You should know this abstract algebra exercise

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[@tcmxiyw]

Herstein was used in my undergraduate Abstract Course (>50 y.a.). In two semesters we covered it almost cover to cover. Over the years, I’ve enjoyed going through it over and over again. My marginal notes provide an interesting glimpse of my undergraduate self. “Topics in Algebra” is so well written that it almost felt that he was co-teaching the course.

[@piaoyugexia]

XD and the textbook is still used today, right now I’m my Algebra class.

[@gucker]

Nice theorem! I always wondered how can someone come up with such an artificial proof :) Thank you for this video!

[@rainerausdemspring894]

A surprising generalisation: Let R be a ring such that any r in R has a natural number r(n) > 1 *depending on r* such that r^r(n) = r, then R is commutative. However, to prove this is really hard.

[@justanotherman1114]

Do you know any reference for this fact?

[@toddtrimble2555]

@@justanotherman1114 Herstein proved this and much more. See section 2 of core.ac.uk/download/pdf/82130311.pdf, first paragraph, for a very general result along these lines.

[@justanotherman1114]

@@toddtrimble2555 Thanks

[@tarikeld11]

Are you sure you didn't mean "for any *n* in R [...] depending on *n*?

[@toddtrimble2555]

@@tarikeld11 He said it correctly. The statement is [(for all r in R) (there exists n in N) n > 1 and r^n = r] => [R is commutative].

[@JaybeePenaflor]

I remember this problem! This appeared in one of my problem sets for my post-graduate Abstract Algebra subject!

[@MrRyanroberson1]

11:00 8 r = 4r^3 + 4r^3 = 2r (4r^2) = 2r (2r (2r)) = (2r)^3, alright that checks out. just had to be rigorous, even if only loosely.

[@enzogiannotta]

Fact: it is a theorem that every ring R with unit such that for every element x there exist a natural numbert n>1 such that x^n=x, then R is comutative. This comes from the book: "A First Course in Noncommutative Rings", Chapter 4, §12, in particular the Jacobson-Herstein Theorem (12.9), p. 209: A (unitary, associative) ring R is commutative iff for any a,b∈R one always has (ab−ba)n+1=ab−ba for some n∈N (n generally depending on a,b).

[@russellsharpe288]

What do you mean? If R is commutative then for any such a, b, ab-ba = 0, so how can any (ab-ba)n+1 = ab-ba? The LHS is 1 and the RHS is 0.

[@williamturner8257]

@@russellsharpe288 he meant (ab-ba)^(n+1)=ab-ba.

[@russellsharpe288]

@@williamturner8257 Ah, OK, thanks. I did wonder about that but thought writing n+1 for some n∈N when he could have written n for some n>1 would be a bit odd. In any case in my experience N usually means {0,1,2...}, not {1,2,3..}

[@Luxaray2000]

4:53 That was some sick chalk-fu!

[@goodplacetostop2973]

0:10 So I guess that Herstein was… the lord of the rings 😎 19:46 Good Place To Stop

[@johnunbehaun6638]

Nice! Tried this problem back in the 60’s.

[@arekkrolak6320]

really cool proof and also fun to watch!

[@atonaltensor]

A really interesting elementary proof, but I’m have a hard time figuring out what to take away from it. Is there a general principle at work here? Why did the proof work? And can we generalize

[@panadrame3928]

Well i'd say this proof works well bcs 3 is rrlatively small number, like it's only 2 away from 1 so you can reduce everything to self or square elements and then you play around with it. to generalize it is impossible since it deeply comes from the number of orders an element can get to, but maybe you can find some interesting properties with 4 or 5 (but with a lot lot more work)

[@lazbn90]

Interesting question. IMO, it's absolute abstract nonsense. Some proofs just work and people in math don't stop to think about why they do. Especially in Abstract Algebra, you will find tons of exercises like this one, and non of them seem related, as they used different algebraic manipulations. There is no way to tell, if the exercise is non-trivial enough, which identities and operations will lead to a solution.

[@MK-13337]

@@panadrame3928 Any ring R in which for all r in R there exists an n (may be depending on r) such that r^n = r is commutative. So any ring where r^4 = r for all r is commutative, any ring where r^5 = r is commutative, and those are just trivial-ish examples of that generalization.

