I am not sure about these two definitions. For example, if we take the power set of A={1,2,3} with the partial order of inclusion. What are the maximal ideals and what are the maximal filters? For example, can a subset of P(A) be a maximal ideal without containing the empty set as an item?

closed as too localized by Robin Chapman, Pete L. Clark, Wadim Zudilin, Martin Brandenburg, Mariano SuárezAlvarez♦ Aug 2 '10 at 2:44
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To get the actual answer out of the way: the usual definition of ideal implies that any ideal contains the empty set  an ideal $I$ (on a set $X$ / the power set of $X$) is nonempty, closed under taking subsets and under taking finite unions (and of course $I\subseteq P(X)$). The first two should convince you that it contains the empty set. The notion of filter is dual  nonempty, closed under taking finite intersections and supersets. Ideals correspond to filters by mapping each element $A\in I$ to its complement $X\setminus A$. Maximal ideals correspond to maximal filters. Just in case this question was only due to a confusion of ideals and filters, let me add: A proper ideal by definition does not contain the 'full' set $X$ (e.g. in your example $X = \{ 1,2,3 \}$). Similarly, a proper filter does not contain the empty set by definition. The 'improper' cases of these definitions coincide  both the improper filter and the improper ideal are just the full power set (as is clear from being closed under subsets and supersets respectively). Usually, filter/ideal means proper filter/ideal, but for notational or technical convenience, it sometimes seems nice to allow the improper case  for example in the StoneCech compactification of the natural numbers, $\beta \mathbb{N}$, proper filters (on $\mathbb{N}$) correspond to closed nonempty subsets and the improper filter to the empty set. But I think the general preference (as Joel David Hamkins pointed out in the comments) is not to do this since no convenience outweighs the confusion caused by the improper case. In your example and (as mentioned by Robin Chapman) for any finite set $X$ , the maximal (proper) filters (or ultrafilters) are the principal filters, i.e. those of the form $\dot{x} = \{ A \subseteq X:\ x \in A \} $ for some $x\in X$. To see this just partition $X$ into singletons  a finite partition by assumption on $X$  every maximal filter contains exactly one part of the partition. The maximal ideals are again the dual. 

