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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

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    $\begingroup$ Your question is not clear. What do you mean by "all solutions to each $f_1,\dots,f_s$"? The variety $V$ is defined as the common zeros of all the $f_i$, i.e. solution to the system of simultaneous equations $f_1=f_2=\cdots=f_s=0$. $\endgroup$
    – Pedro
    Commented Dec 7, 2015 at 23:16
  • $\begingroup$ So, say $f=x^2-y$. I trust the (affine) variety $V(f)=V(x^2-y)$ would contain, ordered pairs $(x,y)$ such that $x^2-y=0$. By "ALL" solutions I meant the usual sense i.e. in this case, every ordered pair that allows $f(x,y)=$ say, in $\mathbb{R}$. For instance, I can have $(0,0)$ and $(1,1)$ as an example of a solution, but, my question was, can $V(f)=\{(0,0)\}$ or $V(f)=\{(1,1)\} \leftarrow$ (partial solution set) etc and not $\{(0,0),(1,1),(-1,1)\} \leftarrow$ (ALL solutions set)? I think "no" i.e. has to be the full solution set but I don't see that explicitly in the definition above. $\endgroup$
    – Kydo
    Commented Dec 7, 2015 at 23:20

2 Answers 2

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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 "$f_i(a_1,...,a_n)=0, \forall 1 \leq i \leq s$".

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  • $\begingroup$ Thank you very much for a straightforward and clear answer, that got me back on track. I now see where in the definition that "all" solutions come to play. I appreciate it $\endgroup$
    – Kydo
    Commented Dec 8, 2015 at 15:02
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Usually $k$ is assumed to be algebraically closed, so that there is a correspondence between varieties and systems of equations (or rather the radical ideal defined by the polynomials). For non-alg-closed fields or rings $K$ you talk about the $K$-points of the variety but the set of points is not considered to be (or to define) the variety.

For example, high degree polynomial equations $P(x,y)=0$ tend to have have an empty set of points over $K=\mathbb{Q}$, but they still can be thought of as curves and attributes of curves such as automorphisms, points at infinity, self-intersections, etc, can be useful in understanding the equation. One does not think of the variety corresponding to the equation as "the empty set", but a field- or ring-dependent set of points that happens to be empty in some special cases.

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    $\begingroup$ This is true, but isn't at all what the question is asking, and in the context of the question is potentially misleading. $\endgroup$ Commented Dec 8, 2015 at 0:06
  • $\begingroup$ There wasn't a well-defined question. I took part of the confusion to be whether (in more precise language) a variety is determined by a subset of its points. I can't imagine what you consider potentially misleading. $\endgroup$ Commented Dec 8, 2015 at 0:53

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