Cosmology

[YesNo]

That example should be completely factored. This Wikipedia article provides a good review: https://en.wikipedia.org/wiki/Irreducible_polynomial

Whether a factorization of a polynomial is reducible or irreducible depends on the ring the coefficients belong to.

There a distinction between primes and irreducibles, however, they are the same if one is in a "unique factorization domain" like the integers. This is what makes unique factorization important and not obvious even though it seems obvious. All of these distinctions can be either a pain in the rear or a delightful puzzle that one cannot put down.

Edit: In addition to not being able to be factored further, a prime p has the property that if p divides n = ab, then either p divides a or p divides b. If a natural number n could be factored into irreducible positive integers in more than one way, that would not be the case. I wonder if it is proper to define a prime by this property or to use this property to define a unique factorization domain?

[desiresjab]

A fellow before my last sleep was discussing in his advanced mathematics video how computers will change formal mathematics. He says some of the infinite operators in math of the past few centuries will be defined differently in this century in finite terms that can be computed. You canot compute something requiring infinite operations to get the job done.

This fellow calculated the roots of unity of x¹⁵-1 without any recourse to problematical Complex numbers which will not factor nicely. He makes use of something he calls Quadrance. He does not want to use its usual name of Norm because that is his name and Norm is used in another part of mathematics with a completely different meaning.

It appears mathematicians have been troubled over The Fundamental Theorem of Algebra since Gauss came up with it, kind of like Euclid's parallel postulated bothered great mathematician's for centuries. It turns out they had good reason to be bothered, since the postulate is not universally true. I wish I were going to be around to observe the eventual fate of The Fundamental Theorem of Algebra.

According to this fellow, ideals went a great ways toward eliminating some of the conflicts in the field by allowing their own arithemetic in which complete factoring was accessible. I will have to make sure of that last statement.

[YesNo]

Interesting observation about the Fundamental Theorem of Alegebra. I'm trying to see how it is like Euclid's parallel postulate, but it may be.

What is the link to the video about the change away from infinite operators in formal mathematics?

[desiresjab]

Interesting observation about the Fundamental Theorem of Alegebra. I'm trying to see how it is like Euclid's parallel postulate, but it may be.

What is the link to the video about the change away from infinite operators in formal mathematics?

Actually, he did not use the word operator, that was my bad handoff, he said away from infinite processes. I like this guy. He knows the mathematics really well, well enough to have deep convictions about the way math should be fundamentally reorganized and taught. He is a real nut in a way, a Kronecker throwback who even teaches something called Rational Trigonometry and eschews most theories in mathematics that smack of the infinite. He at least tries to make it smack less. Out of all the math lecturers I have watched on the internet he is the best, most organized, easy to see, hear and understand with the best laid down presentation. He has a ton of videos and he got me hooked. I can lead you to the right YouTube menu, and maybe the right video.

In the integers Z all ideals are principle ideals, because all ideals are a multiple of the unit. Sometimes it is hard to distinguish if they mean algebraic integers when they say integers. If integers have only this multiple type of ideal, then it must be the algebraic integers in complex numbers that allow other types of ideals. Anyway...

https://www.youtube.com/watch?v=H8xBlLWdzBE

https://www.youtube.com/watch?v=GMZoXXaOFeQ

If it was neither of those it was probably in one of his two videos on Galois theory, though later on he talks about it more extensively.

[YesNo]

I think he mentioned it in the first of those two lectures. He does not believe in infinite sets nor in processes that involve an infinite number of operations. The Axiom of Choice would be needed to assume one could do an infinite number of steps or choices.

I noticed in his two lectures that he called things "prime" that I would have only called "irreducible" and not prime. He did provide an example of a ring without unique factorization in the second lecture. It is good to know that such things exist because it makes one value the Euclidean algorithm for division.

I wonder why he does not like having infinite sets? In my case, I do not assume that the universe has to contain infinitely many integers for this to be true. I wonder if there is more to his objections than that.

[desiresjab]

I think he mentioned it in the first of those two lectures. He does not believe in infinite sets nor in processes that involve an infinite number of operations. The Axiom of Choice would be needed to assume one could do an infinite number of steps or choices.

