Understanding the morphism of schemes involving nilpotents induced by: $\mathbb C[x] \to \mathbb C[x]/(x^2)$ where $x \mapsto ax \pmod {x^2}$ From Vakil's FOAG:

The following imprecise exercise will give you some sense of how to visualize maps of schemes when nilpotents are involved. Suppose $a \in \mathbb C$. Consider the map of rings $\mathbb C[x] \to \mathbb C[x]/(x^2)$ given by $x \mapsto ax$. Recall that $\operatorname{Spec} \mathbb C[x]/(x^2)$ may be pictured as a point with a tangent vector (§4.2). How would you picture this map if $a \ne 0$? How does your picture change if $a=0$?


When $a \ne 0$, I believe the ring map $\mathbb C[x] \to \mathbb C[x]/(x^2)$ given by $x \to ax$ induces the morphism of schemes $\operatorname{Spec} \mathbb C[x]/(x^2) \to \operatorname{Spec} \mathbb C[x]$ given by $(x)/(x^2) \mapsto (x)$. I think this is true because if $a \ne 0$, then $(x)/(x^2)=(ax)/(x^2)$ and the preimage of $(ax)/(x^2)$ is $(x)$.
When $a=0$, then we also have morphism of schemes $\operatorname{Spec} \mathbb C[x]/(x^2) \to \operatorname{Spec} \mathbb C[x]$ given by $(x)/(x^2) \mapsto (x)$ because the preimage of $(x)/(x^2)$ is the set of all polynomials with zero constant term, i.e., the prime ideal $(x)$.

Is the above correct? I am still not sure how to visualize this morphism.
What is this exercise trying to get across?

Pointwise, you are correct.
This is my rookie view of why he says that the tangent vector should be “crushed”. Suppose you have a function $f(x)$ on $\operatorname{Spec} \mathbb C[x].$ Then under the map $\operatorname{Spec} \mathbb C[\epsilon]/(\epsilon^2) \to \operatorname{Spec} \mathbb C[x]$ it pulls back to $f(a\epsilon).$ Now, for $a \not =0$ having the function $\epsilon \mapsto f(a\epsilon)$ in $\operatorname{Spec} \mathbb C[\epsilon]/(\epsilon^2)$ means knowing the value of $f$ plus its derivative. Thus, the tangent vector accounts for the fact that you can take the derivative along it. However, when $a=0$, $f(x)$ pulls back to $f(0)$ and your information about the derivative is lost.
