How to turn elements of a ring $A$ into functions on $\text{Spec}A$?

Solution 1:

To first answer your questions in the order asked:

1. This is fine (and probably the most intuitive way) as long as you're fine not being too meticulous with the domain in which something is defined. That is to say, if you label the monomorphism $D \hookrightarrow K(P)$ as $\iota$, by $\overline{a} \in K(P)$ the author really replaces $\overline{a}$ with $\iota(\overline{a})$. The remainder of your definitions are correct.

2. As some in the comments have mentioned, the field of fractions is an example of localization (at a prime ideal $P$). To give a brief overview, if $P \subset A$ is any prime ideal, we may define the localisation at $P$ as $A_P := A \times (A - P) / \sim$ where $ (p, q) \sim (r, s)$ iff $\exists t \in A - P$ with $$t\cdot (ps - qr) = 0$$ (the whole reason for the $t$ is when we have non-zero divisors in $A - P$ to ensure the only place division by zero has significance is in trivial rings, most of the time you just want $t=1$ so that our equivalence relation is the same as it is for $\mathbb{Q}$). The field of fractions is when your prime ideal is $P = (0)$ — you should basically interpret this as "we consider $(0)$ to be the geometric origin (consequently, if there are zero-divisors, they will be in $P-A$ so you really do want $A$ to be a domain). Short story long, when $A$ is a domain you can do the same things with $\frac{f}{g} \in K(P)$ that you can in $\mathbb{Q}$ (and you really should think of this as a fraction in the full sense, as $\mathbb{Q}$ is realized as the fraction field of $\mathbb{Z}$).

3. As mentioned in the comments, you really have $K(P) \cong \mathbb{C}[x, y] / (x- a, y- b)$ for the same reason you say $\mathbb{C}^2$, which corresponds to $\mathbb{C}[x,y]$ (c.f. Hilbert's Nullstellensatz). To see where the map $\overline{a} \mapsto a(c,d)$ comes from, using your notation in part (1) you have $\overline{a} = a + P = a(x, y) + (x - c, y - d)$ so we may define a map $ \textrm{ev}_P : K(P) \to \mathbb{C}^2$ (called the evaluation map at $P$ or $(c,d)$) by $$a(x,y) + (x - c, y - d) \mapsto a(c, d) + (c- c, d- d) = a(c,d)$$