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The only number in the whole field which is not rational is the field extension itself. But I must confess I am at a total loss understanding the difference between extending to sqrt (-3) and extending by √-3i or something like that. I truly do not understand this difference or whatever advantage might accrue to it. This issue is one of the three or four items I still must bring under control. You addressed it once but I still could not understand.
Another thing which still throws me is--it is just like me these days to forget what it was before I have the sentence finished, dammit!
Oh, the other thing which confuses me is how 1+√-3 et al got into the picture at all. We were considering √-3, for instance, and suddenly here we are with 1+√-3. What justifies that 1 out in front? I understand it helps because it has a conjugate and I am not certain if mere √-3 does. Still, I have a hard time justifying its presence. Do you understand what the justification is?
Suddenly, I think I see how a Quotient Ring works.
Take the polynomial x2+3=0
If we subtract x2+3 from any other polynomial, the difference either is or is not a multiple of x2+3. If it is a multiple, we call that result equivalent to x2+3 itself. We may even call it equivalent to the second polynomial. I am not sure about that point. Probably not, actually. Well, hmmm..., I don't know. More thought.
This is really only modular arithmetic using x2+3 as the modulus.
If I take (x2+4)-(x2+3)=1,
that seems to imply that to me that 1 and x2+3 are equivalent, since it does seem that everything (and this includes x2+3) is a multiple of 1.
Yet it definitely puts x2+4 in the residue class (equivalence class?) of 1, where x2+3 is in the 0 class mod itself, so it is hard for me to see them as equivalent.
The above indicates to me that 1 and x2+4 are in the same residue class (equivalence class?), not that 1 and x2+3 are, which is impossible when x2+3 is the modulus.
It would be possible for polynomials A and B to be in the same residue class. In this case their difference would be, too. A, B and B-A would all have the same residue class.
Each residue class (equivalence class?) must form an ideal. All polynomials leaving a residue of 1, for instance, would form an ideal. All polynomials leaving a residue of 2 would form another ideal, etc. All polynomials leaving a residue of technically 0 would be in the same class as x2+3.
I think sqrt(-3) = sqrt(-1)*sqrt(3) = i*sqrt(3).Quote:
Originally Posted by desiresjab
The a + b*sqrt(-3) has two terms because it is generated by 1 and sqrt(-3) over the rationals. So one has a linear combination of both generators. If we let a and b be in Q, then a*1 + b*sqrt(-3) can take any value in the field extension Q(sqrt(-3)).
The equation at the top is an identity, so it should work for any x. However a polynomial is 0 for only a few roots.Quote:
Originally Posted by desiresjab
If x2+3 is the modulus then any polynomial times that one have those two roots in common. It is like a prime, say 3. Any integer times 3 would be in the ideal generated by 3.
Okay. I am back. I think I see everything so far except the fundamental difference between √-5 and √-3 and why a different treatment is necessary, since we were missing the treatment of √-3 in the article we have linked to several times. That excerpt speaks of that treatment but never shows it, though it gives a hint for the enlightened.
As I see it now, I don't think the treatment from the perspective of ideals is different. In both cases, when we look at ideals the ideals factor into prime ideals (although those prime ideas are not principal ideals because they have more than one generator). What makes a prime is the property, going back to Euclid, that if a prime p (from whatever ring) divides a product, ab, then p must divide either a or p must divide b. The prime is not allowed to divide some of each as 6 dividing 2*3 does.
When we look at elements of the ring of integers they don't factor into primes. I still don't understand the fundamental theorem of ideal theory so I don't see why unique factorization has to work yet. The examples showing that it does work can be checked, but they are just examples.
There is that other difference in the original article we looked at. I think it referred to extending the Z[sqrt(-3)] ring, however, I don't see it at the moment.
On the difference between treatments of 3 and 5 in the second paragraph of the article, I notice that 5 can be factored in Gaussian integers without the use of a √ sign, and that 3 cannot be, and I wonder if this has any bearing on their primary difference, which I suppose must ultimately be related to the fact that 3 is a Gaussian prime and 5 is not.
We also know that Gaussian primes cannot be represented as the sum of two squares. That means 4n+3 type primes in the integers.
What I am still trying to figure out is what the article means when they say that the quotient field of Z[√5] already contains all the integers under it, implying that the quotient field of Z[√3] does not. And then the article says that is why they cannot simply add more numbers to the extension field of √5 they way they did for √3. Of course that mysterious work comes in a previous part of the book we did not get to see.
