Nice! With primes being extended to the realm of complex numbers, the question arises: can we extend the notion of primes to the realm of quarternions (and octonions)..?
at 12:57 I believe you've left out a case. if we take the composite number 15, which is 3 mod 4 N(15) = 225 = 9 * 25 = N(3)*N(5). Unless an implicit assumption has been made to prevent this (in which case such assumption should be mentioned), this may break the proof by contradiction.
Great video. Recently I read a book about number theory and it covered topics like non-unique factorization domains such as Z[√-5],and it introduced ideals to tackle with the problems in such structures.I found the concept of ideal rather abstract,so I'd be very appreciated if you could visualize them in a video in a similar manner as in this one.
Some things to note at the end of the video: A very significant part of the proof was skipped when you looked at the multiplicative group of units modulo 13 and showed that it is cyclic, in particular generated by 2. But you then at 16:28 just stated that this holds for all primes. It is not at all trivial that these groups are always cyclic for prime moduli. In fact, in this version of the proof, this is the most important part of the whole proof; the rest is basically trivial. To illustrate the importance, note that modulo 8 for example, the mulitplicative group of units (namely, {1+8Z, 3+8Z, 5+8Z, 7+8Z} under multiplication) is not cyclic - it is isomorphic to V4. And so even though it has 4 elements (order divisible by 4), there is no q such that q^2 = -1 as would be required for the Gaussian integer version of the proof. So it is important that we are talking about prime moduli (although it also holds for odd powers of primes among a few other cases), but the proof is not immediately straight-forward. For primes, the proof is a fairly simple group theory exercise: it can be shown from considering the number of elements of each order, and using the fact that Z/pZ is a field. Let G be the multiplicative group of Z/pZ. Then if a is an element of G with order d, let H be the subgroup generated by a, so |H| = d. Also note that all elements of H satisfy x^d = 1, which can be viewed as an equation in Z/pZ. Since this is a field, the polynomial x^d = 1 has at most d roots. Since |H| = d, H must contain *all* of the roots of x^d = 1, and in particular, there are no other order d elements of G other than those in H (any order d elements in G satisfies x^d = 1, and H covers them all by size). Now we recall that a cyclic group of order d has phi(d) elements with order exactly d, and therefore G contains exactly phi(d) elements of order d, since they all lie in H which is cyclic (generated by a). The only d for which there could exist an element a with this order in G are the divisors of |G| by Lagrange's theorem, but recalling that the sum of the Euler totient function phi(d) over all divisors of |G| is exactly |G|, we conclude that there cannot be any divisors missing when we count all elements by order (otherwise the total number of elements would be less than |G|). In particular, |G| is a divisor of |G|, so there must be phi(|G|) elements in G with order |G|, that is, a non-zero amount, so G is cyclic. Secondly, it's worth noting that there is no need for Dirichlet's theorem at all. No part of the proof required q to be prime. The only place we needed to be careful is where we said (q+i)/p and (q-i)/p are not a Gaussian integers, and (q+w)/p and (q-1-w)/p are not Eisenstein integers, but these can be checked by just looking at the non-real parts, which are 1/p in all cases, and clearly not an integer.
Thank you, was trying to figure out why this seemed so simple. It was assumption that the group of p-1 elements is cyclic mod p with a there exists generators that can creating the entire group.
One thing to note, that at around 18:32 after you found a Gaussian number. The number you found does not need to be prime itself, just that one of the factors of that number is also a factor of the prime you are looking for. For example if you use this for the number 13, the q in that case is 8 since 8 *8 = 12 mod 13, so your two factors are 8+i and 8-i. But 8+i = (2-i)(3+2i) and the first is a factor of 5, the second is a factor of 13. Note that 5*13=65 which is also the value of 8*8+1*1 which is the norm.
Very well explained, clear and we feel the hard work and passion you invest into your videos. Really cool to watch, and interesting! Thank you so much for the explanations 💖💖
We're only taking points from the unit circle that are Eisenstein integers. We could represent any complex unit using a + bω, but there are only six points for which a and b are integers
@@TheGrayCuber I think the question was if we can extend this concept to infinitely many units. We would need a definition of divisibility for reals (or at least rationals or something), which seems doable tbh
As a challenge: can you define the Eisenstein numbers (not necessarily integers) without any pre-assumption of complex numbers, and then define complex numbers in terms of Eisenstein numbers?
