I'm trying to work out some problems from Ideals, Varieties, and Algorithms, and I've stumbled on one that I'm unsure of how to start:
Let $f,g \in \mathbb{C}[x,y]$ be nonzero. In this exercise, you will prove that:
$V(f,g)$ is infinite $\iff f,g$ have a nonconstant common factor in $\mathbb{C}[x,y]$.
- Prove that $V(f)$ is infinite when $f$ is nonconstant. Hint: suppose $f$ has positive degree in $x$. Then the leading coefficient of $x$ in $f$ can vanish for at most finitely many values of $y$. Now use the fact that $\mathbb{C}$ is algebraically closed.
- If $f,g$ have a nonconstant common factor $h \in \mathbb{C}[x,y]$, then use part (1) to show that $V(f,g)$ is infinite.
- If $f,g$ have no nonconstant common factor, show that $Res(f,g,x)$ and $Res(f,g,y)$ are nonzero, and conclude that $V(f,g)$ is finite.
I'm struggling with the hint in part (1); I don't know what to make of it. I think I can manage parts (2) and (3) on my own, but I'd appreciate any help with part (1). Here's my current attempt at it:
Suppose each term in $f$ has a term $x$ with degree $n≥1$: $$f=a_nx^{s_n}y^{t_n}+a_{n-1}x^{s_{n-1}}y^{t_{n-1}}+...+a_0x^{s_1}y^{t_1}$$ One can simply factor out an $x$ accordingly: $$f=x(a_nx^{s_n-1}y^{t_n}+a_{n-1}x^{s_{n-1}-1}y^{t_{n-1}}+...+a_0x^{s_1-1}y^{t_1})$$
Setting $x=0$ would imply that $y$ could be anything, and we have our result.
What if not every term had $x$ with degree $n≥1$? As long as $x,y$ appear once in $f$, we could express one in terms of the other, thus implying an infinite amount of solutions (for example, if $f=x+y,\quad f(x,y)=0 \implies x+y=0 \implies x=-y \implies V(f)$ is infinite).
I know I might be missing a step (or two, or three) or could be totally off. Any suggestions? Thanks in advance!
You may misunderstand something. Here is the answer. Rewrite $f$ as $$f(x,y)=g_n(y)x^n +g_{n-1}(y)x^{n-1}+\cdots+g_1(y)x+g_0(y),$$ where $g_0(y),g_1(y),\dots,g_n(y)$ are polynomials in $y$. Then you can choose infinitely many $y$ such that $g_n(y)\neq 0$. Because $\mathbb{C}$ is algebraically closed, for every such $y$ you have at least one $x$ such that $f(x,y)=0$, that's all.