There had ought to be 3/2 as many primes in the Gaussians as there are in the integers, since every 4n+1 prime in the integers has two prime factors in the Gaussians, is what I meant by density.
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There had ought to be 3/2 as many primes in the Gaussians as there are in the integers, since every 4n+1 prime in the integers has two prime factors in the Gaussians, is what I meant by density.
I think one could define something like that density if one asked how many primes are less than the norm (or absolute value in integers). There might be more than 3/2 primes in the Gaussian integers less than a certain norm n since they are contained in a plane rather than just a line, but maybe not.
I just meant to give a rough idea that there are more, or at least seem to be.
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Well, it is very interesting that ideals somehow manage to recover unique factorization (though only in some limited cases, according to Wildberger, he called them quadratic instances, or something like that), so now the aim has to be to produce a specific example of how ideals managed to recover uinique factorization. After that, it is on to Carmichael numbers and the understanding of how they have anything to do with ideals. That is the plan anyway and as far as I can see right now. Something always gets in the way of direct progress.
I don't know how ideals do that either. That does seem like a major justification for considering them.
Ideals seem to eliminate the difference between 4n+1 primes and 4n+3 primes. That is, every ideal generated by a prime is a prime ideal regardless of its type to begin with. This is really cool, even if it only provides some unique factorization domains. The fact that any part of unique factorization has been recovered has got to be one of the greatest acheivements of mathematics.Quote:
Originally Posted by YesNo
The Gaussian integers are a unique factorization domain already. So, the 4n+1 or 4n+3 primes are not a problem. We just get different primes than we might have expected in the Gaussian integers.
But the integers with the sqrt(-5) are not a unique factorization domain. So here is where the ideals should help, but I don't see how at the moment.
In red, are you talking about something like 2+i and 2-i as primes?Quote:
Originally Posted by YesNo
There is a bit of mystery here. I am wondering why if the Gaussians are a UFD is there more than one way to factor numbers such as 6 or 5 within it? I can almost trust I am overlooking something.
Anyway, key things get remembered. Another one is
Quotinet rings divided through by a maximal ideal produce a field.
Since the field seems to consist of only 0 and 1, I can't see yet why that is so important, but it seems to be.
Here is another key to hold onto:
Ideals are to rings as normal subgroups are to groups.
That shouts: Go study groups, doesn't it?
I hope that is correct. There is a lot of talk of cosets, too, and I think that may be more group theory in spite of the name recalling set theory.
* * * * *
My experience with being slow tells me a clear understanding of ideals is in front of my face, unrecognized, while I acclimate my brain to something new. All these little brealkthroughs will eventually amount to sort of an ephiphany.
The idea of principal ideals is pretty clear--they are just multiples. Like rings, ideals need 0 and 1 in the set--maybe some unit that stands in for 1, since the set of even numbers, for instance, cannot have a 1 in it.
Ideals are presented in additive terminology, though they have multiplicative properties.
I think this idea of "splitting," I keep running into refers to non-commutivity. I think this splitting is related to factorization problems. My present guess is that ideals manage to bypass this problem. I think this problem is related to the fact that 4n+3 primes are also primes in the Gaussians but 4n+1 primes are not, as they can be factored into smaller factors.
I hope I am not too amiss here. I am trying to put my collage together.
Up to units, there is only one way to factor 6 or 5 in the Gaussian integers. For example one can factor 6 = (3)(2) in Z. But one can also factor it as (-3)(-2). One could also reorder the factors as (2)(3). These are all different factorizations, but they are not what unique factorization tries to capture as an idea. The unique factors do not depend upon the order of the factors, nor do they depend on whether one can multiply the factors by 1 and get a different set of factors (associates). Again using 6 in Z, we can write 6 = (2)(3) = (1)(2)(3) = (-1)(-1)(2)(3) = (-2)(-3). In the Gaussian integers there are four units, not two as in Z: 1, -1, i, -1. Their norms are all 1. That challenges our normal intuition about what a unit should be (not just 1 or -1) and what an associate factor would be (not just multiplying the factor by -1).Quote:
Originally Posted by desiresjab
I am not clear about all of this either and I am finding it interesting to get a better understanding. In the Wikipedia article, https://en.wikipedia.org/wiki/Unique...ization_domain , there is a class chain:Quote:
Originally Posted by desiresjab
commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ GCD domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields ⊃ finite fields
There should be examples of non-unique factorization in structures to the left of "unique factorization domains" but none to the right. There should also be examples of a unique factorization domain that is not a principal ideal domain which is where the prime ideal questions we are discussing seem to be most important. But I am still unclear about how to formulate the questions.