Nbd but the norm is the square root not the square At 11:44 are we assuming p is a prime number in the integers? If so that's cool but you never stated that so it's a bit confusing. In general switching between gaussian, Eisenstein, and integer primes without clarification was a bit confusing as well
You really should look into cleaning the audio a bit, with an headsets the background noise it's incredibly distracting. Amazing content though! Strange thing: When watching the video at x2 speed every time the animation changes there is a deep rumble, it's quite strange. Probably it's tapping on the laptop trackpad? I cannot hear it at x1 speed but at x2 it's very noticeable.
Why does it feel like a lot of this video is just retreading water that has already been explained in the previous one? There is nothing wrong with referring to a previous video and skipping a bunch of explanations, when it comes to maths videos.
Yeah, I considered skipping some things, but I really wanted this video to work as a standalone so that the viewer doesn't need to have seen the prior video. Especially since this is a SoME entry
Most of this video was a presentation of a proof based on the definitions from the last video, which I thought did a great job of showing the reason for which gausian integers are prime. I thought the proof was very well presented and made to be accessible and clear and added to the content in the last video.
Slight nitpick: Your N(z) is certainly an example of *a* norm, but by no means is it *the* norm. The most common meaning of the word "norm", without further/prior specification, would probably be the euclidean length itself.
@@pl412 I'll grant you it is a fairly commonly used norm, especially in the context of complex numbers, like here. It was more that the way he introduced it made it seem like "norm" means "squared magnitude".
so, for integers a and b, the gaussian integers are of the form a+bi (primitive square root of unity) and the eisenstein integers are of the form a+bω (primitive cube root of unity). do any of the things you can do with these guys generalize to other roots of unity, or does it only work for i and ω? if the latter, what's so special about those ones?
Did I overhear the mentioning of the exceptions? sqrt(2) for Gauss and 2, sqrt(3) for Eisenstein? There are reasons, why they are not 1 (mod 4), 5 (mod 6), and 1 (mod 6), respectively! I didn't expect you to go into details, but I think, you just ignored these facts...
@TheGrayCuber, any plausible relation between how 1 mod 24 entails all even powered primes (>3), and that being the composite of both the Eisenstein and Gaussian prime mod basis?
I suppose there can be a class of primes for any collection of roots of unity. Such as 5th roots, so there would be 5 different units and each complex number in this class would have 5 associates. Granted, it wouldn't form a regular grid like you get with squares or hexagons.
@@tcaDNAp You can certainly tile a hyperbolic plane with heptagons, and you can use 7th roots of unity to define a class of "root-7 primes", but the plane of numbers would remain planar, not hyperbolic. The pattern of the primes would have heptagonal symmetry.
I was wondering why the last video wasn't a #SoMEπ submission because most sequels take months to come out! If the motivation of a pretty graph wasn't enough, the previous video had me hooked too... I'll see you all in the finalists (and part 3)!
Thanks for the time stamp, I wanted to watch that part again to understand it! I figured it out with an inverse example: another circle (mod 7) would have six points and it would not divide into quarters 🍕 But if it is 1 (mod 4), the number of points on the circle is divisible by 4, so we can show that there is a nice number at the top (and bottom) of the circle.
My jaw dropped with the circle of multiplication btw, so I think this is my sign to finish the rest of the cyclotomic polynomials series! I'm excited to go into the next video ready for Dirichlet's Theorem 🤞
Maybe a not entirely relavent question... but looking at the Gaussian Primes, and the norms, is there any simple relation to the L series derived from them? ie the L:k series. Would there be (and 'm guessing there is...) a corresponding series for the Eisenstein integers? This is all way above my head, but I'm trying to get a flavour of what's going on here!
14:25 I came up with a simple proof that a² + b² - ab ≡ 2 mod 3 can never hold. There are 3 cases: If a ≡ 0 mod 3, then b² ≡ 2 mod 3, but this is not possible. If b ≡ 0 mod 3, then a² ≡ 2 mod 3, and this is also not possible. If neither hold, then a² + b² - ab ≡ 1 + 1 - ab ≡ 2 - ab mod 3, so ab ≡ 0 mod 3. This is not possible since in mod 3 arithmetic, this implies a≡0 mod 3 or b≡0 mod 3.
