Dtyjlui

I am a beginner in algebraic geometry and I have a doubt over the rings/fields involved in algebraic varieties.

Consider this definition:

An affine algebraic variety over the field $mathbb{F}^n$ is the zero set $V (p_1,...,p_k) ⊂ mathbb{F}^n$ of a finite number of polynomials $p_1, ..., p_k in mathbb{F}[x_1,...,x_n]$

I realise there are types of varieties other than affine ,but even looking ahead in Wikipedia here my question is this:

In algebraic geometry, are the polynomial variables (hence the zeros) always required to live in (a cartesian product of) the field of the coefficients ?

For example, do we also study cases where, say, the coefficient ring is $mathbb{Z_4}$ and the variables (hence the zeros) live in the field $mathbb{C}$ or vice-versa ?

The fields considered must be ''compatible'' in the sense of field extension (of course with same characteristic). For instance, the polynomial $x^2+1$ has the coefficients in the field of rational numbers, but the zeros lie in the field of complex numbers

The definition you've written in your post is an older, not-modern definition not especially suited to working over rings that aren't algebraically closed fields. Using this definition, the polynomials are in $Bbb F[x_1,cdots,x_n]$ and the $x_i$ represent the standard coordinates on $Bbb F^n$ and therefore take values in $Bbb F$.

A more modern concept, created especially to work well with rings which are not algebraically closed fields, is that of a scheme. Locally, schemes look like the spectrum of a ring - that is, the set of all prime ideals of some ring. If one considers $R[x]$, we recover the behavior of the earlier definition: if $R$ is an algebraically closed field, all the maximal ideals (corresponding to the typical points you might expect) are of the form $(x-x_0)$ for $x_0in Bbb F$, so we can say that all of these points, $x$ takes it's value in $Bbb F$.

If $R$ is a field which isn't algebraically closed, more interesting things can happen. Consider $R=Bbb R$ and the ring $Bbb R[x]$. Here, $(x^2+1)$ is a maximal ideal - the quotient $Bbb R[x]/(x^2+1)cong Bbb C$ is a field. But the evaluation map sends $x$ to either $i$ or $-i$, so $x$ takes on a value not in $Bbb R$.

Things can get even more interesting for other choices of a base ring. For example, if we take $R=Bbb Z$ and think about $Bbb Z[x]$, lots of interesting things can happen. For example, let's take the maximal ideals of the form $(p,x^2+1)$ for $p$ a prime number. Depending on what $p$ you pick, the value of $x$ may be considered to be in $Bbb F_p$ for $p=2$ or $pequiv 1 mod 4$ or $Bbb F_{p^2}$ for $pequiv 3mod 4$.

So the answer to your question is a qualified yes, and a push to look at schemes eventually to see what algebraic geometry looks like over more complicated base rings.