Editing Zorn's lemma (section)

=== Every nontrivial ring with unity contains a maximal ideal ===

Zorn's lemma can be used to show that every nontrivial [[ring (mathematics)|ring]] ''R'' with [[Unital ring|unity]] contains a [[maximal ideal]].

Let ''P'' be the set consisting of all ''proper'' [[ideal (ring theory)|ideals]] in ''R'' (that is, all ideals in ''R'' except ''R'' itself). Since ''R'' is non-trivial, the set ''P'' contains the trivial ideal {0}. Furthermore, ''P'' is partially ordered by set inclusion. Finding a maximal ideal in ''R'' is the same as finding a maximal element in ''P''.

To apply Zorn's lemma, take a chain ''T'' in ''P''. If ''T'' is empty, then the trivial ideal {0} is an upper bound for ''T'' in ''P''. Assume then that ''T'' is non-empty. It is necessary to show that ''T'' has an upper bound, that is, there exists an ideal ''I'' ⊆ ''R'' containing all the members of ''T'' but still smaller than ''R'' (otherwise it would not be a proper ideal, so it is not in ''P'').

Take ''I'' to be the union of all the ideals in ''T''. We wish to show that ''I'' is an upper bound for ''T'' in ''P''. We will first show that ''I'' is an ideal of ''R''. For ''I'' to be an ideal, it must satisfy three conditions:

# ''I'' is a nonempty subset of ''R'',
# For every ''x'', ''y'' ∈ ''I'', the sum ''x'' + ''y'' is in ''I'',
# For every ''r'' ∈ ''R'' and every ''x'' ∈ ''I'', the product ''rx'' is in ''I''.

'''#1 - ''I'' is a nonempty subset of ''R''.'''

Because ''T'' contains at least one element, and that element contains at least 0, the union ''I'' contains at least 0 and is not empty. Every element of ''T'' is a subset of ''R'', so the union ''I'' only consists of elements in ''R''.

'''#2 - For every ''x'', ''y'' ∈ ''I'', the sum ''x'' + ''y'' is in ''I''.'''

Suppose ''x'' and ''y'' are elements of ''I''. Then there exist two ideals ''J'', ''K'' ∈ ''T'' such that ''x'' is an element of ''J'' and ''y'' is an element of ''K''. Since ''T'' is totally ordered, we know that ''J'' ⊆ ''K'' or ''K'' ⊆ ''J''. [[Without loss of generality]], assume the first case. Both ''x'' and ''y'' are members of the ideal ''K'', therefore their sum ''x'' + ''y'' is a member of ''K'', which shows that ''x'' + ''y'' is a member of ''I''.

'''#3 - For every ''r'' ∈ ''R'' and every ''x'' ∈ ''I'', the product ''rx'' is in ''I''.'''

Suppose ''x'' is an element of ''I''. Then there exists an ideal ''J'' ∈ ''T'' such that ''x'' is in ''J''. If ''r'' ∈ ''R'', then ''rx'' is an element of ''J'' and hence an element of ''I''. Thus, ''I'' is an ideal in ''R''.

Now, we show that ''I'' is a ''proper'' ideal. An ideal is equal to ''R'' [[if and only if]] it contains 1. (It is clear that if it is ''R'' then it contains 1; on the other hand, if it contains 1 and ''r'' is an arbitrary element of ''R'', then ''r''1 = ''r'' is an element of the ideal, and so the ideal is equal to ''R''.) So, if ''I'' were equal to ''R'', then it would contain 1, and that means one of the members of ''T'' would contain 1 and would thus be equal to ''R'' – but ''R'' is explicitly excluded from ''P''.

The hypothesis of Zorn's lemma has been checked, and thus there is a maximal element in ''P'', in other words a maximal ideal in ''R''.