Looking at the axis lines of the Eisenstein numbers and the Gaussian numbers just SCREAMS Linear Transformation. Is there anything useful in using a linear transformation to take a Gaussian Prime and finding its "Eisenstein Equivalent" or vice versa? Or would that just be a fun but useless curiosity?
at 2:19 the definition is not quite correct. the definition on screen (divisible only by 1 and itself) is for irreducible numbers, while prime numbers are defined [if p divides a*b then p divides either a or p divides b]. while in the given number system primes and irreducible numbers are the same set; there exist number systems where irreducible numbers are not prime.
Why aren't these the 6th roots of unity? @2:53 there's 6 numbers raised to the sixth power, all equalling 1. I read the wiki and Wolfram article and apparently it's the cube roots of unity so I think I'm missing something basic here... thanks!
What about 2? It's neither 1 or 3 mod 4, but it isn't a Gaussian prime one one hand, and on the other, it's neither 1 or 5 mod 6, but it is an Eisenstein integer. I know 2 is weird, but it was on the screen for so long I was hoping you'd at least mention it in passing.
In both cases the sieve gave us the answer of whether or not 2 was prime. The purpose of the more general proof was to apply to all primes since we can't do all infinite steps of the sieve. For a more satisfying answer: there is a Gaussian integer 1 + i with a norm of 2, so 2 is not a Gaussian prime. There is no Eisenstein integer with a norm of 2, so 2 is an Eisenstein prime
Me a nuclear engineer and somebody dabbling in Quantum Mechanics wait I love the complex plane. Did you just say complex primes what a way to open a video I’m freaking hooked.
Could you create a set of primes for every integer division of the complex plane (/polygon)? Or just the even ones? I would assume you can use odd integers as this all started with just the positive integers, which divides C by 1
so there are four types of rings 1. Each element has 2 associates, itself and its negative 2. Gaussian 3. Eisenstein 4. Each element has infinite associates. Minkowski ring
There are lots of examples of rings with more than 6 but still finitely many units. For example, in the integers mod n, Z/nZ, an element a is a unit exactly when it is relatively prime to n. So Z/11Z has 10 units, for example.
1. Yes, and a field is a kind of ring, so this is a valid example of a ring with 10 units! 2. If you want a non-field, there are still plenty of examples, eg Z/15Z and Z/16Z have 8 units each
0:40
"Or something similar"
Oh my god, was that a pun 😭
Lmao I missed that completely 😂
I don't get it...
@@mismis3153 Similar triangles
I didn't know about Eisenstein, only about Gaussian primes. Now... I know.
I always wanted to know about both, but I was too scared to read the Wikipedia page! Now... I want to read the sources 😂
Your content is a breath of fresh air
Anyone misread Einstein primes?
Perfect pacing. I like these videos ☺️
Nice! With primes being extended to the realm of complex numbers, the question arises: can we extend the notion of primes to the realm of quarternions (and octonions)..?
Sort of, though things have to change a bit with the loss of commutativity.
@@TheGrayCuber even worse, with octonions even associativity...
at 12:57 I believe you've left out a case.
if we take the composite number 15, which is 3 mod 4
N(15) = 225 = 9 * 25 = N(3)*N(5).
Unless an implicit assumption has been made to prevent this (in which case such assumption should be mentioned), this may break the proof by contradiction.
This is under the assumption that p is a prime. You're correct that the argument would break down if we try to extend it to composites
Great video.
Recently I read a book about number theory and it covered topics like non-unique factorization domains such as Z[√-5],and it introduced ideals to tackle with the problems in such structures.I found the concept of ideal rather abstract,so I'd be very appreciated if you could visualize them in a video in a similar manner as in this one.
