Let k be the field and $k[y] \to k[x]$ maps $y \mapsto x^2$. Correspondingly, $f : \mathbb{A}^1_k \to \mathbb{A}^1_k$ morphism of affine varieties (schemes). Lets the $\alpha$ be a point in $\mathbb{A}^1_k$ (prime ideal $(x-\alpha)$). Consider image $f(\alpha) \in \mathbb{A}^1_k$. This is corresponds some prime ideal in $k[y]$. I guess: $(y-\alpha^2)$; but how to prove?
I need to compute preimage of the ideal $(x-\alpha)$. So, I need to know which polynomials $P(y)$ such that $P(x^2) \in (x-\alpha)$. Let $P(y) = \sum_i a_i y^i$, then $P(x^2) = \sum_i a_i x^{2i}$. Suppose $P(x^2) = (x-\alpha)(\sum_i b_i x^i)$.
Then I compute $P(y)$ is
$$P(y) = -\alpha^2 b_1 + (b_1-\alpha^2b_3)y + (b_3-\alpha^2 b_5)y^2 + \cdots$$
Hence $P(y) = (y-\alpha^2)(b_1+b_3 y + b_5 y^2 + \cdots)$. Hence, preimage of $(x-\alpha)$ equals $(y-\alpha^2)$. Which means, $f(\alpha) = \alpha^2$.
I think my solution correct.
But very ugly.
How to give better computation? I want to generalize to general map $k[y_1,\ldots,y_m] \to k[x_1,\ldots,x_n]$, some arbitrary map $y_i\mapsto P_i(x_1,\ldots,x_n)$.
Here is what might be a simpler (or clarified?) way to do this:
Note that $(x-a)$ is the kernel of the ring map $\phi: k[x]\to k, \ x\mapsto a$. This is because if a polynomial $p(x)$ is such that $p(a) = 0$, then $p(x) = (x-a)q(x)$ for some $q(x)$.
Now for the map $\psi: k[y]\to k[x], y\mapsto x^2$. The pullback $\psi^{-1}(x-a)$ is indeed just the kernel of the map $k[y]\to k[x]\to k$ given by $y\mapsto x^2\mapsto a^2$. By the same reasoning as above, the kernel is $(y-a^2)$.