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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.
In a sense the task is large, but like a huge jigsaw puzzle. Once one gets a piece in place, it's there. Remembering is easier after one has forgotten it.
My confusion with the earlier link was that it talked about Z[sqrt(-3)], but that is an example of a ring that is not a Dedekind domain. It shows that they exist. One needs to start with Q, the rationals. Then note that Z is the ring of integers, OQ, in Q. An algebraic integer in an algebraic number field is the root of a monic polynomial with integer coefficients. In the case of Q, the monic polynomial is x - n = 0 where n is an integer. We get the expected ring of integers, Z.
What is true for 3 and -3 is true for all 4n+3 type primes, correct? That would make things a bit easier to organize.
The -3 is a 4n+1 type. If you add 4 to -3 you get 1. The quadratic algebraic integers, those that are roots of monic polynomials of degree 2, separate into two groups depending on mod 4 (except for 2 which is handled separately).
Two things come to mind:
(1) How does one tell which are the integers in a quadratic number field, like Q(sqrt(-3)) or Q(sqrt(-5))?
(2) Is the ring of integers, such as OQ(sqrt(-3)), a unique factorization domain?
The first question is resolved by checking if the number is congruent to 1 or 3 mod 4. The second is resolved by using ideals, so unique factorization is possible in the Dedekind domains with ideals. I think the number of UFDs for algebraic numbers that are square roots of negative integers are finite, if one does not use ideals. I will have to check that.
Yes, only 3 is a 4n+3 type.
I am confused by this notation: OQ(sqrt(-3)).
Can you say it in English, please?
* * * * *
What I meant to say is that what is true of one 4n+3 number is true of another. Rather, I meant to ask if that is strictly true.
The OQ(sqrt(-3)) is the notation for the ring of integers from the field of algebraic numbers Q(sqrt(-3)). I think the O notation comes from Dedekind. Sometimes that ring of integers can be represented by something like Z[sqrt(-3)], which is a ring of a + b*sqrt(-3) where a and b are in Z or where a and be are what we normally think of as integers, or "rational integers". Sometimes it is not, as in this particular case, which is why I think Z[sqrt(-3)] was used as an example. Because -3 is congruent to 1 mod 4, we also have integers that look like this (a + b*sqrt(-3))/2. So the notation using O gives the ring of integers from a field which could be different from just extending Z by the same algebraic number. In the case of Q, the rationals, OQ is just Z. So for the field Q we don't need new notation because there is no difference from what we would expect the ring of integers to be.
One other notation that can be confusing: If one is extending a field the notion of "vector space" is used which is a special kind of "module" and one uses parentheses, such as, Q(sqrt(-3)). What makes it special is Q has all of its inverses. If one is extending a general ring (not necessarily a field) one uses the more general concept of a "module" and the notation changes to brackets such as Z[sqrt(-3)]. Here's some discussion of that difference: https://www.quora.com/What-is-the-di...le-over-a-ring
Mod 3 there is no object which when squared equals -1 (mod 3), for -1 is equivalent to 2 (mod 3).
Mod 5 the case is different. We do not even have to adjoin √-1 to our ring, because -1 is already there, since (√-1)2 is equal to 4.
This simple truth hung me up for quite a while.
When something snags me I find it almost impossible to move foreward until I have resolved the problem. Finally I have resolved this one, partly at least. I think you are well beyond me now, I hope not out of yelling distance. A new plateau of understanding may now pour over me quickly.
The article said its quotient field contained all numbers within itself. There is nothing I can square (mod 5) to get 3, however, so it still appears to matter what one is trying to adjoin.
I think it is only mod 4 that is of interest here, not mod 3 or mod 5.
For quadratic algebraic number fields, that is fields where the numbers are roots of quadratic equations like ax2+bx+c=0, the way to tell what algebraic numbers are integers is given by the rule that if we extend Q by the square root of a negative integer, then if that negative integer, -t, is congruent to 3 mod 4 then the algebraic integers are what we would expect them to be, numbers like a+b*sqrt(-t). If -t is congruent to 1 mod 4 then we have to also include a/2+b/2*sqrt(-t) as integers. This comes from using the quadratic formula to find x = (-b+-sqrt(b2-4ac)/2a. That 2 in the denominator does not cancel out in this case. The a = 1 because that is required for an algebraic number to be an algebraic integer. It has to be monic.