i thought i was going to be waiting a lot longer for this
Same
The timing for a sequel has never been more perfect, but now the teaser for part 3 has me ravenous again! 🔥
Some things to note at the end of the video:
A very significant part of the proof was skipped when you looked at the multiplicative group of units modulo 13 and showed that it is cyclic, in particular generated by 2. But you then at 16:28 just stated that this holds for all primes. It is not at all trivial that these groups are always cyclic for prime moduli. In fact, in this version of the proof, this is the most important part of the whole proof; the rest is basically trivial. To illustrate the importance, note that modulo 8 for example, the mulitplicative group of units (namely, {1+8Z, 3+8Z, 5+8Z, 7+8Z} under multiplication) is not cyclic - it is isomorphic to V4. And so even though it has 4 elements (order divisible by 4), there is no q such that q^2 = -1 as would be required for the Gaussian integer version of the proof. So it is important that we are talking about prime moduli (although it also holds for odd powers of primes among a few other cases), but the proof is not immediately straight-forward. For primes, the proof is a fairly simple group theory exercise: it can be shown from considering the number of elements of each order, and using the fact that Z/pZ is a field. Let G be the multiplicative group of Z/pZ. Then if a is an element of G with order d, let H be the subgroup generated by a, so |H| = d. Also note that all elements of H satisfy x^d = 1, which can be viewed as an equation in Z/pZ. Since this is a field, the polynomial x^d = 1 has at most d roots. Since |H| = d, H must contain *all* of the roots of x^d = 1, and in particular, there are no other order d elements of G other than those in H (any order d elements in G satisfies x^d = 1, and H covers them all by size). Now we recall that a cyclic group of order d has phi(d) elements with order exactly d, and therefore G contains exactly phi(d) elements of order d, since they all lie in H which is cyclic (generated by a). The only d for which there could exist an element a with this order in G are the divisors of |G| by Lagrange's theorem, but recalling that the sum of the Euler totient function phi(d) over all divisors of |G| is exactly |G|, we conclude that there cannot be any divisors missing when we count all elements by order (otherwise the total number of elements would be less than |G|). In particular, |G| is a divisor of |G|, so there must be phi(|G|) elements in G with order |G|, that is, a non-zero amount, so G is cyclic.
Secondly, it's worth noting that there is no need for Dirichlet's theorem at all. No part of the proof required q to be prime. The only place we needed to be careful is where we said (q+i)/p and (q-i)/p are not a Gaussian integers, and (q+w)/p and (q-1-w)/p are not Eisenstein integers, but these can be checked by just looking at the non-real parts, which are 1/p in all cases, and clearly not an integer.
Thank you, was trying to figure out why this seemed so simple. It was assumption that the group of p-1 elements is cyclic mod p with a there exists generators that can creating the entire group.
Could there be a pentagonal one and would it be related to the golden ratio?
With luck and more power to you.
hoping for more videos.
Would love to see a video/series on how you built the interactive visualization web app!
One thing to note, that at around 18:32 after you found a Gaussian number. The number you found does not need to be prime itself, just that one of the factors of that number is also a factor of the prime you are looking for. For example if you use this for the number 13, the q in that case is 8 since 8 *8 = 12 mod 13, so your two factors are 8+i and 8-i. But 8+i = (2-i)(3+2i) and the first is a factor of 5, the second is a factor of 13. Note that 5*13=65 which is also the value of 8*8+1*1 which is the norm.
Very well explained, clear and we feel the hard work and passion you invest into your videos.
Really cool to watch, and interesting! Thank you so much for the explanations 💖💖
Great!
Would be really cool to expand the grid to 1000, 10k+ just to see how the pattern would continue. How rare the primes would become on the grid
yes, i would also love to see this
I just pushed an update with a zoom out setting
given this is a cyclotomic extension, this left me wondering why unique factorization fails for cyclotomic extensions for some larger roots of unity
what's stopping us from defining the entire unit circle on the conplex plane as the units that divide the primes?
We're only taking points from the unit circle that are Eisenstein integers. We could represent any complex unit using a + bω, but there are only six points for which a and b are integers
@@TheGrayCuber I think the question was if we can extend this concept to infinitely many units. We would need a definition of divisibility for reals (or at least rationals or something), which seems doable tbh
@@mskiptr That is my question
What's term for the 'series' with 120 degree rotation..
As a challenge: can you define the Eisenstein numbers (not necessarily integers) without any pre-assumption of complex numbers, and then define complex numbers in terms of Eisenstein numbers?
Yes, you can define ω to be a cube root of 1 that does not equal 1. From there you can set i = (1 + 2ω)/√3
To generate some Eisenstein primes:
(.75*(x^2))+(1.5*x)+23 =mostly Eisenstein primes when x is an even number.
Amazing content!
Nbd but the norm is the square root not the square
At 11:44 are we assuming p is a prime number in the integers? If so that's cool but you never stated that so it's a bit confusing. In general switching between gaussian, Eisenstein, and integer primes without clarification was a bit confusing as well
You really should look into cleaning the audio a bit, with an headsets the background noise it's incredibly distracting.
Amazing content though!
Strange thing: When watching the video at x2 speed every time the animation changes there is a deep rumble, it's quite strange.
Probably it's tapping on the laptop trackpad? I cannot hear it at x1 speed but at x2 it's very noticeable.
Why does it feel like a lot of this video is just retreading water that has already been explained in the previous one? There is nothing wrong with referring to a previous video and skipping a bunch of explanations, when it comes to maths videos.