[@JamesLewis2]

Of note is that Herstein does not require associativity in the definition of a ring in Topics in Algebra, so most of the interesting properties are stated for "associative rings"; I had figured that his definition was like that so he could later say that an algebra is a module that is also a ring, but he doesn't quite say that (instead he uses a slightly more restrictive definition, a *vector space* that is also a ring, in which scaling is compatible with vector-multiplication).

[@EquuleusPictor]

Rings are wild. Another good one: if a ring R is also invertible wrt multiplication and finite, then it's commutative. The proof is very hard and non-elementary.

[@norbi275275]

Love your channel, keep up the good work

[@bappaichotu]

Enjoyed after struggling with herstein 30 years ago.

[@timelsen2236]

Note 6r=0 means mode 6, then verify r^3=r mode 6 and 3(r^2=r)=0 as well, mode 6. Funny 6 wasn't mentioned. Really liked the proof. Perhaps 6 would have trivialized the great proof. Showing the ring is mode 6, would cause many to quickly loose interest. I want further application to something like what would usually be non-commutative like matrices, say becoming commutative for the scalar ring mode 6. Either try that or suggest another possibility.

[@ojas3464]

👍The same author in his Noncommutative rings has an exercise, if for all a in a Ring there exists a b (depending on a) such that aba = b, then Ring is commutative. I maybe missing some details, but is that exercise is already answered by parts of this video? Thanks☺

[@JM-us3fr]

What a simple condition! That’s quite remarkable.

[@HershO.]

Even as a sophomore in HS (admittedly a bit ahead of my curriculum), this was pretty comprehendible! You're a great teacher.

[@brian8507]

You need to give up on math and smoke weed and skip school. Do not goto college. I need less competition

[@HershO.]

@@brian8507 as a musician on the side this is a viable career option lmao.

[@LIVIU2003S]

Problem suggestions, similar to this one: R is considered a ring as in your video with the addition that it has a multiplicative identity. a) Prove R is commutative if r^12=r b) Find R if r^4=1 for any r != 0 c) 6r=0 implies r=0; a-b,b-c,c-a are idempotent. Prove that a=b=c

[@ultimatedude5686]

Doesn't b have several solutions? i.e. Integers modulo 2 - 6, 8, 10, 12

[@LIVIU2003S]

@@ultimatedude5686 2^4=16 which is not 1 modulo 6, 8, 10 or 12. But yes, it does admit multiple solutions

[@ultimatedude5686]

@@LIVIU2003S Thanks for correcting me. The only correct solutions in my answer were the integers mod 2, 3, and 5. I suppose the answers I gave would only be correct if we assumed that r wasn't a zero divisor, but that was not the question. I am curious how many solutions there are. There must be at least four: the three I listed above and the zero ring.

[@LIVIU2003S]

@@ultimatedude5686 Those are in fact the only solutions (isomorphism wise). The trick here is that R is actually a field. r^4=1 implies r*r^3=r^3*r=1, so r always has an inverse (r^3) and since R is a ring, we get R is a field.

[@alexandersanchez9138]

@@LIVIU2003S This can't be? The quaternions are a ring for which, e.g., i^4 = 1 and i != 0. However, the quaternions aren't a field.

[@winniedobrokot]

Beautiful fact! But how to solve this or similar exercises without prior memorization every step and lemma, specific to this exercise? Can this fact to be used as tool for rings where commutativity is not evident? The beauty of this fact fades, because I don’t see how I can use either the result either the proof techniques

[@drjohnsmith5282]

A fascinating exercise. Might have been better structured if those two Lemmas had been included as 'part a and part b' before the commutativeity proof was part c. I often try and do these exercises before watching Michael, and without those Lemmas there was no foundation. Still, a great video. Learned a lot.

[@drdca8263]

The first proof I thought of for lemma 1 assumed that R was a unital ring, and I was pleasantly surprised when the proof shown did not assume R to be unital (I.e. did not assume that R has a “1”)

[@cannot-handle-handles]

Ah, good point! I wondered why we couldn't just expand (1+r)^3, now I see why.

[@matthewpugh5965]

For the second lemma is it valid to prove it from the first lemma by substituting the multiplicative identity? Like: L1: 0=3(r+r^2) sub 1: 0=3(e+e^2)=3(e+e)=6e 0•r = 6e•r 0 = 6r

[@stighemmer]

What multiplicative identity? We haven't established that it exists.