Speaking of prime factorization in the Gaussians, you must be right. But in the Complex numbers is 2 even a prime? I can factor it as
(1+i)(1-i).
5 =(2+i)(2-i). Those are Gauusian integers, are they not (for a and b are both integers)? I believe those factors are not units, either. Does this make 5 not a prime in Gaussian integers. I believe no 4n+1 prime is a prime in the Gaussian integers, but I could be confusing Gaussians with the Complex numbers in general.
Now for 6 I really do not understand why 2x3 would be a prime factorization in the Gaussian integers, since I do not even believe 2 is a prime in that set. Isn't this the prime factorization?
(3)(1+i)(1-i)=6
What am I not getting about units for asking this question?
I have been looking around for an example of a non principal ideal. Of course it has been sitting in front of my nose.
All x+1 seem to form a non principal ideal. A non principal ideal is the kind I suspect Carmichael ideals to be.
In the complex numbers there aren't any primes. All are units since every complex number has a multiplicative inverse. So 2 is not a prime in the complex numbers. It is a prime in the regular integers denoted by Z, but it is not in the Gaussian integers because of the factorization you mentioned above. In the Gaussian integers, unlike the complex numbers, 2 is not a unit because 1/2 = 1/2 + 0i does not exist in the Gaussian integers since 1/2 is not an integer in Z.Quote:
Originally Posted by desiresjab
In the Gaussian integers you have factored 5 into two other Gaussian integers. So 5 is not a prime in the Gaussian integers. In the complex numbers 5 is not prime either, but that is because it is a unit. It has a multiplicative inverse 1/5 + 0i in the complex numbers, but that inverse is not in the Gaussian integers because both a and b in a + bi have to be normal integers that one has in Z.Quote:
Originally Posted by desiresjab
Right, in the Gaussian integers 6 factors as you mentioned. In the normal integers or Z, 6 = (2)(3). I was using that factorization to show what an associate was. In Z 2 has an associate -2. In the complex numbers 2 has -2, -2i and 2i as associates, since I just multiplied 2 by all the units in the Gaussian integers (1, -1, i, -i).Quote:
Originally Posted by desiresjab
A non-principal ideal for this polynomial ring over the integers would be (x, 2), that is, the ideal generated by x and 2. This is not the whole polynomial ring. That would be generated by a unit, say 1, but it contains those polynomials where the constant term must be even. It cannot be reduced to a principal ideal because then we would need some polynomial that divided both x and 2 (besides a unit like 1) so we could get x and 2 as a product. If such a thing existed it would be both an integer > 1 and a polynomial of first degree in x which is a contradiction. So this can't be reduced.Quote:
Originally Posted by desiresjab
This link might be useful in terms of making sense how ideals resolve the problem of not having unique factorization. These are just notes used with a larger text that popped up from an internet seach and I don't know who the author is, but the first few pages seemed to summarize the problem: http://www2.math.ou.edu/~kmartin/nti/chap11.pdf
That link was helpful in that it knocked a few chips off a mountain of marble and raised more questions than it answered. I must say the notation and terminology is (to resort to Trumpian irreducibles) very bad; very, very bad, and the people who devised it were some of the worst people.
