Prove that the upper half plane of Cartesian plane is not an affine variety

Solution 1:

If you want to prove some set is not affine varieties, there would be two main steps: First, find out what would it be if it vanishes in the set: In this process, you will need to use the property that a nonzero polynomial can only have finitely many distinct roots, while there are infinitely many roots in the upper half plane, the only possibility is that it is the zero polynomial. Therefore, it would be the whole plane of real numbers. Second, find out the contradiction: there are so many counter examples in the other half plane, which should be in the plane but excluded. In short, we just say that R is not equal to R^2.

Solution 2:

As per Exercise 6b. of Chapter 1, Section 1 (page 5) of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms, the notation $\mathbb{Z}^n_{M+1}$ denotes the set of all points of $\mathbb{Z}^n$, all of whose coordinates lie between $1$ and $M+1$. So, for instance, $(1,1,2)$ lives in $\mathbb{Z}^3_{1+1}$, but $(0,1,2)$ does not (so, $M=1$ and $n=3$ in this example). Cox, Little, and O'Shea ask the reader to prove that if $f$ vanishes at all points of $\mathbb{Z}_{M+1}^n$, then $f$ is the zero polynomial. The exercise you would like to solve asks you to apply this result.

For each of the polynomials $f_i$ in the vanishing set $R=V(f_1,...,f_s)$, let $M_i$ denote the largest power of $x$ or $y$ appearing in $f_i$. Then, by the aforementioned exercise, since each $f_i$ vanishes on each point of the upper half plane, each $f_i$ vanishes at the integer lattice points of $\mathbb{Z}_{M_i+1}^2$, and so each $f_i$ is the zero polynomial. But this a contradiction, since we're assuming they only vanish on the upper half plane!