Don't worry about being out of reach. If I can't explain it, then I don't really understand it well enough and I don't understand this myself all that well. Also, I might have some of this wrong.
I have been traveling again, and I will have to do even more later this week or early next. While I am away I have no computer to do research on, so I concentrate on organizing as many details as I can remember on these math subjects, or what I can put down on paper.
Ideals and field extensions are parts of the classical theory of pure mathematics. I seriously doubt that ideals have found "many," applications outside of conducting more pure research. Also I would be only slightly more surprised if ideals had not already found "some," applications outside of pure mathematics.
Almost nothing is known about fields beyond quadratic fields. It was only for some quadratic fields that ideals recovered unique factorization. In the future I expect much more to be known about higher order fields. Perhaps the complexity will eventually be unraveled by quantum computers. At such a time more applications would come into being. Some of them might be more reliable and secure encryption techniques, as happened with the congruence theory of Gauss, a math language which did not find its big application outside of pure mathematics for a full 200 years.
Right now I feel like tackling more math. The harder it gets, the less I feel like diving in, so I had better take advantage of every time I feel like pushing my boundaries.
There is probably a lot that is not known also because the questions haven't been asked, but I don't know what the limits are. I ran into this article on "monogenic" fields which have an example of a cubic field that is not monogenic: https://en.wikipedia.org/wiki/Monogenic_field
So here is another technical term, "monogenic", and also, "power integral basis". These terms are more pieces in the jigsaw puzzle.
What I need are a few specific examples worked out with all the algebra. If I can see it just once I can figure out how the discriminant figures into this. The problem with any examples I have seen is that they suddenly introduce new variables to make the job harder. Often I cannot see which step to take. For instance, when they say it is obvious so and so is the minimum equation for so and so, I will not see why or how they got the answer.Quote:
Originally Posted by YesNo
A few examples geared just for me would make everything clear, I am convinced, but that luxury usually does not exist in math.
Of course I understand that if the discriminant is negative then we intorduce the complex numbers through a field extension. I cannot reproduce the algebra involved. I would have to see it once.
Here is a video describing the discriminant for a quadratic equation: http://www.virtualnerd.com/algebra-1...ant-definition You may already know this.
If the square root of the discriminant is not an integer, then the root (which could be either a real or a non-real complex number) is not a rational number. That is, it is not in the field of rational numbers, Q. We could extend this field by creating an object that has all of Q plus one of these non-rational roots. That would also be a field, and its structure would be a two-dimensional vector space. One of the dimensions would have 1 as the base and the other would have this non-rational root, r. The base would look like this: (1,r). The field extension of Q is written like this Q(r). If a and b are arbitrary rational numbers, that is elements of Q, then a number in this new field would look like this: a + br.
Now suppose we don't extend Q but instead extend Z, the ring of integers of Q. Z is not a field, but the extension would be similar and called a module. This is not always a Dedekind domain with ideals forming a unique factorization domain. However, we still could extend Z and see what we get. If we are extending a ring that is not a field the notation would be modified to Z[r], but the idea of the extension is the same. If a and b are integers, that is elements of Z, then a + br would be in Z[r].
Since the ring of integers OQ(r) may be larger than Z[r], we need some way to tell when they are not the same. The discriminant does this. If the discriminate is congruent to 1 mod 4, then we have to include the root that has a 2 in the denominator which is what the earlier link referenced when it discussed Z[sqrt(-3)].
Very good. It seems like I have most of the theory under control. Cannot look at the link yet, for I must travel again. Yes, I am no stranger to the quadratic formula. I just have to see exactly how it fits into all this business.
