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.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.
Last edited by YesNo; 03-31-2017 at 10:28 AM.
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
Last edited by YesNo; 04-02-2017 at 01:58 AM.
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.
pα(x)=x2−2ax+(a2−Db2)
Last edited by desiresjab; 04-03-2017 at 05:47 PM.
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)
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.
I seem to have carelessly lost a long post that was supposed to go here.
Last edited by desiresjab; 04-04-2017 at 08:31 PM.
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.
Last edited by YesNo; 04-08-2017 at 10:41 AM. Reason: made clearer (hopefully)
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