Yeah, I considered skipping some things, but I really wanted this video to work as a standalone so that the viewer doesn't need to have seen the prior video. Especially since this is a SoME entry
Most of this video was a presentation of a proof based on the definitions from the last video, which I thought did a great job of showing the reason for which gausian integers are prime.
I thought the proof was very well presented and made to be accessible and clear and added to the content in the last video.
Lost in a sea of symbols, visuals are great
5:05 aaaaaaaaaaa!!!!11!!!!1!!1!!
(that lack of parentheses was actually painful)
New order of operations rule: colors are implied parentheses.
Slight nitpick: Your N(z) is certainly an example of *a* norm, but by no means is it *the* norm. The most common meaning of the word "norm", without further/prior specification, would probably be the euclidean length itself.
the context is specification enough no? pretty standard
@@pl412 I'll grant you it is a fairly commonly used norm, especially in the context of complex numbers, like here. It was more that the way he introduced it made it seem like "norm" means "squared magnitude".
so, for integers a and b, the gaussian integers are of the form a+bi (primitive square root of unity) and the eisenstein integers are of the form a+bω (primitive cube root of unity). do any of the things you can do with these guys generalize to other roots of unity, or does it only work for i and ω? if the latter, what's so special about those ones?
5:30 What's going on with the missing brackets in your videos? They were already missing in your last video on complex primes.
it's all just notation. colors are brackets now :D (only half joking)
17:17 question: why is it important that q is prime? it seems like the proof would work even if it wasn't
Hah! Yeah I guess it's not important. That is a great point, thank you
Excellent. Well presented.
Did I overhear the mentioning of the exceptions?
sqrt(2) for Gauss and 2, sqrt(3) for Eisenstein?
There are reasons, why they are not 1 (mod 4), 5 (mod 6), and 1 (mod 6), respectively!
I didn't expect you to go into details, but I think, you just ignored these facts...
@TheGrayCuber, any plausible relation between how 1 mod 24 entails all even powered primes (>3), and that being the composite of both the Eisenstein and Gaussian prime mod basis?
How wide is the Eisenstein moat?
I suppose there can be a class of primes for any collection of roots of unity. Such as 5th roots, so there would be 5 different units and each complex number in this class would have 5 associates. Granted, it wouldn't form a regular grid like you get with squares or hexagons.
You beat me to it! I had the exact same thought.
I've been playing too much HyperRogue, so I immediately wondered if you could do it with regular heptagons on the hyperbolic plane!
@@tcaDNAp You can certainly tile a hyperbolic plane with heptagons, and you can use 7th roots of unity to define a class of "root-7 primes", but the plane of numbers would remain planar, not hyperbolic. The pattern of the primes would have heptagonal symmetry.
I was wondering why the last video wasn't a #SoMEπ submission because most sequels take months to come out! If the motivation of a pretty graph wasn't enough, the previous video had me hooked too... I'll see you all in the finalists (and part 3)!
I wonder what Eisenhorn primes would look like. O_O
I didn't quite follow the argument at 16:49, why is there only four different numbers modulo p? Also is q just an integer or is it a Gaussian integer?
There are more than 4, we just only care about those four (and really just the 3 I mention) during the argument. q is a natural number.
Thanks for the time stamp, I wanted to watch that part again to understand it!
I figured it out with an inverse example: another circle (mod 7) would have six points and it would not divide into quarters 🍕
But if it is 1 (mod 4), the number of points on the circle is divisible by 4, so we can show that there is a nice number at the top (and bottom) of the circle.
My jaw dropped with the circle of multiplication btw, so I think this is my sign to finish the rest of the cyclotomic polynomials series! I'm excited to go into the next video ready for Dirichlet's Theorem 🤞
roots of unity
Maybe a not entirely relavent question... but looking at the Gaussian Primes, and the norms, is there any simple relation to the L series derived from them? ie the L:k series. Would there be (and 'm guessing there is...) a corresponding series for the Eisenstein integers?
This is all way above my head, but I'm trying to get a flavour of what's going on here!
14:25 I came up with a simple proof that a² + b² - ab ≡ 2 mod 3 can never hold. There are 3 cases:
If a ≡ 0 mod 3, then b² ≡ 2 mod 3, but this is not possible.
If b ≡ 0 mod 3, then a² ≡ 2 mod 3, and this is also not possible.
