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?
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1$\begingroup$ Actually this is not the right site for your question (this is devoted to research questions), and in fact your question is going to be closed soon. Please try e.g. en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics where you will certainly be satisfied. $\endgroup$– Pietro MajerCommented Aug 1, 2010 at 12:19
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$\begingroup$ Of course, in the power set of a finite set all ultrafilters and maxmal ideals are principal. $\endgroup$– Robin ChapmanCommented Aug 1, 2010 at 12:21
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$\begingroup$ Robin, don't you mean "mximal" instead? ;-) $\endgroup$– Wadim ZudilinCommented Aug 1, 2010 at 13:08
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$\begingroup$ Thanks. I was just wondering if a maximal ideal must contain the empty set.. $\endgroup$– taliCommented Aug 1, 2010 at 13:21
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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 non-empty, 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 -- non-empty, 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 Stone-Cech compactification of the natural numbers, $\beta \mathbb{N}$, proper filters (on $\mathbb{N}$) correspond to closed non-empty 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.
answered Aug 1, 2010 at 14:29
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1$\begingroup$ In many contexts, however, including much of set theory, people use the words filter and ideal to mean what you call proper filter and proper ideal. $\endgroup$ Commented Aug 1, 2010 at 15:06
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$\begingroup$ Thanks. That is what I meant with 'usually the interest lies' :) But I will try to make this clearer. $\endgroup$ Commented Aug 1, 2010 at 15:36
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$\begingroup$ Thanks a lot Peter! The confusion indeed was caused by the omitting of the word proper in some places. Actually the interest in this terms aroused from trying to understsand the topology on the Stone-Cech compactification \beta(X) of a topological space X. For example, in your example of the space \beta(N), who would be the basic open sets for this topology? $\endgroup$– taliCommented Aug 1, 2010 at 16:17
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1$\begingroup$ Glad I could help. I remember that it confused me when I met filters/ideals for the first time -- to use 'improper' is really improper as Joel David Hamkins pointed out :) $\endgroup$ Commented Aug 1, 2010 at 16:17
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1$\begingroup$ Thinking of $\beta \mathbb{N}$ as the set of all maximal filters, i.e., ultrafilters, on $\mathbb{N}$, the so called Stone topology is generated by basic clopen set of the form $$\bar{A} = \{ F \in \beta \mathbb{N}: \ A \in F \}$$ for every $A\subseteq \mathbb{N}$. $\endgroup$ Commented Aug 1, 2010 at 16:21