How many maximal ideals of $\mathbb{C}[x,y]$ contain $(f(x,y))$? Consider $\mathbb{C}[x,y]$, and let $f(x,y)$ be irreducible. My question is how many maximal ideals $(x-a,y-b)$ contain $(f(x,y))$? I am trying to visualize the curves in $\mathbb{A}_{\mathbb{C}}^2$ as (finite union of) the zariski closure of $(f(x,y))$. I want to see how many zero dimensional points lie on a one dimensional point.
 A: Note $(f(x,y)) \subset (x - a, y - b) \Leftrightarrow f(a,b) = 0$. Indeed, by changing coordinates, can assume $a,b = 0$, then $(f(x,y)) \subset (x, y) \Leftrightarrow f(x,y) = x \cdot p_1 + y \cdot p_2,$ where $p_1, p_2 \in \mathbb{C}[x,y] \Leftrightarrow f(x,y)$ has no constant term $\Leftrightarrow f(0,0) = 0$.
Now $f(x,y)$ has infinitely many solutions, since for every $x = a$, you get a polynomial in one variable over $\mathbb{C}$, which always has a solution (if there are no terms containing $x$ or $y$, you can easily modify this).
A: I think there are infite maximal ideals containig $f(x,y)$ and You can see it in a lot of ways, one of these could be:
Let $a$ be an arbitrary complex number and consider the the polynomial $p(y)=f(a,y)$.
If $y-b$ is a linear factor of $p$ (it exist because $\mathbb{C}$ is algebrically closed), the assert is equal to show that $m=(x-a, y-b)$ contain $f(x,y)$.
But this is easy to see by algebraic geometry: we have chose b such that $0=p(b)=f(a,b)$ and then $V(m) \subseteq V(f(x,y))$ but this implies $(f(x,y)) \subseteq m$ as required.