There is always some unpleasant algebra if one wants to view these things in detail. That will be my last step. We are almost done with ideals and the whole 19th century business of settling the theory of equations. Be back in four or five days this time.
You probably know what is in the link. It just talks about the discriminant for a quadratic polynomial. However, there are a lot of questions. It is worthwhile solving questions at the end of chapters in a text book. I have been reading Saul Stahl, "Introductory Modern Algebra: A Historical Approach". It is an undergraduate level survey of algebra with many questions. If we get a common text with problems that might be a way to go deeper into this subject. What books are you reading?
Here's a question I found interesting about quadratic equations associated with the idea of "algebraic expressions", that is, the ability to write the roots of an equation as an algebraic expression using the coefficients of the equation: Given two numbers, r and s, and the quadratic equation, x2-(r+s)x+rs = 0, show that r and s are the two roots of that equation.
We know we are on the lookout for UFD's (Unique Factorization Domains). It is nice to know then that all PID's (Principal Ideal Domains) are UFD's and all ideals in Z are Principal, for nice consolidation.
Irreducible ideals and prime ideals are not always the same in ideal language, but irreducible maximal ideals are always prime ideals.
Without becoming a grinding algorist, one can then follow the language of ideals as written in math the way a non-native pidgin speaker pieces together a newspaper bulletin in a foreign language.
The last paragraph of your second to the last post spills a lot of light on the significance of the discriminant in the theory: It is the discriminant itself whose value (mod 4) we are looking at to make our determination of what can be regarded as an algebraic integer.
Are some numbers with an irreducible 2 in the denominator then algebraic integers, or do they become mere algebraic numbers because of their denominator, causing a domain switch as a result?
That's how I follow it also, in bits and pieces. I get stuck, I stop and look for clues elsewhere.
I have heard that the reason to look at algebraic number fields and their rings of integers is to help us understand better the rational integers, Z. I wonder what insights have been gained? Perhaps insights into Fermat's Last Theorem? Just one of the many questions I have at the moment.
I think they would be algebraic integers, not just algebraic numbers. They don't look like integers because they can't be written as r + st where r and s are in Z, but they are integers because they are roots of a monic polynomial, that is, one where the coefficient of the x2 term is 1.Quote:
Originally Posted by desiresjab
But do we even need to look at the discriminat for that? Can't we just determine the value (mod 4) of the number originally adjoined to our field? Doesn't this accomplish the same thing while being much easier to perform?
What is different about the quotient fields (rings?) of √-5 and √-3 mentioned early in one of the articles we have been referencing back and forth?
I believe numbers of the form a/2 + br/2 can be algebraic integers though they do not appear to be. It may come down to what they can do. Can they help recover unique factorization? Do they qualify as an algebraic integer, or does their denominator represent a weakening which demotes these numbers to mere algebraic numbers but which are still useful for the purposes of ideals, i.e., which still qualify to help?
I woke up this morning thinking about why integers are defined as roots of monic polynomials. I probably don't have the full picture.
I can see how the integers Z can be defined using first degree monic polynomials. (A monic polynomial is a polynomial over the rationals Q, all coefficients are rational numbers, that when written as a polynomial over Z by getting rid of any denominators the highest nonzero coefficient is 1.) A first degree polynomial looks like this x+3/2 = 0. That has 1 as the coefficient of the highest term, x, but it is not monic because 3/2 is not an integer. It could be written as 2x + 3 = 0 which shows the coefficient of the highest term, 2, is not 1. So the root of that polynomial, 3/2, is not an integer, which is what we expect. If we look at 4x + 12 = 0 and divide by 4, we get x + 3 = 0. That is a monic polynomial since all coefficients are integers and the highest term has 1 as a coefficient. The root is 3 which we know is an integer.
If we try to do the same thing for second degree polynomials we find that the roots of some of these monic polynomials look like a/2 + br/2. That is because the quadratic formula has a 2 in the denominator. Sometimes this 2 cancels out. Whether it does or not depends on the discriminant. If the discriminant is -3 it doesn't cancel out so if we consider the ring of integers in Q(√-3) some of them will look like a/2 + b√-3/2 where a and b are integers. So the ring of integers of Q(√-3), written OQ(√-3), has more elements than the ring formed by extending Z with √-3, that is, Z[√-3].
