Real Quadratic Integers

[@TheGrayCuber]

IMPORTANT: this video is a discussion specifically of real quadratic integers WITH UNIQUE FACTORIZATION. Unfortunately I neglected to include that as a disclaimer within the video. There are some domains such as O(√10) that are not UFD, for which some of the discussion in this video will not apply.

[@parzh]

Olgebraic Onteger, I died

[@realmless4193]

Now we need hypercomplex primes.

[@lordeji655]

Hey ! I really love your series, it really expanded my view that was limited to gaussian integers. Juste a simple question, if possible, could you, in the description, link some ressources (pdf, videos, etc) to delve DEEPLY into the subject please ?

[@TheGrayCuber]

staff.fnwi.uva.nl/t.j.dekker/PrimesPaper/Primes.pdf This paper is a good read!

[@lordeji655]

@@TheGrayCuber thank you so much !!

[@ahoj7720]

Very impressive! Quadratic residues/no residues in F_p(i) also produce nice patterns! (If p = 3 mod 4)

[@Arcticgdnmore]

Will there ever be a new no peeking episode? I would love to see the axis cube or polaris cube in one.

[@TheGrayCuber]

No I do not plan to make any more cubing content

[@StentorCoeruleus]

I can’t wait for the next video

[@sclearDevelopment]

amazing videos!!!

[@luke-wg8gi]

4:28 i wrote a program in rust and checked all values up to 4 trillion and still nothing for factors of 2 with sqrt 86 :(

[@lagomoof]

If we assume the form of two factors (not necessarily prime) of 2 in O(√86) are (a±b√86) and thus their product is a²-86b², there is a problem: There is no real integer solution to a² - 2 = 0 mod 86. There are two solutions for O(√94) and 1464 fits one of these. Also, 2 is a zero divisor mod 86, thus ½ mod 86 does not exist, and so making 'a' a half-integer doesn't seem to be the way to go here either. (Wouldn't be surprised to find I've missed something obvious here.)

[@COArSe_D1RTxxx]

the two factors can be different

[@sigmaoctantis5083]

In the real quadratic case the norm can be negative. Hence we also have to look for solutions of N(a±b√86)=-2, and there is one: N(102±11√86)=102^2-86*11^2=-2.

[@TheGrayCuber]

@@sigmaoctantis5083 Thank you for pointing this out!

[@PefectPiePlace2]

Great series of videos! Keep it up!

[@jakobr_]

12:44 Would it be faster to get to 50 by building up from left to right? Start with 86 (1), square (10), times 86 (11), square x3 (110, 1100, 11000), times 86 (11001), square (110010).

[@makagyngrimm3392]

Ye

[@TheGrayCuber]

Yes this is a great point!

[@Patrickoliveirajf]

Amazing content !

[@purplenanite]

4:33 I had trouble finding a factorization as well but if 2 has a factorization, it must contain some units followed by two numbers of norm 2. but I don't think you can make numbers of norm 2 in O(sqrt(86)). a^2 - 86b^2 = 2 mod 86, a^2 = 2 (mod 86) which has no solutions

[@TheGrayCuber]

Yes unfortunately I neglected the fact that this factorization could also use numbers with norm of -2

[@purplenanite]

@@TheGrayCuber ah, i made the same mistake! so, something like (102+11sqrt(86))(102-11sqrt(86))* -1?

[@TheGrayCuber]

@@purplenanite Exactly!

[@popop614]

6:49 I think something might be off with this argument. Consider x^2 - 37y^2 = 11. Since 2^2-37 * 1^2 = -33, here is a solution thats divisible by 11 but not 11^2, hence 11 should mot be prime. You've argued that there are elements in Z[sqrt(37)] such that ab = 11, and N(a) = N(b) = 11. But actually its known that such a solution does not exist. The issue is that p prime --> p|ab means p|a or p|b is sometimes false! (so-called Euclid's Lemma.) Consider for example, in Z[sqrt(-5)], (1 + sqrt(-5))(1 - sqrt(-5)) = 2 * 3. Then is 2 prime? If it was, then assuming Euclid's lemma we wpuld get 2 dividing one of the left hand side. But this isnt true, so 2 should have some prime factors. Except you can show then that x^2+5y^2 = 2 does not have a solution by size reasons. So something deeper is going on here.