That is still my gripe about mathematics--the same notation and terminology mean different things in different areas of math, sometimes closely related areas! The job of revision is too big for me, but I may get stubborn and refer to ideals like this <1+√5> in the future, instead of with parentheses which are so easily confused with standard multiplication. Later on, the author of that paper admits the terminology and notation in ideal theory has evolved in an unfortunate manner.
I wonder how many graduate students have sat in upper level math classes praying for a concrete example using almost all numbers and a minimum number of letters in place of numbers, just so they can establish through a concrete example what the hell is being generalized, and how so, in the first place?
* * * * *
Some of the pieces I am chipping off are above my current understanding and stored away until such time as they fit into a total picture. The idea of division in ideals is completely queer to me as I view it presently. It doesn't even seem like division. Division in ideals seems like the reciprocal of division as I know it but not multiplication.
What I want to do is go back to a more basic level.
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Back to 2Z, the good old even integers. Is this the set that forms the ideal of 2Z?
{0, 1, 2, 4, 6, 8,...}.
I am not sure if I have to notate it somehow as 0 and 1 adjoined to this infinite set of even numbers or not. From everything I have read I am led to think that
{0, 1, 2, 4, 6, 8,...}
would be the complete set of ideals for 2Z, 0 being the 0 ideal (additive identity), 1 being the unit (multiplicative identity), and 2 being the generator or principal ideal of the set. Each of the elements is called an ideal, 2 being the principal ideal.
So what is the entire set itself called? Is it called the set of ideals? If so, then I suppose the set of ideals of 2Z, or something like that? I am a bit confused on these several points, so I hope you can answer them.
Once I have these issues straight, I can move on to my next issues.
I have to go to a Chinese New Year celebration, so I'll get back to you either this evening or tomorrow.
I agree that examples are always good. The simpler the better.
The ideal (2) in the integers is the set of all even numbers, positive, negative and 0 with the operations that one has in the integers. Are you referring to the finite field containing two elements 0 and 1 which can be viewed as equivalence classes of even and odd integers?
Since an ideal is a set of elements from the ring with ring operations, division of these objects would be viewed as set inclusion.
We could speak in general of the ideals generated by the ring of integers Z, where Z is really 1Z where the 1 has been dropped. The numbers generated when 1 is used as the generator are just 1∙1, 1∙2, 1∙3, 1∙4,..., in other words just the full set of integers, which contains within itself all subsets of multiples of every integer, since those multiples are just integers, too.
The ideal 2Z is the set of all even integers, plus 1 and 0, because an ideal always has to have 1 and 0 in its set.
We could have 4Z, where the ideal would be all multiples of 4 from the integers, plus the elements 1 and 0 again, adjoined or however it might be expressed in the case of 0.
An ideal is not maximal if all the elements of the ideal are found in a larger subset of Z which is smaller than Z itself. In the case of 4Z, all integers which are multiples of 4 would be contained in the larger set of merely even numbers 2Z, which lies strictly between the multiples of 4 and the multiples of 1. Therefore 4Z is not a maximal ideal, because all its elements can be found in 2Z. However, 2Z is a maximal ideal because no subset smaller than Z itself "contains" it. This use of contains actually means divides in the language of ideals, according to the link recently posted by Yes/No.
It turns out that maximal ideals are generated by ordinary prime numbers, and only by them, no distinction being made between 4n+1 primes and 4n+3 primes.
I believe all maximal ideals are prime ideals, but 0 is a prime ideal, too, so all prime ideals are not maximal.
The ring of integers (mod p) where p is a prime is actually a field, a finite field. These fields are cyclic. I am not sure what precisely makes it a field instead of a ring, since I have been limping along under the impression that all four arithmetic operations had to be fully defined for fields. Well now, it occurs to me based on what Yes/No recently said about inverses, that as long as the modulus is prime every element of the residue system will have an inverse. Since every element has an inverse, that must be what makes it a field. They call that every element being a unit, which is a new one on me.
Under a modulus that is prime where every element has an inverse, I suppose that is enough to fully define division, which also would make finite fields under a modulus fields instead of rings. Somehow, I think they are fields and rings at the same time, if the definitions are not strictly mutually exclusive, which they could very well be.
