# Intersection condition for polynomial ring and maximal ideals

In ring theory, there is interest in a condition known as the intersection condition. There is a brief comment in McConnell-Robson along these lines: Consider the ring $R = k[x,y]$ where $k$ is a field. Let $I$ be the ideal $(x, y)$. For any prime ideal of $R$, say $P$, denote by $C(P)$ the set of elements of $R$ that are regular modulo $P$. Then for any maximal ideal $M \neq I$ of $R$, $I \cap C(M) \neq \emptyset$. However, $I \cap (\cap(C(M)) = \emptyset$, where the inner intersection is over all maximal ideals of $R$ not equal to $I$. This is provided as an example of a set of prime ideals that do not satisfy the intersection condition.

My proof of this fact is a bit cumbersome, and I wonder if there is a better proof. Here's mine:

1. Let $X$ be the set of maximal ideals of $R$ that are not $I$.
2. $\cap_X C(M) = k \setminus \lbrace{ 0 \rbrace}$.
3. $(x, y) \cap k = \lbrace{ 0 \rbrace}$.

So the step to prove is that $\cap_X C(M) = k \setminus \lbrace{ 0 \rbrace}$. Suppose that $f(x, y)$ is a polynomial in $R$ of degree at least 1 in $x$ or $y$. Then, by the principal ideal theorem, any minimal prime ideal containing $f(x, y)$ has height one. Each height one prime ideal, $P$, in $R$ is contained in infinitely many maximal ideals. Consequently, $f(x, y)$ is contained in a maximal ideal in $X$.

So, I have two questions:

First, I have to justify my claim that every height one prime ideal is contained in infinitely many maximal ideals: this seems unnecessarily heavy for the intended purpose. Is there a better approach?

Second, (and perhaps more interesting): Let $Y$ be a set of maximal ideals of $R$. When does $\cap_Y C(M) = k \setminus \lbrace{ 0 \rbrace}$?