My problem asks me to show that if $f$ is non-constant, then $\mathbf{V}(f)$ is finite. Assume that $f \in \mathbb{C}[x]$.
If $f$ was an ideal, this would be straightforward; however, $f$ is merely a polynomial described above. I suppose by "non-constant" the prompt means "a polynomial with max degree greater than or equal to $1$."
I tried proving this through the contrapositive: if $\mathbf{V}(f)$ is infinite, then $f$ must be constant. $\mathbf{V}(f)$ is infinite implies that every point in the affine space must be a zero; if every point is a zero, then $f(x)=0$, which is a constant function.
Is this logic valid? Am I missing something fundamental?
The reason I am doubting myself is the following: if $f=c$ where $c ≠ 0$, wouldn't $\mathbf{V}(c)= \emptyset$? I'm not sure that $\emptyset$ and infinity is the same.
Thanks in advance for any useful clarification/hints.
I assume you mean $f \in \mathbb{C}[x]$, not $f \in \mathbb{C}[x, y]$, because otherwise $V(f)$ is a curve consisting of infinitely many points for any nonconstant polynomial $f$.
Now, if $f \in \mathbb{C}[x]$, you have the following:
The contrapositive of "$f$ non-constant implies $V(f)$ finite" is "$V(f)$ infinite implies $f$ is constant." In particular, it does not include the statement that $V(f)$ is infinite whenever $f$ is constant.
Now, as for your argument, it is true that $V(f)$ infinite implies that every point of $\mathbb{A}^1$ is a zero of $f$, but I'm not convinced from what you've written that you understand why. Why can't I have some polynomial with zeroes at 1, 2, 3, 4, 5, ... but nowhere else?