Cosmology

[desiresjab]

Broken back to English, it seems to go like this for ideals in commutative rings:

1 Prime ideals are of the form nZ, where n is a prime.

2 Primary ideals are composed of powers only of a prime element. This means n, the prime element, is also primary, even with the lowly power of 1.

3 Semiprime ideals are combinations of more than one prime, but which are also square-free.

I heartily agree with Wildberger that more good examples are needed. The Wiki-pega article on Semiprimes is valuable just because it gives a specific example, which cuts off a lot of questions at the pass. The article notes with the required specificity for dummies that 30Z would be a semiprime ideal, where as 12Z would not be. Mathematicians act like specific examples are going to kill them or lower their princely standards. The example makes it clear that 30 is semiprime because its factors are no more than single powers of primes.

For myself, adjusting to the language of ideals will take further familiaization to become entirely comfortable. Experts often talk somewhat loosely among themselvs, and tend to continue this trend in their expositions. Most mathematicians are poor expositors when it comes to bringing their abstract notions out of the darkness for laymen or even interested amateurs.

There is a reason for this: the second job is more formidable. With another expert, talking over concepts is easy. As Wildberger notes somewhere, it is basically Santa Claus to the Easter Bunny power, a pure manipulation of symbols. When I am done, I will be able to make the notion of ideals and their ramifications clear to an interested person. If it is clear to me, I should be able to do that.

By the way, I am looking at 12Z. It is not a prime, it is not primary, and it is not semiprime either. It is white on the diagram. It must play the role of a strict composite in ideals. I don't know, I am just guessing. I will overcome many impasses and wrong notions as I continue to chip away. In the end I have to be able to perform the arithmetic of ideals as easily as I can perform modular arithmetic in integers.

[YesNo]

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.

An ideal is a subset of a ring. If G is a generator of an ideal and F is any element in the ring then GF is in the ideal although F may not be in the ideal. In particular since G is in the ring, then GG is in the ideal. So the square of G is in the ideal, just as one would expect the square of any prime to be in the set of multiples of that prime in the integers.

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?

I think every Gaussian integer would be the root of a polynomial with integer coefficients. Let a + bi be a Gaussian integer, where a and b are regular integers. Note that (a + bi)(a - bi) = a² + b², an integer. Multiply together (x - (a + bi)(x - (a - bi)) to see if this forms a polynomial with integer coefficients. I get x² - 2ax + a² + b², unless I made a mistake. So the arbitrary Gaussian integer a + bi is the root of a polynomial with integer coefficients.

A principal ideal is an ideal generated by a single element: https://en.wikipedia.org/wiki/Principal_ideal

I am asking myself if it is true that every ideal in the Gaussian integers is a principal ideal. I think it is, because of what you mentioned, but I will have to find a proof.

[YesNo]

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.

Yes, the diagram was only partial. The purple row contains all the primes in the full diagram which can't be written out. However, now that you pointed it out, that diagram mentions a bunch of terms: prime ideals, semi-prime ideals and primary ideals. They apparently mean different things and have some use value, but now I am trying to clarify in my own mind what those are.

Edit: I just saw your recent post. I think you have clarified these terms.

[desiresjab]

Yes, the diagram was only partial. The purple row contains all the primes in the full diagram which can't be written out. However, now that you pointed it out, that diagram mentions a bunch of terms: prime ideals, semi-prime ideals and primary ideals. They apparently mean different things and have some use value, but now I am trying to clarify in my own mind what those are.

Edit: I just saw your recent post. I think you have clarified these terms.

If a principal ideal is generated by a single element, then primary ideals must also be principal ideals, since they are powers of a single element. Keeping all the lingo straight in order to go farther apparently comes with the territory. Specific examples are the color gold in an otherwise black and white setting. All prime ideals are primary, but obivously all primaries are not prime. I still do not know the official classification of 12Z.

Primes are both semiprime and primary, two different branches, but which makes sense because they are the geneators of everything after all (not sure about 0, however), so they should generate all the branches, it seems intuitively.

* * * * *

It has diverted my attention whether all integers defy the Fundamental Theorem of Arithmetic over the Complex. I saw 5 factored two different ways. The same technique should work for any prime--just use the conjugate. Since a composite can be broken into prime factors (which themselves defy unique factorization), then the composite has more than one factorization as well. The number of ways to combinatorially compute the division algorithm would be a simple extension of adding in more factors, but some of which do not work together. Hence, all integers defy unique factorization over the Complex field.

Excuse that little aside. I needed that. I am that rusty in areas.

[YesNo]

I was thinking about the different terms today as well while walking. This is how I see it.

If one has a field, a special kind of ring where all elements (except 0) have a multiplicative inverse, then there are only two trivial ideals: the whole ring and the set containing only 0. We can forget about fields except as sources of examples.

So a ring has to have elements that do not have multiplicative inverses for ideals to be interesting. The integers would be an example of such a ring as well as polynomials.

