My textbook states that an affine variety is
Definition. Let $k$ be a field and let $f_1,...,f_s$ be polynomials in $k[x_1,...,x_n]$. Then we set $$V(f_1,...,f_s)=\{(a_1,...,a_n) \in k^n : f_i(a_1,...,a_n)=0, \forall 1 \leq i \leq s\}$$
We call $V(f_1,...,f_s)$ the affine variety defined by $f_1,...,f_s$.
(Ideals, Varieties and Algorithms by Cox, Little and O'Shea)
My question is, simply, must an affine variety contain "ALL" the solutions to each $f_1,...,f_s$? So say, for simplicity I try to find a variety $V=V(f)$ for some $f$ then, can $V$ still be referred to as an "affine variety" while containing maybe only partial solutions?
I think the answer is no, and it MUST contain all solutions(otherwise, a particular variety will not be unique) but it's just that the definition above doesn't seem to (to me) make that very obvious. It seems like there's no problem letting it be the set with only "some" solutions too.
Can anyone point out which part of the definition makes this clear? Thank you in advance
Yes, $V(f_1,\dots,f_s)$ contains all the solutions. That is exactly what its definition means. When you define a set $V$ by writing $V=\{x\in X:P(x)\}$, this means that $V$ is the set of all elements $x\in X$ such that $P(x)$ is true. In this case, $X=k^n$, $x=(a_1,\dots,a_n)$ is written as an $n$-tuple, and $P(x)$ is the statement "$f_i(a_1,...,a_n)=0, \forall 1 \leq i \leq s$".