What are the differences? It turns out the ring of integers of Q(√-3) is a unique factorization domain, but Z[√-3] isn't even a Dedekind domain.
If we look at Q(√-5), the discriminant, -5, is congruent to 3 mod 4 and so there are no integers in the ring of integers of Q(√-5) that look like a/2 + b√-5/2. The 2 in the denominator of the quadratic formula for roots of monic polynomials with -5 as the discriminant cancel out. They all look like a + b√-5. So the ring of integers of Q(√-5) is the same as Z[√-5]. The problem is this ring of integers does not have unique factorization. We will need to use ideals (special subsets of elements) to recover unique factorization rather than individual elements from that ring of integers.
For the first question, looking at the number originally adjoined to the field Q is almost the same thing as looking at the discriminant. For example when we adjoin √-5 to Q to get the algebraic number field Q(√-5), -5 is the discriminant. It just has a radical sign over it. The discriminant is the part under the radical sign in the quadratic formula.
The discriminant does not tell us if the ring of integers will be a unique factorization domain or not. It only tells us if there are integers of the form a/2 + br/2 in the ring of integers of Q extended by the square root of the discriminant. I don't know how they determined which of these algebraic number fields have unique factorization. That would be another piece of the puzzle for me to find out.
i like
I believe the root of 4x+12 would be -3 not 3.
That polynomials must be reduced to a minimum polynomial which has to be monic to be of use to us in recovering unique factorization, is understood. However, it seems to me that finding the minimum polynomial for even simple equations is not particularly simple and can involve a lot of algebraic labor.
I still do not see enough to relate this subject to quadratic reciprocity, which I find disturbing. I keep asking myself if the extra factor of 2 in 4n numbers which we observed in the Eisenstein diagram has anything to do with the different behavior of 4n+1 and 4n+3 numbers we observe here. Where does the expression of that extra 2 take place in this arena, if there is any? The only place I can see so far where there is an extra 2 sticking out like a sore thumb is in the denomoinator of those numbers of the form a/2 + br/2, but I do not see clearly if and how they relate.
And I still cannot decide if OQ(√-3) represents a quotient ring, i.e., the kind often represented by R/I. Those rings usually involve division by an ideal which contains a complex number, from what I have seen.
For instance, the possible remainders when dividing by 3i (which involves the Ideal of (3)), are 0, i, 1, 1+i, 1+2i, 2, 2i, 2+i, 2+2i, which is an example given in the chapter 11 link.
Having a way to state long chains of mathematical symbols in spoken language is important to me.
For something like R[√-3], I simply say R adjoined to the square root of -3.
I say it the same way for Q(√-3), and I am wondering if there is a more appropriate way to speak it once we reach the Rationals.
Yes, you are right. It is -3 and not 3. I got it wrong.
For first and second degree polynomials finding the roots are easy. There exist general algebraic formulas for third and fourth degree polynomials, so the general case is relatively easy there as well, but I think it stops with fifth degree polynomials. No general formula exists. So, I agree it gets harder to find roots of higher degree polynomials.
I don't see where it relates to quadratic reciprocity at the moment either. Given a monic polynomial of second degree, one has something like x2+bx+c where b and c are integers. Then use the quadratic formula to find the roots and see why knowing whether the discriminant is congruent to 1 or 3 mod 4 determines if we will be able to cancel out the 2 in the denominator of the quadratic formula. I think the details were given in one of the links for math stackexchange.
The ring of integers OQ(√-3) would not be quotient ring. It would be an infinite ring like the integers, Z. To get a quotient ring one would have to take an ideal and mod out by it: http://mathworld.wolfram.com/QuotientRing.html I suspect a quotient ring would be a finite ring (at least for Z). The ring of integers is an infinite ring containing all the integers in the algebraic number field Q(sqrt(-3)).