[@TheGrayCuber]

Yes there is a big piece missing from that argument - this video is a discussion specifically of real quadratic fields that are UFD, but I failed to point that out. That argument should hold within a UFD. The key thing with your example is that O(sqrt(37)) is not Z[sqrt(37)]. 37 is 1 mod 4, so O(sqrt(37)) also includes (a + bsqrt(37))/2. This does give us (213 + 35sqrt(37))/2 that has norm 11.

[@falnica]

I was waiting for this one

[@benhsu42]

11:45 mark I think the factorization should be 9+sqrt(86) and 9-sqrt(86) (you put the minus in front of the 9)

[@catmacopter8545]

What if for our "basis", we use 1 (as usual), and K = e^i (1 radian around the unit circle)? K's most interesting property is that K to any integer power has a norm (in the sqrt(a²+b²) sense) of 1, and there is no integer power where K to that power is K. Does this result in anything new?

[@Han-b5o3p]

This will be my last deja vu :(

[@rockstonic52]

weve transcended

[@StentorCoeruleus]

Stay rational

[@TheArtOfBeingANerd]

Get real

[@NonTwinBrothers]

Hey that's pretty good

[@05degrees]

🌌

[@tothm129]

Are there quaterion primes?

[@TheGrayCuber]

Sort of: en.wikipedia.org/wiki/Hurwitz_quaternion

[@sheeep7378]

that’s great thank you

[@FrankHarwald]

I'm wondering: isn't it a little weird to even talk about "the norm" in a space which isn't even metric? (hyperbolic spaces aren't metric because the triangle inequality doesn't hold, & the triangle inequality is one of the necessary prerequisites to have a unique way of assigning a positive, real value ("the norm") to elements of a set.)

[@05degrees]

It surely isn’t a norm in a normed vector space fashion, but it should be called something...

[@mskiptr]

Could we achieve an infinite number of units by having them be on the complex unit circle and then swapping integers for rationals or reals?

[@AdrienZabat]

In general, there will be an infinite number of units if the degree of the number field is large (for example, adding the root of a large degree polynomal)

[@mskiptr]

@@AdrienZabat I mean, my question is more like "can we generalize Gaussian and Eisenstein primes not by going to quadratic integers, but somehow packing more and more units on the complex unit circle".

[@AdrienZabat]

Yes, those are the cyclotomic fields, where you adjoin roots of unity to the rationals. In general there will still be units outside the unit circle too though.

[@mskiptr]

@@AdrienZabat But can we somehow define primality there?

[@AdrienZabat]

Yes primality isn't hard to define, you just say p is prime if p divides ab implies that p divides either a or b (or both). It's unique factorization that requires tweaking as was done in last videos.

[@gaier19]

So in the rational numbers all numbers( minus 0) are unit so no unique factorization. But if you allow negativ exponents at the "Integer" Primes it "works". So how would you define this "unique factorization".... can not do it with units... (same works in Gaussian rational and Gaussian primes) ... can not find "good" info, only yeah it works but no "defintion"

[@05degrees]

Rational numbers, and any field, _have_ unique factorization. Uniqueness is defined up to units (because we indeed can’t do better). So each unit is factorized as just itself, times an empty product of prime powers.

[@gaier19]

@@05degrees ah ok ... but again how would you change the definition for Rational numbers because you can use the "normal" prime numbers again ^^ it works with negativ exponens

[@05degrees]

​@@gaier19 I’m not sure I follow this time! Regardless: rationals are way more “fixed” than reals because it’s just the smallest field that contains integers; and integers are also a ring that can be sent by a homomorphism into any other ring in a unique way. In a sense, integers are what you get when you try to collect all expressions which use 1, 0, +, −, × and for which we are allowed to use ring axioms to treat some of those expressions as same (for example, (1−(1+1)))+1 and 0; or (1+1)+1 and 1+(1+1) - the latter two and all others equivalent to them via axioms get the convenient name “3”). In any other ring, we can make the same expressions too (we have 1, 0, +, −, × in any ring!), and if two expressions were equal by ring axioms, they would necessarily evaluate to the same ring element (but different expressions can also evaluate to the same element, the extreme case being a zero ring in which there’s just a single element 0 ≡ 1). In this way, there’s image of integers in every ring, though different integers can get conflated. We can categorize this conflation via “characteristic” of a ring. For fields it ends up very important: there are no morphisms between two fields if they have characteristics, say, 0 and 2. So there are no morphisms between finite fields and fields like ℚ, ℝ and ℂ which is why finite fields may look very weird in many aspects. But I digressed.

