# Why an affine algebraic plane curve is the zero locus of exactly one polynomial

I just started to read Smith's Invitation to Algebraic Geometry. On one of the first pages it states:

An affine plane curve is the zero set of one complex polynomial in the complex plane.

By definition, an algebraic variety is the zero locus of a collection of polynomials.

For the purpose of this question let's assume the affine space is $\mathbb C^n$ or $\mathbb R^n$.

It is not clear to me why the dimension of the variety should equal the number of polynomials and whether or not the degrees matter.

(For example, the zero locus of two polynomials that produce two unit circles touching at the origin seems to me is a curve also. What am I missing?) (<-- this is a bad example that I came up with because I don't yet understand the subject well.)

Please could someone help me understand why an affine algebraic variety has dimension one if and only if it is the zero locus of exactly one polynomial of two variables (this is how I interpret what the book states)?

Is it impossible to start with an affine space, say $\mathbb R^3$ or $\mathbb R^4$ and then produce two polynomials so that their zero locus is a curve?

• Polynomial rings are UFDs. – Mohan Mar 3 '16 at 3:05
• In your example, I think you are saying the zero locus of $f$ union the zero locus of $g$ is the zero locus of $fg$. It's not clear to me what you are asking. There is the question of showing that every dimension 1 variety in the affine plane is cut out by exactly one question (instead of more), which uses that $k[x,y]$ is a UFD. It's no longer true for example, a curve in $\mathbb{P}^3$ is always cut out by exactly two equations (you might need more). However, I think you're asking something else. – DCT Mar 3 '16 at 3:21
• @Dtseng Thank you for your comment. I am trying to understand why an affine algebraic variety has dimension one if and only if it is the zero locus of one polynomial in two variables. This is how I interpret the sentence in the book. I know that polynomial rings are UFDs but I don't understand the implication. Please can you help me? – student Mar 4 '16 at 0:11
• @Dtseng No, I think you understand what I am asking correctly. (In your comment perhaps you meant "polynomial" rather than "question"?) – student Mar 4 '16 at 0:20
• Is your definition of a variety include being irreducible? In that case, the variety is the zero locus of $\mathfrak{p}\subset k[x,y]$ for some prime ideal $\mathfrak{p}$. Since $k[x,y]$ is a UFD, $\mathfrak{p}$ is principal, which is what you wanted. – DCT Mar 4 '16 at 5:31

The way to think about these things is, your initial surface is the whole space, say, a vector space of dimension $n$ (e.g. a plane of dimension 2). Then, cutting out each equation is sort of identical with reducing dimension of your surface by 1. So if your dimension was 2, introducing one equation will give you something of dimension 1 (a curve), and introducing two equations will give you dimension 0 (i.e. a set of points). In particular, the locus of two polynomials which produce circles touching at a point will be just that point that's common to both of them, i.e. something of dimension 0.
The reason it works like that is, because each equation can be considered as a function from your surface of dimension $n$ to $\Bbb C$ of dimension 1. The zero of that equation is the inverse image of $\{0\}$ under that function. So you are losing one dimension as you throw away all the points that map to a non-zero value under that equation.