I use the following words which might not be correct terminology: R[√-3] is a module over the ring R extended with the square root of -3.
Q(√-3) would be a vector space (or module) over the field (ring) of rational numbers, Q, extended with the square root of -3. It is an algebraic number field. https://en.wikipedia.org/wiki/Algebraic_number_field
Here is the stackexchange article: http://math.stackexchange.com/questi...ed-in-that-way See Adam Hughes response to the question. This is where the mod 4 criteria is helpful. There is nothing about quadratic residues in the answer that I see.
However, reading this again I now picked up on the idea of "integral closure" which might be another key to understanding this better. Here is a definition of an "integral element": https://en.wikipedia.org/wiki/Integral_element As I see it at the moment, using this term, the reason why Z[√-3] is not the way to get to the ring of integers of the algebraic number field Q(√-3) is because Z[√-3] is not integrally closed in Q(√-3). It needs more integral elements from Q(√-3).
I don't understand this either, but that it seems to make sense makes it worth trying to understand better.
The article says that below is the minimum polynomial for a+b√D. I am not sure how they got that. I realize it is a simple concept and manipulation, but I am unable to make this small leap. It would help a lot if you performed the manipulations that got it to the form below. I am not sure how to get a involved.Quote:
Originally Posted by YesNo
pα(x)=x2−2ax+(a2−Db2)
I say the following as, "Z mod six Z," taking my example from the abstract algebra course I watched last year.
Z6/6Z
If a+b√D is a root, then so is a-b√D a root. Multiplying the two linear polynomials together we get (x - (a+b√D))(x - (a-b√D)) = x2 - (a+b√D)x - (a-b√D)x + a2-Db2 = x2−2ax+(a2−Db2)Quote:
Originally Posted by desiresjab
If one lets r and s be two roots of a quadratic equation, multiplying linear polynomials together leads to a general solution: (x - r)(x - s) = x2 - (r+s)x + rs The coefficient of the x term is negative the sum of the roots and the unit term is the product of the roots.
Alrightee, that is clear enough. I got it. I should have had it before, but sometimes merely hearing someone you trust say something clears it up better than texts. Of course I am supposed to know what you had to write out, having already seen it multiple times, but somehow it did not completely stick. As long as I finally comprehend things, I cannot concern myself with the time it takes, since that slows me down even more.Quote:
Originally Posted by YesNo
I seem to have carelessly lost a long post that was supposed to go here.Quote:
Originally Posted by YesNo
The minimum polynomial when extending a field by √5 is
x2=5, or x2-5=0. Applying the quadratic formula:
-0+-(√02-4·1·-5)/2= +-√4·5=2√5
For √-5, which is a 4n+3 number, the minimum polynomial is
x2=-5, or x2+5=0 Applying the quadratric formula:
-0+-√-20, which merely reduces to 2√-5
I do not see the real difference in terms of 5 being a 4n+1 number and -5 being a 4n+3 number. In both cases above, the 2 beneath the discriminants factors out. What simpleton mistake have I made this time? I have to be looking at something major wrong.
Your questions make me wonder what is meant by a minimal polynomial. I think it means that given a root, such as √5 or (1+√5)/2, the minimal polynomial of that root is a polynomial with rational coefficients with minimal degree. If we have square roots of non-square integers, these would have a second degree polynomial as their minimal polynomials. Rational numbers would have first degree polynomials as their minimal polynomial. One can always multiply that polynomial by some other linear polynomial, say x-7, to get a larger degree polynomial. The minimal polynomial is associated with specific algebraic numbers, the roots of that polynomial. (The term of the minimal polynomial with the largest degree should have 1 as its coefficient. This guarantees uniqueness of that polynomial. It can be found since the coefficients of the minimal polynomial are over a field such as the rationals, Q.)
What one has with Q(√5) are all of the algebraic numbers that can be written as a + b√5 with a and b being rational numbers. All the rational numbers are in this field because b could be 0.