Here are the definitions:
1) Ideal, a subset of a ring generated by a finite set of elements.
2) Principal Ideal: an ideal that can have the set of generators reduced to one element.
3) Zero ideal: the ideal generated by the 0 element and containing only 0.
4) Unit ideal: the whole ring generated by a unit such as 1.
5) Prime ideal: a principal idea generated by an element p such that if ab is in the ideal then either a is in the ideal or b is in the ideal. For example, the ideal generated by 6 would not a prime ideal since 36 = 4 * 9, but neither 4 nor 9 are multiples of 6 and so they are not in the ideal. The ideal generated by 6 would not be a prime ideal, as expected.
6) Semi-prime ideal (radical ideal): is an ideal generated by a square-free integer. Here the ideal generated by 6 = 2 * 3 would be example and the ideal generated by 12 would not be an example.
7) Primary ideal: is an ideal generated by the power of a prime.

I am sure there are other critical definitions and then one needs to find out how these work in many different rings.

In the Gaussian integers 5 is not a prime because it can be factored since 5 = 1² + 2² = (1 + 2i)(1 - 2i). This is true of all primes in the regular integers that are congruent to 1 mod 4. But primes in the regular integers that are congruent to 3 mod 4 cannot be represented as a sum of squares and so they are prime even in the Gaussian integers since they are irreducible. The Gaussian integers are supposed to be a unique factorization domain which means irreducibles are primes. To construct the Gaussian integers add to the regular integers i = sqrt(-1). The example that did not have unique factorization was when one added sqrt(-5) to the regular integers. The unique factorization failed in that case, but these are not the Gaussian integers.

[desiresjab]

I was thinking about the different terms today as well while walking. This is how I see it.

If one has a field, a special kind of ring where all elements (except 0) have a multiplicative inverse, then there are only two trivial ideals: the whole ring and the set containing only 0. We can forget about fields except as sources of examples.

So a ring has to have elements that do not have multiplicative inverses for ideals to be interesting. The integers would be an example of such a ring as well as polynomials.

Here are the definitions:
1) Ideal, a subset of a ring generated by a finite set of elements.
2) Principal Ideal: an ideal that can have the set of generators reduced to one element.
3) Zero ideal: the ideal generated by the 0 element and containing only 0.
4) Unit ideal: the whole ring generated by a unit such as 1.
5) Prime ideal: a principal idea generated by an element p such that if ab is in the ideal then either a is in the ideal or b is in the ideal. For example, the ideal generated by 6 would not a prime ideal since 36 = 4 * 9, but neither 4 nor 9 are multiples of 6 and so they are not in the ideal. The ideal generated by 6 would not be a prime ideal, as expected.
6) Semi-prime ideal (radical ideal): is an ideal generated by a square-free integer. Here the ideal generated by 6 = 2 * 3 would be example and the ideal generated by 12 would not be an example.
7) Primary ideal: is an ideal generated by the power of a prime.

I am sure there are other critical definitions and then one needs to find out how these work in many different rings.

In the Gaussian integers 5 is not a prime because it can be factored since 5 = 1² + 2² = (1 + 2i)(1 - 2i). This is true of all primes in the regular integers that are congruent to 1 mod 4. But primes in the regular integers that are congruent to 3 mod 4 cannot be represented as a sum of squares and so they are prime even in the Gaussian integers since they are irreducible. The Gaussian integers are supposed to be a unique factorization domain which means irreducibles are primes. To construct the Gaussian integers add to the regular integers i = sqrt(-1). The example that did not have unique factorization was when one added sqrt(-5) to the regular integers. The unique factorization failed in that case, but these are not the Gaussian integers.

I pretty much have most of that. But I forgot something critical, which I was supposed to know, which I put in blue; and I have completely overlooked something critical, if it is true, which I have put in red.

[desiresjab]

Then again, it seems like I have done exactly the same thing to each of these:

Factorization of 5.

(√4+i)(√4-i)=4-√4 i+√4 i+1=5

Factorization of 6

(√5+i)(√5-i)=5-√5 i+√5 i+1=6

Factorization of 7

(√6+i)(√6-i)=6-√6 i+√6 i+1=7

[YesNo]

In the real numbers or the rationals or the complex numbers, which are all fields, you can factor 5 in many ways. Everything in those sets (except 0) has a multiplicative inverse. They are all units. There are no primes or irreducibles.

It is only when you have a set like the integers or the Gaussian integers or even some set of algebraic integers such as the integers with sqrt(-5) added to them, that you don't have multiplicative inverses for everything. Because not everything (except 0) has a multiplicative inverse there are elements that could be called irreducible, or if the ring is a unique factorization domain, a prime.