I have been studying all day. I do not want to look this latest item up. Yes/No will know the answer. I will let him answer it. I want to watch movies right now.
Ah, yes, finally many things are coming together at once. Maximal ideals are indeed always prime ideals.
And, no, it is not impossible to be a Field and a Ring at the same time. A spoken proof sees that all Fields are Integral Domains, but the definition of Integral Domain explains it as a particular type of Ring.
The integers (mod p) where p is a prime, which we are so familiar with, forms a commutative ring where no two elements (residue classes) multipled together equals zero. The cyclic elements also form a Finite Field, and finite or not, all fields are integral domains, which are also rings.
In the language of ideals the integers mod (n) are called a Quotient Ring. This is what we are most familiar with in different language. Still, it is best not to lean too heavily on this particular understanding, for there is nothing cyclic about ideals in general. It does, however, apply because we would not be amiss in thinking of a quotient ring simply as the integers (mod n), where n is not necessarily a prime, as long as we do not forget that an object such as a Quotient Ring can be and is constructed not only from integers but from polyniomials as well, so that one has to remain aware of which type an article or portion of an article is alluding to.
* * * * *
Rings of polynomials are more difficult to deal with than integers algebraically. The dudes who developed the theory were high-powered intellects at the kind of old style algebraic stuff that mathematicians like Euler and Jacobi were known for--those long painful derivations and such that one amazes anyone ever unpuzzled for the first time. We have not gotten our hands dirty yet with manipulating polynomials. We might have to do that some time, or perhaps it can be avoided.
I hear other areas calling, so I want to get this one settled up as soon as I can. Combinatorics I already have pretty good basic experience and knowledge of in the sense that I played poker for many years and I was the kind who liked to calculate a lot of things out. Fancy counting is a load of fun, as I see it. I want to do more and see how it relates back to some of the stuff in number theory we have been looking at. We know for a fact that every group is a permutation group. I think Cayley proved that.
Yeah. Combinatorics, there is a lot more I want to learn there. In complex combinatorial situations, getting the logic just right is hairy. I love that stuff. Right now I am stuck here with ideals. But that is good. I only get stuck where I have chosen to get stuck already, for any one place that one settles in to confront higher mathematics there will be challenges that block progress for a while. Though I have in the last day made many connections that have avoided me, still I have a long way to go with ideals before I will be satisfied. The final capper will have to be an exposition of Carmichael ideals.
We can relax a bit now and give an entertaining question I remember from somewhere in the past that has to do with probability and game theory.
Three men with rifles are arranged in an equilateral triangle, all an equal distance apart. They are going to fire at each other in turn, until only one remains.
Participant A has a 90% accuracy rate, and participant B has a 70% rate. Poor C is a lousy shot, hits his target only 30% of the time, and is first to fire. What is his best strategy?
That is how I see it as well except for the part "an ideal always has to have 1 and 0 in its set". An ideal must have 0 in it since 0 is in the ring and so the generator times 0 will be 0. So 0 is in the ideal. However, if 1 were in the ideal, then the entire ring would be in the ideal and the ideal would be generated by 1. So, in general 1 would not be in the ideal, however, 0 must be there.Quote:
Originally Posted by desiresjab
A field is a special kind of ring as you mentioned in later posts.
That's is an interesting game theory question. I don't know the answer. If I were C I would aim first at A. If I were A I would aim first at B. If I were B I would aim first at A. This would lead to a pure strategy for A, B and C for their first moves. But I wonder how to solve this in a more general way.
What you say about 1 not being in the set makes sense, however my mind believes it has read a hundred times that 1 must be there. It only has an additive identity without 1 there, no multiplicative identity. Commutative, one-sided ideals always have inverses, do they not? If 1 is not supposed to be there I need to clear it out mentally, but I need to understand how I have mistread a hundred times.