Could it be that the failure of 3 to factor nicely without stepping out of Gaussian integers the way 5 does into (2+i)(2-i) is involved?
I am sure there is a simple division process to demonstrate this. I am not sure how to do it. But I do have a crude idea.
If the √ of anything is part of the quotient field, I see how dividing by it could mess things up. Not too helpful, I know, but a small insight. You would not get nice integers anymore.
It hadn't ought to be that hard to construct quotient fields for 3 and 5, should it, and demonstrate how the latter already contains all the integers under it and the former does not?
I am having trouble understand that same distinction mentioned in the article. Dedekind introduced the language of ideals, but Kummer came before him and introduced ideal numbers. I think they are referring to Kummer's approach rather than Dedekind's, but I don't know yet. The sqrt(-5) is supposedly the critical one leading to the results in algebraic number theory, not sqrt(-3), but I don't see it yet.
Later on in the article, they say on page 4, ex. 11.5:
Z[√-3] is not a Dedekind domain, (which I believe Z[√-5] is). It is not a Dedekind domain because it does not contain every integer in its quotient field.
It appears from the article we may have to pass to the ring of ξ, or whatever that strange symbol is they keep using. This seems to be another level of abstraction to tackle. Happily, it may be the last great hurdle on this particular journey.
I see that part. I am puzzled by it. I would have thought that Z[√-3] would have all the integers from Q[√-3]. Apparently it is not "integrally closed". That means there must be "integers" in Q[√-3] that are not in Z[√-3]. It looks like one also has to include (1+√-3)/2 = (1/2) + (1/2)√-3. But 1/2 is not an integer in Z.
Whatever is in the quotient fileds of √-3 and √-5 only gets there after the proper division, right? Maybe there are some "objects," in the quotient field of Z[[-3], but what the article says is that Z[√-5] contains all the integers in its quotient field, suggesting that Z[√-3] does not. We have exactly the same problem at this point.
If x2+3 is our polynomial, it is also our quotient field modulus, isn't it, or do we simply use √-3?
This only goes to show how great a paucity there is of clear, specific examples in this literature. The phd candidates who write these posted papers would not dare insult those judging their dissertations by anything so lowly and nasty as specific numerical examples all worked out here and there.
Mathematicians write only for those who are 100% up on the language. Even professors do this teaching their courses. They expect everyone to already understand and be intimately familiar with whatever is being studied. Good examples are for neophytes.
I have aired this gripe before. Of course most math articles are not written with me in mind. Nonetheless, I think mathematicians are generally bad writers and bad teachers.
I agree it isn't clear. I did find a Wikipedia article that is making sense to me at the moment although it is not completely clear: https://en.wikipedia.org/wiki/Quadratic_integer
The quadratic integers are roots of monic quadratic equations. Looking at Q(sqrt(-3)) we can find the ring of integers OQ(sqrt(-3)) which are integrally closed because we take all quadratic integers from Q(sqrt(-3)), but this is larger than the ring Z[sqrt(-3)]. There we are starting with the ring Z not the field Q when we make the extension. I think that is where part of the confusion comes from. What we need to extend the rational integers Z by is not sqrt(-3) but (1+sqrt(-3))/2 to get all the roots of all monic equations where only the sqrt(-3) appears as an algebraic number. However if we are looking at a field like the rationals Q, which is the field of fractions of Z, we can simplify this to Q(sqrt(-3)) because 1/2 is in Q already. It is not in Z unless we put it in there some how.
The monic quadratic equations containing sqrt(-3) in the root somewhere would be x2 + 3 = 0 and x2 + x + 1 = 0. The second is a factor of x3 - 1 making that root one of the cubic roots of unity.
The challenge is to understand the concepts and then maybe write a clearer exposition.
Edit: Here's another link. I like the first answer and the way the question was formed: http://math.stackexchange.com/questi...ed-in-that-way
That article seems to have about everything in it that is needed for understanding the subject the way I would like to. It is a very tough piece. I have stayed away from it because it hurts my brain and I have not been feeling that well. But now I feel okay and I need to go after it. The fine points are somewhat explained. But of course there are always things you wish they had cleared up in any article on the subject.
In order to properly understand ideals all these related areas like field extensions and polynomial rings must be comprehended. So in the end it is a much bigger task to learn than quadratic reciprocity was.