I noticed in his two lectures that he called things "prime" that I would have only called "irreducible" and not prime. He did provide an example of a ring without unique factorization in the second lecture. It is good to know that such things exist because it makes one value the Euclidean algorithm for division.

I wonder why he does not like having infinite sets? In my case, I do not assume that the universe has to contain infinitely many integers for this to be true. I wonder if there is more to his objections than that.

He talks about his objections more extensively in other videos. He feels that mathematics has gone the way of complex convolutions hardly anyone can understand. I only partially agree. But if there is an easier way to approach many topics as he claims, I am interested. His claim that e and i and pi are not needed to find the roots of unity of x¹⁵-1, in other words cyclotomic monic polynomials, is quite appealing. He does it on the video. I only partially followed the reasoning, but I will go back for another helping.

Something he complains heavily about is the lack of good examples in these highly abstract areas such as infinite sets. Though he does a fine job I could level the same complaint his way. Where is my example in detail of an ideal other than the simple principle ones of the integers? After extensive investigation I still have no idea how a Carmicheal number can have anything to do with an ideal. A third grader could understand a principle ideal, now show me how these other objects can be ideals when they are not multiples of a generator, or whatever. I am a little peeved at the math industry myself, you might say. I have put in enough effort. Are these idiots lost in their abstractions hiding the answer or am I simply too thick? Certainly it takes time, for there are times I will read right over a key statement multiple times without noticing it holds the answer. I do not know what I am overlooking this time.

[desiresjab]

Continue with my own whines...

It was very neat, for instance, how they "said," (and that is about all they did) all equations that have a constant of 0 form an ideal. Okay. Neat. How is that? Are they multiples of one another? I think not. Then in exactly what way does having a constant of 0 unite them into something we call an ideal? I don't get the parameters. What qualifies to make something an ideal, and what does not qualify, for that matter?

[YesNo]

I might be misunderstanding this, but regarding polynomials that have a constant term 0, they would be generated by the polynomial x. Take any polynomial p out of the ring of polynomials, say p = a_nxⁿ+...+a₁x + a₀ and multiply p by x. You will get another polynomial that has the constant term 0 because the polynomial p from the ring, which might have had a constant term a₀ not equal to 0, was multiplied by x making the constant term of the product, xp, equal to 0.

I don't see, at the moment, how Carmichael numbers relate to ideals, but I suspect they are.

[desiresjab]

I might be misunderstanding this, but regarding polynomials that have a constant term 0, they would be generated by the polynomial x. Take any polynomial p out of the ring of polynomials, say p = a_nxⁿ+...+a₁x + a₀ and multiply p by x. You will get another polynomial that has the constant term 0 because the polynomial p from the ring, which might have had a constant term a₀ not equal to 0, was multiplied by x making the constant term of the product, xp, equal to 0.

I don't see, at the moment, how Carmichael numbers relate to ideals, but I suspect they are.

So the constant term of x is zero but we multiply by it as if it were there anyway. It is there (invisibly) in the constant term which has to be multiplied by every term in p, so that particular row of the polynomial multiplication will be all zeros. Is this correct?

If I am not misinterpreting this, you are saying that every polynomial without a zero constant can be multiplied by any polynomial that has a zero constant and the result will be a polynomial that has a constant of zero. Is this correct?

But would not this operation put any polynomial at all in the ring?

And couldn't any polynomial with zero constant force every polynomial into its particular ring?

[YesNo]

I assume we want a set, an ideal, that contains all the polynomials with 0 in the last constant term. One way to get that would be to take all the polynomials and multiply them by x + 0. Some polynomials would not be in this ideal. For example the polynomial x + 1 would not be in the ideal. It has 1 as a constant term. But we could use it to get (x + 0)(x + 1) = x² + x + 0 which is in the ideal.

If we used x² + 0 as the generator we would miss x + 0 in that ideal. Not all polynomials with 0 in the constant term would be in that ideal.