I think a commutative ring must have both 0 and 1 in the ring. In Birkhoff and MacLane, "A Survey of Modern Algebra", one of the axioms of a commutative ring requires that there exists an element that is the multiplicative identity. That would be 1. The ring must also have the additive identity or 0. However an ideal need not be the full ring. For example the ideal in Z generated by 3 has all the multiples of 3. This would include 0 but not 1 which is not a multiple of 3. The ring must have 0 and 1, but not the ideal.
The ring itself is one of the ideals of the ring, which might make this confusing. That ideal which equals the ring itself must contain 1 since that ideal contains everything in the ring. But all the other ideals do not contain 1.
The following has a highly interesting opening paragraph.Quote:
Originally Posted by YesNo
https://en.wikipedia.org/wiki/Subring
Many (in fact the majority) of mathematicians seem to require that a ring contain a multiplicative inverse, but there are some who dispense with the notion.
The next to last sentence in that paragraph is a killer. It says the subring may have a multiplicative identity that is different from the one for R. What the...?
Anyway, slowly this thing is making sense. I always find afterwards that I have read over the truth many times before it had enough meaning for me to be included in my picture. This theory is rife with details that mean everything. If so much information were not included in these precise abstract algebraic formulations mathematicians would not be able to handle quite complex statements with a few swipes of the chalk, as I have seen them do. The propositions at this level are literally crammed full of detailed information upon which the whole content rests. Unless you understand the details perfectly, you will get the content wrong. That is why I have to be so nit-picky right now and question everything I do not firmly understand and believe--not just until I receive the right answer from someone else, but until I understand perfectly for myself. So when I sto piss questioning of everything, you will know I have probably understood or died.
That short Wiki-peja article I linked to is full of powerful statements. Of course some of those powerful statements are not entirely clear at this point in the journey, but are quite intriguing as future references to lean on.
* * * * *
Your daughter might find interesting the answer to the question I posed about A, B and C shooting at each other in trun from an equal distance. I could not remember the exact numbers of the question, so used numbers that were safe. The general idea is there.
C shoots first with only a 30% chance of striking his target, against opponents who have respectively 90% and 70% chances of hitting their target at this distance.
C must fire into the ground. If he were to accidentally kill one of the opponents, the next shot would be fired at him from an opponent who is far more accurate. His best case scenario is that B (70% shooter) kills A (90% shooter) with the next shot. That way C gets to at least fire the first shot at his superior opponent.
That makes sense that C should deliberately miss the target. I'll try to see how that game fits in with the game theory I've been reading about.
A ring without a multiplicative identity would not fit Birkhoff and MacLane's assumptions, but it shows that there are other ways to organize these algebraic structures. They mentioned that the assumption of the existence of 1 not equal to 0 was to eliminate examples of rings consisting of only the 0 element. A ring, by their definition, has to have at least two elements: 0 and 1. If the ring doesn't have an identity then the ideals would be subrings as the article mentioned whereas I would not think of ideals as rings except in the trivial case where the ideal (generated by 1) equaled the whole ring.
There seems to be some confusion even among the experts as to what is called what in the theory.
I think I am getting fairly close to a decent understanding. It really helps to work out these little inconsistencies and get everything in its proper place.
One thing that makes ideals so fun to study is that the word is not used elsewhere in mathematics with a gross of other meanings. A brand new word for a brand new idea. One could improve mathematics considerably by combing through an English dictionary and finding suitable terms to replace those that are overworked to the point of ambiguity.
Oooh! Oooh! I got something else straight that has been mystifying me, and which shows my guess was wrong about "splitting fields" being connected with 4n+1 and 4n+3 primes.
Splitting refers to factoring a polynomial all the way down to linear factors. Even if you factor it you have not split it unless all the factors are linear.
A Splitting Field involves something called Field Extensions, which are literally what they say. Mathematicians take a field like the rational numbers and "Adjoin," enough non-rational numbers to it to enable them to Split a particular ploynomial or class of polynomials into linear factors. The trick is to Adjoin just enough numbers to the original set to get the job done, instead of ending up with a much larger set which indeed gets the job of splitting done but also contains many superfluous elements. You want to adjoin the minimum number of elements to the set that enables splitting. This involves a lot of complex techniques but the basic idea is not that hard, though I imagine the idea of adjoinment in this fashion was quite revolutionary when it was first proposed. The seeds may have originated in group theory.
