Cosmology

[YesNo]

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).

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.

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.

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.

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?

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).

[YesNo]

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.

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.

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

[desiresjab]

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.

* * * * *

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.

[YesNo]

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.

[desiresjab]

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.

[desiresjab]

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.

[desiresjab]

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.

[desiresjab]

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.

[desiresjab]

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?

[YesNo]

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.

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.

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.

[desiresjab]

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.

[YesNo]

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.

[desiresjab]

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.

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.

[desiresjab]

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.

[YesNo]

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.