Ring and Filters I am working on the following problem:
Suppose that $R$ is the direct product of a family of fields indexed by a set $X$ (and hence is easily seen to be a ring).  Show that $I \mapsto F_I := \{U \subset X : \exists f \in I$ s.t.  $f(x)=0 \iff x \in U \}  $ defines a bijection between the set of proper ideals in R and the proper filters $F$ on $X$.  Also show that the following are equivalent:
(i) $R/I$ is an integral domain in which $0 \neq 1$
(ii) $R/I$ is a field
(iii) $F_I$ is a proper ultrafilter
All I have shown so far is that $F_I$ is indeed a filter.  However, I am not sure about the rest of this problem.  I would greatly appreciate help with this.  Thank you.
 A: $\def\Pot{\mathscr P}\def\ideal#1{\left\langle#1\right\rangle}$For a Filter $F \subseteq \Pot(X)$, define an ideal $I_F$ by 
$$
I_F := \ideal{\chi_U \bigm| U \in F} 
$$
where $\ideal{}$ denotes the generated ideal and $\chi_U\in \prod_x F_x$ (one minus) the characteristic function of $U$, that is $\chi_U(x) = 0_x \iff x \in U$ and $\chi_U(x) = 1_x\iff x \not\in U$. $I_F$ is an ideal by definition and it is not proper if $1 \in J_F$, that is there exist $f_i \in R$, $U_i \in F$ with $1 = \sum_{i=1}^n f_i \chi_{U_i}$, that is for every $x$
$$  1 = \sum_{i=1}^n f_i(x)\chi_{U_i}(x) $$
this gives $x \not\in \bigcap_{i=1}^n U_i$. As $x$ was arbitrary, $\emptyset = \bigcap_{i=1}^n U_i \in F$, so $F$ wasn't proper. 
Now let's show that the operations $I \mapsto F_I$ and $F \mapsto I_F$ are inverse to each other, we have:


*

*Let $U \in F_{I_F}$, the there is an $f \in I_F$ such that $x \in U$ iff $f(x) = 0$. As $f \in I_F$ we can write $f = \sum_i f_i \chi_{U_i}$ for finitely many $U_i \in F$. If $x \in \bigcap_i U_i$ we have $f(x) = 0$, so $\bigcap_i U_i \subseteq U$. As $F$ is a filter $U \in F$.

*Let $U\in F$, then $\chi_U \in I_F$ and hence $U \in F_{I_F}$ for we can take $\chi_U$ as $f$ in the definition of $F_{I_F}$.

*Let $f \in I_{F_I}$, then there are $f_i \in R$ and $U_i \in F_I$ with $f = \sum_i f_i U_i$. Now for each $U_i$, there is an $g_i \in I$ with $g_i(x) = 0$ iff $x \in U_i$. Define $h_i\in R$ by $$ h_i(x) =\begin{cases} g_i(x)^{-1} & g_i(x) \ne 0\\ 0 & g_i(x) = 0\end{cases} $$
then $\chi_{U_i} = h_i\cdot g_i \in I$ and hence $f = \sum_i f_i\chi_{U_i} \in I$.

*Let $f \in I$, let $U = \{x : f(x) = 0\}$, we have $\chi_U \in I_{F_I}$, but $f = f\chi_U \in I_{F_I}$.


So $I \mapsto F_I$ is bijective.
(i) $\to$ (ii): Let $f \in R$ with $f \not\in I$. Define $g \in R$ by $$ g(x) = \begin{cases} f(x)^{-1} & f(x) \ne 0\\ 0& f(x) = 0\end{cases}
$$
Then $f\cdot g \cdot f = 1 \cdot f$, and as $R/I$ is integral $g\cdot f + I =1 + I$, that is every $f + I\in R/I$, $f \not\in I$ is invertible and hence $R/I$ is a field.
(ii) $\leftrightarrow$ (iii): Note that $F \mapsto I_F$ is monotone, hence bigger filters give bigger ideal and $I\mapsto F_I$ is also. As $R/I$ is a field iff $I$ is maximal, this gives $R/I$ is a field iff $F_I$ is a maximal proper filter, that is an ultrafilter.