I don't know much about splitting fields, but the idea of adding only what one needs to the base field in order to factor a polynomial over the rationals sounds interesting.
A pair of confusing terms for me is "prime" and "irreducible". I assume primes only exist in a unique factorization domain, otherwise what one gets are irreducibles which are as far as one can factor an object in the algebraic structure. At least there is a factorization even though it is not the only one.
Yes, one would think every irreducible would be a prime, but that maybe is not the case. I do not yet know enough to satisfy myself but I am running out of ideas of how to proceed. I have not seen a bit of use for primaries and semi-primes, but I suppose that lies ahead.Quote:
Originally Posted by YesNo
Once ideals are understood fully, that, I believe, is the major portion of higher arithmentic that occupied great minds in the 20th century and late 19th. Going beyond ideals may require a measure of inventiveness . A full understanding of Artin's work along with ideals would be much of what one needs. I am not shooting for Artin's work, though. He is too difficult. I would be nowhere near ready for Artin at this point.
Dr. Salomone talks about that in this video. This guy lectures at light speed, and I like that. Subjects other professors take 45 minutes to discuss he dispatches of in ten minutes. No wasted time.
According to him the only cases where irreducibility and primality do not coincide are cases that are special and "not nice." I notice such special cases are usually shoved back and relegated to a later timetable when one is supposedly more advanced. The class is here is Abstract Algebra II.
I figure I won't understand anything fully, but some things will be understood enough. Eventually you should be able to read Artin's work. At the moment I don't think I would understand it either. But if we kept searching for clues, it should eventually make sense. I suspect it would take less time and effort to understand Artin than to understand Joyce's Finnegans Wake.
And this is the road to Artin. It forks ahead, if anyone wants to go there. Never say never. The mathematician seeks isolated clues about the structure of the universe, the artist builds an artificial one out of the materials at hand. There is some connection. It is a long ways off for the mathematician.Quote:
Originally Posted by YesNo
The right junction could present itself. There may be a turnoff toward Brocard. For this reason squares and anything about them is good to pick up. There is so much juice of squares left in field extensions, I have to stay a while longer. I also need now to go back to that difficult paper you linked to a few weeks ago, and see if I can better understand the operation of multiplication of ideals presented there.
I am getting more secure with ideals, and the idea of a subgroup within a ring that is only reachable through the subgroup itself, which precisely corresponds to what an ideal is, I hope.
Rings are to ideals as normal subgroups are to groups. Does this mean the friendly rings referred to by Salomone are the normal subgroups of group theory, as opposed to some other kind of subgroup that is not normal, like a non-commutative one, perhaps? I am going to stop guessing, but it is an addictive habit, and seems to serve me even when I am wrong, by keeping me asking questions until I am sure about something. Right now I am not sure what I am supposed to try to be sure about next, which is less fun than knowing where you should look, which is less fun than suspecting where you should look. At the moment I neither know or suspect.
I hope my old pal Yes/No is all right. It is not like him to leave posts unanswered. I wanted to say that division of ideals is not so challenging as it seemed at first glance. They call the ideal that does the dividing larger (as in terms of a larger set), for theoretcially its elements have greater density along the number line than the ideal it is dividing, such as 2Z dividing 4Z. The first may have more "elements," per unit distrance, but we realize both sets are actually infinite so there is no real difference in their cardinality but only in their density of occurence along the number line. However, it is plain to see that 2Z has a smaller generator than 4Z of which 4Z is a multiple, so naturally 2Z can divide it evenly.
As for multiplication of ideals, I am still trying to locate a paper I recently glanced at which explains it.
