Minimal prime ideal in a ring of functions

Question

Suppose $X$ is an infinite set, $k$ a field and $R = \{f \mid f:X\to k\}$. Define $P_x :=\{f \mid f(x) = 0\}$. We know that $P_x$ is maximal and hence prime, but I am unsure how to show that $P_x$ is a minimal prime ideal.

So far I've got: Let $P$ be another prime ideal s.t $P \subset P_x$ and let $f \in P_x \setminus P$ such that $f$ differs minimally from all functions in $P$ ie. the set $\{a \ \mid f(a) \neq g(a), a \in X, g \in P\}$ for some function $g \in P$ is minimal.

Suppose $g \in P$ is a function in $P$ such that $f(a) \neq g(a)$ in the fewest possible terms.

If $g(a) \neq 0$ then say $f(a) =b$ and $g(a) = c$. As $R$ is all functions we may choose a function $h:X \to k$ with $h(a) = c^{-1}b$ and $h$ sends all other elements to $1$. Then $gh(a) = b = f(a)$ and otherwise $gh = g$. As $P$ is an ideal $gh \in P$ and, by minimality we have $f \in P$.

However, at the moment I cannot solve the case when $g(a) =0$. Any suggestions?

Answer 1

So far I've got: Let $P$ be another prime ideal s.t. $P\subsetneq P_x$ and let $f\in P_x\setminus P$

This is the right way to start, namely to assume that there is a prime $P\subset P_x$ and an element $f\in P_x\setminus P$ separating $P$ from $P_x$. Note that $f(x)=0$, since $f\in P_x$. Now define $g\in R$ so that $g(x) = 1$ and $g(y) = 0$ for $y\in X\setminus \{x\}$. We have $f, g\notin P$ and $fg=0 \in P$, showing that $P$ is not prime.

Answer 2

Clearly $R$ is a reduced ring, a ring which has no nonzero nilpotent element, and in a reduced ring a prime ideal $P$ is minimal if and only if for every $s\in P$ there exists $r\in R-P$ with $rs=0$. It is easily seen that $P_x$ has this property, and so is a minimal prime.