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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 $y_i\mapsto P_i(x_1,\ldots,x_n)$.

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    I do not understand how you conclude the equality of the two ideals. Anyway yes, this generalises and if you have understood this particular case, the general case is basically the same (this is why my first sentence above). There are two main steps: 1) if $p$ is a polynomial, then $p(x)-p(a)=(x-a)q(x)$ for some polynomial $q$; 2) you should get an inclusion $(y-p(a))\subset \varphi^{-1}(x-a)$ and hence equality by maximality (here $\varphi$ is the homomorphism $y\mapsto p(x)$). – 2017-02-09

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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 $(y-a^2)$.