[desiresjab]

I assume we want a set, an ideal, that contains all the polynomials with 0 in the last constant term. One way to get that would be to take all the polynomials and multiply them by x + 0. Some polynomials would not be in this ideal. For example the polynomial x + 1 would not be in the ideal. It has 1 as a constant term. But we could use it to get (x + 0)(x + 1) = x² + x + 0 which is in the ideal.

If we used x² + 0 as the generator we would miss x + 0 in that ideal. Not all polynomials with 0 in the constant term would be in that ideal.

Yes, I am finally beginning to see. Any equation with 0 as a constant will generate from any other equation a multiple of itself. It can generate every multiple of itself on any other equation, but not the same multiples that most other 0 constant equations can generate of themselves, for they are different multipliers unless their difference is a trivial one of not being reduced fully.

Now, just what is the supposed infinite set here, the multiples of itself that a 0 equation can generate from one other equation, or what it can do to all equations? It must be the latter from what you say. Any equation with 0 as constant will spread that effect on multiplication. I am hesitant to use the word class right now because I know that word is probably used in a precise manner in the language of ideals ahead.

So, from all other equations it generates an infinite set by multiplying itself with them. This is the infinite class of ideals associated with this particular equation. Now another distinct equation with 0 as a constant does the same thing to all other equations, and thereby generates its own infinite class of ideals based on being a multiple of itself. That is how I see it at the moment. Next I need to see how Carmichael numbers form ideals, unless I have still got something wrong. I am almost there.

[YesNo]

The infinite set would be the set of polynomials that have the generator as a factor.

If one uses 0 as the generator one would get the zero ideal which has only 0 in it. The ideal with 1 as the generator would be the opposite extreme and contain all the polynomials.

It seems to me at the moment that when one looks at ideals one is looking at all the multiples of a generator. Often I think of a prime p in the integers as a positive integer such that if p divides ab then p divides a or p divides b. That would be looking at a division property of the prime rather than all of its multiples. For a prime ideal P one would rewrite that as if ab are in P then either a is in P or b is in P. https://en.wikipedia.org/wiki/Prime_ideal

[desiresjab]

In other words the generator only makes a single copy of each equation it comes in contact with. If I have a generator equation G and another equation F, then multiplicatiobn produces a multiple of G. It is not exactly a multiple of F, because F did not have 0 as constant and the new equation does. If G multiplied itself with the F more than once it would in effect be squaring itself, which is not part of the deal.

So a generator function acts once upon each equation in the world without a zero constant, and thereby produces an infinite set of equations with 0 as constant because there are infinite equations without 0 as constant to multiply itself with.

To me this is different than a single integer generating an infinite set of intervals (ideals) on the number line, for I see nothing regularly spaced about these new equations generated by G. However, I have certainly read that every ideal in integers or Gaussian integers is a principle ideal. As far as I can tell, this is not true of algebraic integers, which are strictly the roots of equations. Every Gaussian integer is not the root of some equation, is it? There is some confusion still whether expositors are speaking of Gaussian integers or algebraic integers at a given time in a discussion. That is, algebraic integers would not entirely fill the lattice points of the Complex plane as Gaussian integers do. Is there any truth to this or am I misinterpreting something?

[desiresjab]

If I took this new equation generated by GxF and multiplied it times each integer, then I would would have infinite multiples of G from one F, just as I can generate with a plain integer, but that does not seem to be part of the definition (the deal). Here the equations (F) without 0 as constant stand in an infinite line waiting to be multiplied with G. Those single multiplications on each F generate the infinite set of ideals (equations which are all multiples of G).

Name a new multiple of G as H. Now G would divide each H it generates. But each H in turn does not have to divide every other H, anymore than 6 has to divide 9 just because they are both multuiples of 3. Though not multiples (necessarily) of one another, some of the new equations in the infinite set would have to be multiples of some other equations in the new set, just as 6 is not a multiple of 9 but is a multiple of 12. Equations in the infinite set which happen to be exact multiples of one another would form an infinite subset within the set. Whether this subset carries much meaning I do not know.

[desiresjab]

In the vector-ball diagram in the Wiki-peja article you linked to, am I to take it that the top purple row is actually an infinte row of pure primes? Otherwise they would be saying that only 2, 3, and 5 can generate ideals.