Do algebraic curves exclude the whole space? Prove that the locus $y = \sin x$ in $\mathbb R^2$ doesn't lie on any algebraic curve in $\mathbb C^2$

Question

Artin Algebra Chapter 11

Here is the solution of Brian Bi:

Here is the definition of algebraic curve:

Why can't we have the zero polynomial?

Answer 1

You are right - the curve determined by $y=\sin x$ does lie on the curve determined by the zero polynomial. However, this is a mistake on the book author's part - in the definition of an algebraic curve, the zero polynomial should be explicitly excluded.

There are a few ways to justify this, but the simplest one relates to what is said in the paragraph you have copied: in a "good" algebraic curve $X$ determined by a nonzero polynomial $f(x,y)$, we can locally represent the points on $X$ using just a single complex parameter, because the constraint $f(x,y)=0$ shows that $x,y$ are not independent, and in principle one can be computed from the other. But what I've said is not the case when we consider the zero polynomial - then $f(x,y)=0$ puts no condition on $x,y$ whatsoever, and you still need both of those numbers to describe a point (hence it wouldn't be a curve, it would be a two-dimensional surface).