Abstract Algebra 135 Prime and Maximal Ideals

Author:

Patrick Jones

Keywords:

abstract algebra,ideal,ring,factor ring,prime,maximal,integral domain,field

Subtitles:

let's look at a couple of different types of ideals so we're going to find a prime ideal so first of all it has to be a proper ideal it can't be all of our and the whole idea is that if we take two elements of R and when we multiply them together we get something in the ideal then one of those two elements have to be in the ideal let's look at an example let's say I've got my commutative ring our our standard example we're just going to use the integers and I'm gonna say that the ideal generated by 5 is a prime ideal why is that well remember that the ideal generated by 5 is going to end up being basically the set of multiples of 5 and the whole idea is if you multiply two integers and get a multiple of 5 one of those things had to be a multiple of 5 now of course that doesn't work for all integers if I took the ideal generated by 6 same kind of thing 6 adds such that and it's an element of Z that's not necessarily a prime ideal because 2 times 3 is certainly an element of that ideal but 2 & 3 neither one of them is in there okay another Rahil is a maximal ideal a maximal ideal again only working with commutative rings again it has to be a proper ideal can't be all of the ring and the whole idea though is that if you've got an ideal that contains that maximal ideal the only possibility is either the ideal it contains it is just the ideal itself or it's the whole ring there's no thing that is between the ideal in the ring for example let's look at all of the ideals of z24 we can create a sub ring lattice just like we did for subgroups so z24 it has ideals generated by 2 & 3 and again all these things are going to be T is the multiples of these things it's just we're doing it in Z 24 instead of all of Z it said in multiples of six that's a sub ring of both the multiples of 2 and the multiples of 3 said of multiples of 4 come off there but not off of the 6 it's down below these we've got the set of multiples of 12 over here we've got set of multiples of 8 and then off of both of those we have just the zero ideal now looking at this just going by this thing these two things the ideal generated by 2 the ideal generated by 3 are both maximal ideals here there's no ideal in between here and the entire ring both of these ideas the prime ideal in the maximal ideal can be tough to verify in general but there are some interesting properties involving them let's say we have a commutative ring with unity and we've got an ideal of the ring then the factor group are over a is an integral domain if and only if that ideal was a prime ideal okay it's an if and only if statement so there's two directions to prove so let's start and suppose R / a is an integral domain so we're trying to show that a is a prime ideal if we go back to our definition of prime ideal that means that whenever we have two elements of the ring that end up in that ideal when we multiply them together then one of the two things has to be in the ideal so let's go ahead and suppose a and B are in my ring and a B is an element of that ideal a okay well let's just actor ring suppose I have a t plus a times B plus a by definition that's a B plus a but wait a minute ATB is an element of a so that thing is a now a is the zero element of that factor so that that factor ring since that's the zero and these are elements of the factor ring we have two things in there that multiply together to give us the zero element since we're assuming this thing is an integral domain that must mean that we have no zero divisors so that must mean that either a plus a is a or B plus a is a since we have no zero divisors one of those things must be the zero element but again that means if a plus a is the is equal to a that must mean that a is an element of a or similarly over here B is an element of a so there we go if a B or an R and a V is an element of a when R of a was an interval domain it must mean that one of those two things was in the ideal which meant that a was a prime ideal okay so now what about the other direction so let's suppose that a is a prime ideal so we need to show that our /a a is an integral domain so we need to show that our slash a has no zero divisors so let's say I have two elements of my factor ring so I've got some kind of a plus a and some B plus a and I'm multiplying those together to get 0 plus a or just a again it's basically the same kind of argument a B plus a is equal to a so a B is an element of a but a was a prime ideal since a B is an element of a that must mean that a is an element of A or B is an element of a because again that's the definition of a prime ideal and so therefore either a plus a is the 0 or B plus a is the 0 once again so we have two things equaling zero meant that one of the two things had to be a zero so therefore our slash a is an integral domain we've got a similar thing if we have a maximal ideal so if we've got a commutative ring with unity and we've got an ideal then the factor ring is a field that is it has multiplicative inverses if and only if a is a maximal ideal I'm not going to go through the proof for this one but it's an important property to realize that you have