Let {$F_\alpha$}$_{\alpha \in I}$ - fields.
=> $\exists$ bijections between filters over $D$, where $D$ is filter and ideals of $\prod F_\alpha$.
In this case ultrafilter corresponds to max ideals.
I tried to use this question, but I didn't reach any useful conclusions.
Let $U$ be a filter over $I$. Consider the assignment $$U\mapsto \left\{a\in \prod_{i\in I}F_i\mid \left\{i\in I\mid a(i)=0\right\}\in U\right\}.$$
I claim this gives a bijection between filters over $I$ and ideals of $\prod_{i\in I}F_i$. First of all we claim that $\left\{a\in \prod_{i\in I}F_i\mid \left\{i\in I\mid a(i)=0\right\}\in U\right\}$ is an ideal of $\prod_{i\in I}F_i$. So let $b\in \prod_{i\in I}F_i$. Then $(ba)(i)=b(i)a(i)=0$. It follows that $\left\{i\in I\mid (ba)(i)=0\right\}\subset \left\{i\in I\mid a(i)=0\right\}$. By definition of a filter, we have that $\left\{i\in I\mid (ba)(i)=0\right\}\in U$. Hence $\left\{a\in \prod_{i\in I}F_i\mid \left\{i\in I\mid a(i)=0\right\}\in U\right\}$ really is an ideal.
It remains to show that the assignment is a bijection. (Try this, also, you can only filter something w.r.t. an ordered set, do you see which ordered set I'm using?)