I've learned that given any ring homomorphism $\varphi \colon A \rightarrow B$ we may induce a morphism of affine schemes as follows. The continuous map on topological spaces is
\begin{equation}
\text{Spec}_{\varphi} \colon \text{Spec }B \rightarrow \text{Spec }A, \mathfrak{q} \mapsto \text{Spec}_{\varphi}(\mathfrak{q}) = \varphi^{-1}(\mathfrak{q}).
\end{equation}
The morphism on the structure sheaves is
\begin{equation}
\varphi^{\sharp}_{D(f)} \colon \mathscr{O}_{\text{Spec}A}(D(f)) = A_{f} \rightarrow \mathscr{O}_{\text{Spec}B}(D(\varphi(f))) = B_{\varphi(f)}, \frac{a}{f^{n}} \mapsto \varphi^{\sharp}_{D(f)}(\frac{a}{f^{n}}) = \frac{\varphi(a)}{\varphi(f)^{n}},
\end{equation}
for any distinguished open set $D(f)$. My issue with this is that I have not seen an actual example of how this works in practice. For instance, Vakil gives the following example (without details):
Consider the ring homomorphism $\varphi \colon \mathbb{C}[y] \rightarrow \mathbb{C}[x], y \mapsto x^{2}$. I am trying to figure out what Spec$_{\varphi}$ and $\varphi^{\sharp}$ are.
$(i)$ I must be missing something trivial here but why are we specifying the ring map $\varphi$ only how it acts on $y$ and not on arbitrary polynomials $f(y)$? For instance, I would expect something like $\varphi(f(y)) = g(x)$ where $f(y) \in \mathbb{C}[y]$ and $g(x) \in \mathbb{C}[x]$.
$(ii)$ I know that Spec $\mathbb{C}[y] = \{ \langle 0 \rangle \} \cup \{ \langle y – a \rangle : a \in \mathbb{C} \}$ and similarly for Spec $\mathbb{C}[x]$. Thus we get
\begin{equation}
\langle x – a \rangle \mapsto \text{Spec}_{\varphi}(\langle x – a \rangle) = \varphi^{-1}(\langle x – a \rangle) = \{ f(y) \in \mathbb{C}[y] : \varphi(f(y)) \in \langle x – a \rangle \}.
\end{equation}
I expect this to be the prime ideal $\langle y – a^{2} \rangle$ but I don't see why/how? Also I assume that the generic point $\langle 0 \rangle$ gets mapped to $\langle 0 \rangle$.
$(iii)$ It would be beneficial if someone could provide me with some more examples/exercises on how to go from a ring homomorphism $\varphi$ to the morphism on their spectra. Something that I can compute in practice. Also perhaps some reverse computations, i. e. going from the morphisms on spectra to the ones on rings (I know that we have an equivalence of categories between affine schemes and commutative rings with unit).
Best Answer
Homomorphisms on polynomial rings are determined by their behavior on the constants and the variables - this is the universal property of polynomial rings. Universal properties are the first thing you should think about when working with an object (that has a universal property, of course), because it makes life so much easier. In algebraic geometry, the following are essential:
In particular, when we want to define a ring homomorphism on $\mathbb{C}[y]$, we only specify what is happening on $\mathbb{C}$ and with the variable $y$. Here, we are actually only working with $\mathbb{C}$-algebras and their homomorphisms, so it is clear what the homomorphisms are doing with $\mathbb{C}$.
So, $\varphi : \mathbb{C}[y] \to \mathbb{C}[x]$ is completely determined by $\varphi(y) := x^2$. In fact, for every polynomial $f \in \mathbb{C}[y]$ we will have $\varphi(f) = f(x^2)$. For example, $\varphi(y^3-y)=x^6-x^2$.
You easily see that $\varphi$ is injective. Hence, $\varphi^{-1}(\langle 0 \rangle)= \langle 0 \rangle$, which means that the morphism $\mathrm{Spec}(\varphi) : \mathrm{Spec}(\mathbb{C}[x]) \to \mathrm{Spec}(\mathbb{C}[y])$ indeed maps the generic point to the generic point.
All other points in the spectrum of $\mathbb{C}[x]$ have the form
$\langle x - a \rangle = \{g \in \mathbb{C}[x] : g(a) = 0\}$
for some $a \in \mathbb{C}$. Hence, the preimage in $\mathbb{C}[y]$ is
$\varphi^{-1}(\langle x - a \rangle) = \{f \in \mathbb{C}[y] : \varphi(f)(a) = 0\} = \{f \in \mathbb{C}[y] : f(a^2) = 0\} = \langle y - a^2 \rangle$
So indeed $\mathrm{Spec}(\varphi)$ maps $\langle x - a \rangle$ to $\langle y - a^2 \rangle$. In other words, when we identify both spectra with the set $\mathbb{C} \cup \{\eta\}$ (where $\eta$ is the generic point), then the map is just $a \mapsto a^2$ and $\eta \mapsto \eta$. (In classical algebraic geometry, you would actually just say that $\mathbb{C} \to \mathbb{C}$, $a \mapsto a^2$ is a morphism of algebraic varieties, and you can always switch between the classical and the modern language.)
The sheaf part of $\mathrm{Spec}(\varphi)$ just consists of localizations of $\varphi$. For example, take the element $f = y$. Then $\varphi(y) = x^2$, the preimage of $D(y)$ under $\mathrm{Spec}(\varphi)$ is $D(\varphi(f))=D(x^2)=D(x)$. As a set, $D(x)$ consists of all points except for $\langle x \rangle$. Similarly for $D(y)$. The homomorphism $\Gamma(D(y)) \to \Gamma(D(x))$ (where $\Gamma$ denotes the ring of sections) is given by
$\mathbb{C}[y]_{y} \to \mathbb{C}[x]_{x}, \quad y \mapsto x^2.$
Again, the universal properties of the objects involved allow us to describe the whole homomorphism just by specifying the image of $y$ (as long as it is invertible, of course). For example, $y - y^{-1}$ is sent to $x^2 - x^{-2}$.
Let's do another example: $f = y-1$. The preimage of $D(y-1)$ is $D(x^2-1) = D(x+1) \cap D(x-1)$. This is the complement of two points (the intersection of the parabola $y=x^2$ with the line $y=1$). But again, the sheaf part is just $\mathbb{C}[y]_{y-1} \to \mathbb{C}[x]_{x^2-1}$ mapping $y \mapsto x^2$. For example, it maps $\frac{1}{y-1} \mapsto \frac{1}{x^2-1}$.
The open set $\{\eta\} \subseteq \mathrm{Spec}(\mathbb{C}[x])$ is not basic-open, but we can still describe the sheaf part there: it is the homomorphism induced on the fields of fractions:
$$\mathbb{C}(y) \to \mathbb{C}(x),\quad y \mapsto x^2.$$
Since you have asked for exercises along these lines: Describe $\mathrm{Spec}(\varphi)$ for the homomorphisms of $\mathbb{C}$-algebras $\varphi$ below:
In the first four examples, since the prime ideals of $\mathbb{C}[x,y]$ are a bit complicated, restrict to maximal ideals if you like. These are not very interesting morphisms, but will help to understand what is happening.