What the discriminant being congruent to 1 or 3 mod 4 is supposed to tell us is whether there exist algebraic integers in Q(√5) that have a 2 in the denominator or not. The 2 won't cancel in all cases if 5 is congruent to 1 mod 4.
What does it mean to be an algebraic integer rather than just another algebraic number? The minimal polynomial has integer coefficients and the coefficient of the largest non-zero term is 1.
As an example, consider (1+√5)/2.
This is an algebraic number in Q(√5) because (1/2)+(1/2)√5 is of the form a + b√5 where a and b are rational numbers, in this case both rational numbers are 1/2.
To find its minimal polynomial, I used the idea that if (1+√5)/2 is a root then so is (1-√5)/2. (That might be worth trying to prove, but I can't think of the proof at the moment.) If r and s are roots of a quadratic polynomial, then (x - r)(x - s) = x2 - (r+s)x + rs. So, to get the middle term I add the two roots (1+√5)/2 and (1-√5)/2. I get 1 and then subtract it. To get the unit term I multiply those two roots to get -1. So the minimal polynomial is x2-x-1. Using the quadratic formula, I check that (1+√5)/2 is a root of that polynomial.
Is it an algebraic integer? Yes. The coefficients of its minimal polynomial are all integers and the highest term has coefficient of 1.
So Q(√5) has algebraic integers that have a 2 in the denominators as the determinant tells us to expect. That means the ring of integers of Q(√5) cannot be completely represented by Z[√5]. There are algebraic integers in Q(√5) that don't have this 2 in the denominator (such as all the rational integers and √5), but we are only interested in knowing if some of them need that 2 in the denominator.
Great post. Nice amount for me to think about.
Are you saying...well, exactly what are you calling an algebraic integer--x2-x-1, or one of its roots? Do those roots with a 2 in the denominator become classified as algebraic integers whenever they are the roots of monic polynomials?Quote:
Originally Posted by YesNo
Even when you do a great job I still have dumb questions, you see.
The algebraic integers are the roots of monic polynomials with integer coefficients, that is with coefficients in Z. Algebraic numbers in general are defined in a similar way. They are the roots of monic polynomials with rational coefficients, that is with coefficients in Q.
For quadratic number fields sometimes those roots have a 2 in the denominator such as (1-√5)/2 which comes from the quadratic formula with the 2a in the denominator.
Yes, very good. Your explanations are quite acceptable. I take it, then, that an algebraic number with a denominator of 2 does not become an algebraic integer just because it happens to be the root of a monic polynomial with integer coefficients. It is still only an algebraic number, but an important one now on equal footing with an algebraic integer that is a root because it is a root itself.
The denominator of these roots tells us whether the root is an algebraic integer or merely a "rootish," algebraic number.
Multiplying the roots together as you did one can see some action in the denominator. However, working with the quadratic formula in the usual high school fashion, nothing seems to reveal itself with regard to 4n. At that level I do not detect anything about that which we are speaking. I already know there will be an unreduced 2 in the denominator at the end, or there will not be, and I know why it will or will not be there.
I feel I am getting pretty close, but there are still small pieces here and there I do not have in place yet.
When you start out with a field extension you already can see what "type," the root will be, whether it will be a 4n+1 or a 4n+3 number, for instance. When you start out with the quadratic formula you are trying to determine what the roots are. Once you find a root you could always assume a field extension was made earlier. I do not see anything discouraging me from looking at it this way.
Unless I am making a field extension, I have to use the quadratic formula to find roots, and therefore will not know the relationship of my roots to 4n in advance, or not until the formula has worked its root-finding magic.
At least that is my current view of the whole situation. Some of it is bound to be deficient, I suppose, downright incorrect, or short-sighted.
And, by God, I believe I know (at least I hope I am right) that every algebraic integer is the root of some minimum polynomial with integer coefficients. But of course every root is not reciprocatively an algebraic integer (because we know of the existence of some monic polynomials with integer coefficients whose roots are yet of the form a/2 + br/2).
I am just musing out loud to see if I am correct or incorrect on a few ideas. Jigsaw puzzles are completed at exponential acceleration as time increases.