[@gaier19]

​@@05degrees Thx a lot! Sorry ^^ ok next try! What would be a definition to use "natural" prime number with negative exponents to "uniquely express" rational number or ^^ is that the definition? Definition: A rational number can be "split" in to "prim(or Irreducible ... should be the same in this context)" elements that are prime elements of natural numbers with integer exponents, up to units and rearranging. ??? Again sorry yeah the "old definition" only leafs the empty set (if I understand it right)... but you can define a "factorization" that "works" like in the natural numbers... sorry if I dont make sense! So if I understand it right "the definition of unique factorization" in fields does not "work nicely" with Rational numbers ... because you can get a "better" unique factorization with this definition.

[@05degrees]

@@gaier19 Hm hm hm. Okay I don’t think I follow much but I can at least confirm that rationals ℚ indeed behave like integers ℤ wrt to factorization to primes and units _of ℤ_, yeah, but I don’t think this necessarily works if we replace ℤ with an arbitrary unique factorization domain (say) and ℚ with its field of fractions. Or maybe it always works but then when we try to generalize to rings that are weaker than UFD itmight as well start failing. There’s a huge lattice of ring types that are intermediate between ℤ and rings that can be very badly behaved, and maybe you’ll find some help at the Wikipedia page for “Unique factorization domain” (not posting links because the comment can get flagged by YT and be completely deleted). There’s a chain of different ring types (I bet not all of them that are of note to ring theorists but at least there are a few!) which I can paste here with no problem: rngs ⊃ rings ⊃ commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ GCD domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields ⊃ algebraically closed fields -but the article itself has hyperlinks so you could have a better picture what each rung makes possible to compute or have. (Rng is the same thing as a semiring: a ring without additive i-nverses.) So you can see UFD has some generalizations and specializations that are close to the unique factorization property and are still not close to being integers (or another “good” ring) nor they are completely hopeless so one may want to generalize factorization to them if it hasn’t done already, and that might be hard if not.

[@kales901]

what if D=0?? edit sqrt(0)=0, so it is just the integers

[@TheGrayCuber]

𝓞(√0) would just be Z, which has the natural primes and their negatives

[@kales901]

@@TheGrayCuber z???

[@angeldude101]

@@kales901 Z for "Zahlen", German for "numbers". It means the integers.

[@kales901]

@@angeldude101 oh

[@05degrees]

I would actually presume a square root always gets _added,_ so it would be “dual integers” ℤ + εℤ, a subset of dual numbers, where ε² = 0 is a degree-2 nilpotent. The same with adding a square root of another integer which is already a square: I’d expect the result to be a variation on split-complex numbers (adding √1 will give “split-complex integers”). I mean if we define ℤ[√d] as ℤ[x] factored by an ideal containing x² − d, then we surely should get duplicates like that; and ℤ[√0] = ℤ[x] / ⟨x²⟩ _is not_ isomorphic to ℤ, it has a basis of (1 + ℤ[x] x², x + ℤ[x] x²) (or (1, x) for simplicity when we work with arbitrary representatives).

[@jounik]

3:12 As long as a != b√D of course, because then you'd be multiplying with 0/0

[@ManyWaysMA]

If a^2 - b^2 * D = +-1 then a already can’t be bsqrt(D) because 0 != +-1

[@jounik]

@@ManyWaysMA That is true, but it really should be pointed out that if (c-d)(c+d) = ± 1 then _neither_ c-d or c+d can be a zero divisor. It's taken here as assumed.

[@DeadJDona]

-O(√D)*a+O(√D)*b

[@MichaelDarrow-tr1mn]

Here's a hint for your factorization thing: You can check numbers only between the angle 1 is from the origin and the angle [closest non-1 unit with norm 1] is from the origin

[@TheGrayCuber]

Yes, though when I say I checked a billion values, I mean checking each value of a, solving for b, and checking if that b is an integer

[@MichaelDarrow-tr1mn]

@@TheGrayCuber just use the sieve with the new limited area

[@hola-aloh-hola]

First comment