If you multiply two elements from the ideal, the result ends up back in the ideal. The cool thing about ideals is that if you multiply one of its element times one of the elements in the ring that is not in the ideal, the result ends up back in the ideal, too. They call that "absorbing," multiplication. In the case of non-commutative ideals, multiplication will be absorbed from either the left or the right.
I now have to look at some details of multiplication, then actually multiply some ideals together to inspect the results, then I should be done with this part and ready to investigate the relationship of Carmichael numbers to ideals.
I myself must now be gone for a few days traveling. I will investigate ideals and multiplication before I go so I will have something to dwell on while I am away. May you all be here and in good health when I return.
I think I missed your previous to last post. What you say makes sense to me. I don't know how ideals fit in with Carmichael numbers, but I heard they do in some way. I'll see if I can find out more about that.
Here's something on the multiplication of ideals that I thought was interesting:
This one shows that the product of two ideals is not their intersection, in particular, 2Z multiplied by 2Z is 4Z,but that is not the intersection of 2Z with itself: http://math.stackexchange.com/questi...e-intersection
That question also gives a definition of what multiplying two ideals means. Let I and J be ideals and ij the product of one element from I and one from J. Consider the set of all finite sums of these kinds of pairs. That would be the product of two ideas. It has to be finite since an infinite sum would likely take one out of the ring.
That is the problem. I do not see where 2Z or 4Z is finite. Element by element multiplication makes perfect sense for a finite number of elements. How do these two ideals (any two ideals) work out be finite? I do not see that.
One restricts the sums to be over a finite number of products by definition. The definition makes it finite. So we have I = 2Z = {...-4,-2,0,2,4,...}. Suppose we want to multiply the ideal I by itself. We can construct a set that contains every product, 2m * 2n, where m and n are integers. This multiplies every element in I by another element in I. Now take any finite number of those products and add them together. This becomes an element in the ideal that is formed from the product of 2Z * 2Z. The smallest positive integer in that product of ideals would be 4 and that would be the generator of the ideal. This is what we would expect because we get 2Z * 2Z = 4Z = {...-8,-4,0,4,8,...}.
Regarding Carmichael numbers the Wikipedia article says that the idea of Carmichael numbers is extended to other algebraic structures through ideals. Ideals don't help us solve problems about Carmichael numbers in the regular integers. So, ideals help generalize the idea of Carmichael numbers to other algebraic structures: https://en.wikipedia.org/wiki/Carmichael_number
Thanks again. I have to revisit an article to see if I can establish for a fact that the sum of two ideals is simply their GCD. That is very curious. Well, let me see. How about a concrete example? If I took two numbers such as 19 and 12. When I add them, their sum of 31 would actually be 1, in that case. So, yes, that seems curious. It seems modular. I can see where cycling back to 1 occurs in a cyclic ring such as a modulus, but how does that apply here? Of course the GCD of 19 and 12 separately is 1, so in that regard it makes sense. The notion here seems "not usual," and I will have to think about it some more and read some more.
I think one defines the sum of two ideals as what one would get if one added one element of an ideal to an element of another ideal. So if (19) is an ideal in Z and (12) is an ideal in Z, what would be in their sum would be elements like 19a + 12b for integers a and b.
Thinking about this as a greatest common divisor, since we know that the GCD of 19 and 12 is 1, we should be able to write an equation like this: 19x + 12y = 1 for some integers x and y.
But now think of 19x as some element in (19) and 12y as some element in (12). This matches up the result of the GCD operation and the sum of two ideals.
More is becoming clear but the muddled parts still drive me nuttier. There are quite a few items in the following article which confuse me. For instance, I do not "see," why or understand statelments like
Z[√-5] is already the full ring of integers of its quotient field Q(√-5). The examples they use are always √-3 and √-5, because larger ones quickly start to become unweildly. We would like to understand the difference in behavior in this arena of 4n+1 and 4n+3 primes, for apparently there is one, is about what I could get out of page four through page five of
http://www2.math.ou.edu/~kmartin/nti/chap11.pdf
Though I always pick up additional incidental details I do not fully understand or which, on the other hand, make perfect sense to me.