[desiresjab]

I tried factoring the number 6 down further because:

The factorization of 2=

(√1+i)(√1-i)=1-(√1-i + √1-i)+-(i)2=1+1=2

And the factorization of 3=

(√2+i)(√2-i)=2-(√2 i+√2 i)+1=3. Therefore

(√1+i)(√1-i) x (√2+i)(√2-i),

is just another expression of 2 times 3, and does work by my clumsy calculations, but only 8 of the possible 24 permutations of the factors, which we can name ABCD, will produce 6.

(AB)CD, (AB)DC, (BA)CD, (BA)DC and the palindrome (here I mean reverse order) of each works, for a total of eight. Any other order of factors does not produce 6 for me, but those eight orders do. I do not know if this is enough to be defined as a generalization of commutivity in the Complexes, or if it signifies the Complex version of non-commutivity. From an integer standpoint it seems some commutivity and some non-commutivity are involved.

[desiresjab]

In the real numbers or the rationals or the complex numbers, which are all fields, you can factor 5 in many ways. Everything in those sets (except 0) has a multiplicative inverse. They are all units. There are no primes or irreducibles.

It is only when you have a set like the integers or the Gaussian integers or even some set of algebraic integers such as the integers with sqrt(-5) added to them, that you don't have multiplicative inverses for everything. Because not everything (except 0) has a multiplicative inverse there are elements that could be called irreducible, or if the ring is a unique factorization domain, a prime.

I may require a slight perspective change. Water is trying to soak into plastic here. It just may be that I am so used to thinking of commutative rings with a modulus involved, that I need to step back and temporarily release the notion of a modulus ruling the ring, to see ideals more clearly. With a prime modulus in normal integers every element of the ring will have an inverse and no two elements the same inverse, I believe, as well. In a modulus ring of integers there is some notion of divisibility, whereas if one must strictly be in Integers, no inverse could exist. But stick a modulus anywhere and some inverses will appear.

[YesNo]

If one is thinking of a modulus one might be in a field of equivalence classes of integers rather than the integers themselves. For example, Z₅ would contain five equivalence classes or sets of integers. Each would have a different remainder in the integers modulo 5. This would be a field since every equivalence class or element of that field except the 0 equivalence class would have an inverse. There would be no primes or irreducibles in that structure.

It might be interesting listing different fields and then different kinds of rings that are not fields just to get a set of examples to work with.

The fields would contain the following: (1) real numbers, (2) rational numbers, (3) complex numbers, (4) equivalence classes of integers modulo a prime integer. The only ideals here are the zero ideal (0 element) and the unit ideal (entire field).

The rings would contain the following: (1) integers, (2) Gaussian integers, a + bi where a and b are integers, (3) rings of various algebraic integers, like the Gaussian integers but instead of i = sqrt(-1) some other root of a monic polynomial, (4) polynomials with rational coefficients. Rings should provide interesting examples of ideals.

There are probably many others. The above are all commutative and so there must be some non-commutative examples such as matrices.

[desiresjab]

Being a ring or a field depends on the number of defined operations, I believe. Fields, as I take it, usually have four defined operations, rings two or three operations.

[desiresjab]

You said earlier that all interesting examples of ideals were non commutative. I do not see why that is true. I thought there were interesting examples in both, but my vision will develop as I continue. I do need to shed the habit of always thinking in modulus rings.

My thinking was temporarily derailed over the last few days by music. I made a few posts in a music thread and that got me listening again. I have to stay away from music or it derails everything else and takes over my intellectual life. I unfortunately have to segregate all of my interests like that. I can only do one field at a time.

I cannot live on human diurnal schedules. My cycle is about 41 hours instead of 24. My overlap is congruent to 17 (mod 24), with irregularities associated with obligations. Night and day are arbitrary to me. I got that way from decades of marathon poker sessions.

[desiresjab]

(√4+i)(√4-i)=4-√4 i+√4 i+1=5, also 1(5)=5

(√6+i)(√6-i)=6-√6 i+√6 i+1=7, also 1(7)=7

* * * * *

You assert there are more (in fact infinite) factorizations of 5. What are some? At the level above, the difference between 4n+1 and 4n+3 numbers is not obvious. The "splitting," they do must be visible from another vantage point.

[YesNo]

I don't know if the interesting examples of ideals are in commutative or non-commutative rings. At the time I was only able to think of commutative examples.

I think it would depend on the ring or field whether there were infinitely many different elements that could divide into 5. In a finite field there would only be finitely many different elements to divide into 5. In the rationals, one could take every rational number and divide it into 5. These aren't really interesting because in the field 5 is a unit. It is not a prime.

One could say that (1)(5) = 5 is a factorization, but the 1 is a unit. If one restricts the factors of 5 to be irreducibles or primes, then some rings such as the Gaussian integers would be able to factor 5 and other rings such as the regular integers would not.

Edit: Here's a list of algebraic structures. I have not studied most of them, but it is good to see them in one place with links: https://en.wikipedia.org/wiki/Algebraic_structure Maybe one day, I'll look at them more closely.