$-algebra that is not of finite type.  
For $\lambda=0,1$, denote by $e_\lambda$ the $\mathbb{C}$-algebra homomorphism, $$e_\lambda:R_0 \to \mathbb{C}, \ \ e_\lambda(u) = u(\lambda).$$  This maps every element in $\Sigma$ to an invertible element of $\mathbb{C}$.  Thus, there is a unique induced $\mathbb{C}$-algebra homomorphism, $$ f_\lambda:R\to \mathbb{C},$$ with $f_\lambda(u)$ equal to $e_\lambda(u)$ for every $u\in R_0$.  The kernel of $f_\lambda$ is a maximal ideal $\mathfrak{m}_\lambda$, i.e., it is a $\mathbb{C}$-point of $\text{Spec}\ R$.  Of course $\text{Spec}\ R$ is connected, and even integral.  The only prime ideals are $\mathfrak{m}_0$, $\mathfrak{m}_1$ and the zero ideal.  Indeed, the primes of $R$ are in bijective correspondence with the primes of $R_0$ that are disjoint from $\Sigma$.  The complement of $\Sigma$ equals the union of the prime ideals $\langle t\rangle$ and $\langle t-1\rangle$.  By Prime Avoidance, every prime contained in this union is contained in one of these primes.  The only prime ideals of $R_0$ contained in one of these primes are the primes themselves and $\{0\}$.  
For every finite type $\mathbb{C}$-algebra $A$ that is an integral domain,  for the group of units $A^\times$, the quotient group $A^\times/\mathbb{C}^\times$ is a finitely generated, Abelian group.  However, the group $R^\times/\mathbb{C}^\times$ is infinitely generated.  Thus, for every $\mathbb{C}$-algebra homomorphism, $$g:R\to A,$$ the induced map of groups of units must have a kernel, i.e., there exist nonscalar units $v\in R$ such that $g(v)$ equals $g(c)=c$ for some $c\in \mathbb{C}^\times$.  Thus, the kernel of $g$ contains the nonzero element $v-c$.  Since the kernel of $g$ is a prime ideal of $R$, the kernel equals $\mathfrak{m}_0$ or $\mathfrak{m}_1$.  So the homomorphism $g$ factors through $f_0$ or $f_1$.  Thus the $\mathbb{C}$-morphism $$\text{Spec}(g):\text{Spec}\ A \to \text{Spec}\ R,$$ is constant with image $\mathfrak{m}_0$ or $\mathfrak{m}_1$.
More generally, every finite type $\mathbb{C}$-algebra $A$ for which $\text{Spec}\ A$ is connected is a finite union of irreducible components.  For every $\mathbb{C}$-morphism, $$h:\text{Spec}\ A \to \text{Spec}\ R,$$ the restriction of $h$ to every irreducible component is constant with image $\mathfrak{m}_0$ or $\mathfrak{m}_1$.  Thus, the $h$-inverse images of the two open subsets, $$U_\lambda=\text{Spec}\ R \setminus\{ \mathfrak{m}_\lambda \}, \ \ \lambda=0,1,$$ give an open covering of $\text{Spec}\ A$.  Since $\text{Spec}\ A$ is connected, one of these two open sets equals all of $\text{Spec}\ A$, i.e., also $h$ is constant. 
