Prove every ring contains a maximal ideal. ...?

Question

Any hints on this one?

I'm starting in a way that seems awkward and roundabout. Let R be a ring. If R is a field, then <0> is a maximal ideal of R; otherwise R contains at least one nonzero element that is not a unit, say x.

Then if R has a unity element, is a proper ideal of R.

Hints on where to go next or how to start over? I know there are lots of theorems on maximal ideals, but I'm (for now) trying to do this by brute force.

Doug: your answer is wrong by definition of maximal ideal: a maximal ideal must be a *proper* subset of a ring. Otherwise the term would be useless, as the maximal ideal of any ring would be the ring itself.

Answer 1

Order the ideals of the ring R by setting I<=J if I is a subset of J. This makes the ideals of R into a partially ordered set. If C is a chain in this set, i.e. a collection of ideals that is totally ordered, then the union of C is an ideal, so is an upper bound for C. I use the assumption that a ring has an identity here, since 1 is not in any element of C, so it is not in the union of C, so the union of C is proper. Now, Zorn's lemma gives a maximal ideal.

The result need not be true for rings without identity (although you can define 'modular' ideals and get maximal modular ideals). Also, for general rings, some version of the axiom of choice is needed.

Answer 2

Do it from the definitions. If (R,+,*) is a ring, then an ideal is a subring (I,+,*) such that I is a subset of R and for all r ε R and i ε I, ri ε I and ir ε I. (The last is required since, from the definition of R, there is no requirement that * be commutative. Only transitive and distributive)

Now, since (R,+) is, by definition, an Abelian group, so is (I,+) which is a sub-group of (R,+) and, by well-ordering, there must be a maximum such I.

If you have a ring with identity it's real simple since, by definition, if I is contained in R and 1 ε I then for all r ε R
r1 = 1r = r ε R so R is contained in I => R = I which is clearly the largest possible subgroup of R

As an aside (because it sometimes doesn't get said) just because T is a subring of J and J a subring of R inplies T is also a subring or R, it does *not* follow that if I is an ideal of S and S is an ideal of R then I is an ideal of R. This 'transitivity' fails in the same way that it fails for normal subgroups.


Doug

Answer 3

It follows from Zorn's Lemma, which is equivalent to Axiom of Choice. I don't even know if you can prove it any other way.

Answer 4

an outline of the proof is given at the following link(middle of the page)

http://en.wikipedia.org/wiki/Zorn's_lemma