On geometric interpretation of commutative rings

Question

Recently I am reading *Basic Notions of Algebra* by Shafarevich, and have some questions about it. My question is about the interpretation of commutative ring as a geometric object.
More specifically, book claims,

1. For any point $x_0$ on set $X$, one can interpret it as ring homomorphism such as, $x_0: F(X) \rightarrow K$, where $F(X)$ is the functions ring on $X$ and $K$ is a field that corresponds to range of $F(X)$. More precisely, constructed as follow, $$ \begin{array}{ccc} x_0: F(X) & {\longrightarrow} & K \\ {\in} & & \in \\ f & \longmapsto & f(x_0) \end{array} $$

2. Because any point on $X$ has the corresponding homomorphism on field, it can be related to maximal ideal which is isomorphic to kernel of homomorphism. If unfamiliar with this check Wikipedia:Ring homomorphism.

3. Alternatively, one can take any ring and interpret it as geometric object by relating maximal ideals with points on geometric object.

4. For example of this idea, book took $\mathbb{Z}$ and claimed that it should be interpreted as function ring on set of prime numbers. Because maximal ideals of $\mathbb{Z}$ can be written as (p) and this will corresponds to points on base space of function ring $\mathbb{Z}$.

I have 2 questions about it.

1. I do agree that one can relate points on $X$ to ring homomorphism, but relating it to maximal ideal is where I am confused. is it obvious that mapping between each point on X to the maximal ideal is injective?
2. Interpreting $\mathbb{Z}$ as function ring on Set of prime number (I'll denote it as $P$) is also confusing one.

• If corresponding maximal ideals are $(p)$, doesn't that implies range field of $\mathbb{Z}$ is $\mathbb{F}_p$ and so , it is different on each point in P? if so, can this be called "function" on P?
• Also, assuming one can call it as function, function ring made by these kinds of functions P $\rightarrow \bigoplus_p\mathbb{F}_p$ will not be a domain ring. but on other hand, $\mathbb{Z}$ is domain. So, these two can't be isomorphic to each other. Is $\mathbb{Z}$ actually able to interpreted as function ring on $P$?

I hope my question makes sense. thanks.

Answer 1

This is adapted directly from Eisenbud/Harris: Take any prime $p \in \mathbf{Z}$. For any other prime $q$, which is naturally a point* $x \in \operatorname{spec}(\mathbf{Z})$, $p$ defines a function, call it also $p$, by sending it to its residue value in $\mathbf{Z}/q$. So the map here is the natural quotient map $\mathbf{Z} \to \mathbf{Z}/q$, and the value of $p$ on $x$ is the class of $p$ in $\mathbf{Z}/q$, suggestively denoted $p(x)$.

They then ask you to compute the value of $15$ (as a function) on at the points $(7)$ and $(5)$ of $\operatorname{spec}(\mathbf{Z})$, which could be a good exercise if you are completely new to the subject. For more details I urge you to read the text. It is very forgiving for beginners.

*Or rather the ideal it generates, but let us not be precious.