If neither hold, then a² + b² - ab ≡ 1 + 1 - ab ≡ 2 - ab mod 3, so ab ≡ 0 mod 3. This is not possible since in mod 3 arithmetic, this implies a≡0 mod 3 or b≡0 mod 3.
6:23 what if there is multiple of 7 outside the list, and one of these numbers isn't prime??
Ive been working with catenary curves embedded in shapes. this would be interesting to add to the lattice
Looking at the axis lines of the Eisenstein numbers and the Gaussian numbers just SCREAMS Linear Transformation. Is there anything useful in using a linear transformation to take a Gaussian Prime and finding its "Eisenstein Equivalent" or vice versa? Or would that just be a fun but useless curiosity?
IIRR for every set of associates of primes, except those of √-3, one associate is congruent to 2 mod √-3. Can you plot those associates?
is there a set of complex primes who's roots of unity are all tau/8 radians apart (8 total)
yes there is!
at 2:19 the definition is not quite correct. the definition on screen (divisible only by 1 and itself) is for irreducible numbers, while prime numbers are defined [if p divides a*b then p divides either a or p divides b]. while in the given number system primes and irreducible numbers are the same set; there exist number systems where irreducible numbers are not prime.
What would such a number set be?
My algebra is rusty. Aren't the definitions equivalent in a PID?
Yes! This is important and I plan to clear up the distinction in my next video where I introduce a ring that is not UFD
we have 2 variables, why do we not use a square grid???
Why aren't these the 6th roots of unity? @2:53 there's 6 numbers raised to the sixth power, all equalling 1. I read the wiki and Wolfram article and apparently it's the cube roots of unity so I think I'm missing something basic here... thanks!
They are the sixth roots of unity! Three of them are the cube roots of unity: 1, w, and -1-w
@@TheGrayCuber Oh, so the 6th roots of unity are just two 3rd roots of unity put together?
@@erawanpencil The 6th roots of unity are the 3rd roots of unity and the negatives of the 3rd roots of unity
What about 2? It's neither 1 or 3 mod 4, but it isn't a Gaussian prime one one hand, and on the other, it's neither 1 or 5 mod 6, but it is an Eisenstein integer. I know 2 is weird, but it was on the screen for so long I was hoping you'd at least mention it in passing.
In both cases the sieve gave us the answer of whether or not 2 was prime. The purpose of the more general proof was to apply to all primes since we can't do all infinite steps of the sieve. For a more satisfying answer: there is a Gaussian integer 1 + i with a norm of 2, so 2 is not a Gaussian prime. There is no Eisenstein integer with a norm of 2, so 2 is an Eisenstein prime
My birthday is a Gaussian prime 🎉 but not an Eisenstein prime 😢
Neat!
underrated!
14:00 - it seems to me you have only showed that p is a prime, not a Gaussian prime. What have I lost?
Your voice from 0:00-1:55 is so relaxing it's like asmr
yeah I intended to do that the whole time but then I got lost in the sauce
@@TheGrayCuberdo a full video of math explained but asmr. It would be nice for complicated topics be soft spoken
Me a nuclear engineer and somebody dabbling in Quantum Mechanics wait I love the complex plane. Did you just say complex primes what a way to open a video I’m freaking hooked.
the duality problem.
Could you create a set of primes for every integer division of the complex plane (/polygon)? Or just the even ones? I would assume you can use odd integers as this all started with just the positive integers, which divides C by 1
I am feeling a deja vu, I like it
This time it feels... hexagonal ;P
You sound a bit like Presh Talwalkar.
"bro thinks he is einstein" became a real thing
I keep on seeing ω and thinking of the first infinite ordinal number
SAME
Wow
🤍
so there are four types of rings
1. Each element has 2 associates, itself and its negative
2. Gaussian
3. Eisenstein
4. Each element has infinite associates. Minkowski ring
There are lots of examples of rings with more than 6 but still finitely many units. For example, in the integers mod n, Z/nZ, an element a is a unit exactly when it is relatively prime to n. So Z/11Z has 10 units, for example.
@@japanada11 z/11z is a field. all the elements in a field are units
1. Yes, and a field is a kind of ring, so this is a valid example of a ring with 10 units!
2. If you want a non-field, there are still plenty of examples, eg Z/15Z and Z/16Z have 8 units each
You can have a ring with an arbitrary number of units, finite or infinite
first!
That makes you the unit!
Congrats! Now, keep studying hard and some day you may learn counting... 😜