[@matthewpugh5965]

@@stighemmer Don’t rings have multiplicative identity by axiom?

[@stighemmer]

@@matthewpugh5965 No, they do not. For example, the even integers (2Z) form a ring.

[@matthewpugh5965]

@@stighemmer Huh, I didn't realise that, and the definition in Herstein agrees with you. Thanks for your help. Reading around, it seems like there is a debate about whether the definition should require a multiplicative identity or not. I guess I can't just assume it does in future.

[@MuffinsAPlenty]

@@stighemmer Most ring theorists nowadays require rings to have a 1 element. But for the context of this problem, we shouldn't assume a multiplicative identity because the text it is from does not assume it.

[@BethKjos]

@8:00 now wait a minute. How do you deal with multiplication in a ring composed of pairs? I'm assuming FOIL applies, and I know how to deal with Z, but those D components --- They only seem to have one well-defined operation. Maybe D*Z is the same as D+D+D+... Z times, and maybe D*D is the same as a sequence operation, but that leaves D+D to be defined.

[@adamkapilow]

If you're talking about the group ring he mentioned in the beginning of the video, Wikipedia is your friend here, check out en.wikipedia.org/wiki/Group_ring. If you're talking about just how you reconcile the notion of the integers living in an arbitrary ring, then there's a different story. Namely for any ring R we by assumption have elements called 0 and 1, where 0 acts as additive identity (0 + r = r + 0 = r) and 1 acts as multiplicative identity (1*r = r*1 = r). Then with these two special elements, we get a function f : Z --> R defined by f(n) = 1 + 1 + ... + 1 adding 1 to itself n times for positive n. And furthermore there is an element -1 such that -1 + 1 = 1 + -1 = 0 by definition of a ring. Then we can define f(-n) = (-1) + (-1) + ... + (-1) adding this element to itself n times. This function f has three important properties. First is that f1) = 1. Second is that f(n + m) = f(n) + f(m). And third is that f(n*m) = f(n)*f(m). It turns out that f is the unique function Z --> R satisfying these three properties. This shows that some version of the integers lives inside of every single ring, and so any arithmetic facts that are true in Z remain true for the version of Z inside R, as f preserves arithmetic. New facts arithmetic equalities could hold inside R, but we never lose arithmetic in Z. For instance, in the ring in the main content of this video, we get 6r = 0 for all r, and taking r = 1 we get 6 = 0 in this ring, an equality that holds here but not in Z. But equalities like 25 = 2*10 + 5 will remain true in any ring, as if we apply f on both sides we get f(25) = f(2*10 + 5) = f(2*10) + f(5) = f(2)*f(10) + f(5). Those three properties are sensible properties to ask of any function between two rings g : R --> S, and any function satisfying these properties is called a ring homomorphism (meaning same form, or same shape), and is our main method of comparing two rings, as indeed they show that any arithmetic relations that hold in R are preserved by g, becoming arithmetic relations in S. EDIT: I'm noticing that Michael doesn't require rings to have an element 1. This is not my personal preference, but you can still apply the above reasoning in that case. To do this, you start with a ring R, not necessarily having a 1, and form the endomorphism ring of R, denoted End(R). This is the ring of functions g : R --> R which are abelian group homomorphisms, namely we just require that g(r + s) = g(r) + g(s). This is a ring where addition of endomorphisms is defined pointwise, that is g + h is the endomorphism g + h(r) = g(r) + h(r). Multiplication is composition of endomorphisms, so that g*h (r) = g(h(r)). This ring has a unit, namely the identity endomorphism, id : R --> R given by id(r) = r. This all works for the same reason that matrix rings are rings, as matrix rings are almost exactly the same thing. So the above reasoning gives us a homomorphism f : Z --> End(R). Feel free to think about how the maps Z --> R and Z --> End(R) are related in the case where R has a unit.

[@BethKjos]

@@adamkapilow I think the point is you lost me at "group ring". Very cursory look suggests I'd need another four-year degree to follow the article. It looks like we treat the group-elements as elements of a basis vector and the ring-elements as vector coefficients, but that's confusing because dihedral-group elements (rotations and flips) are not linearly independent like we normally ask of a basis vector. And I thought matrices formed a ring because you can add or multiply them sensibly and they have zero and identity elements.

[@dannycrytser7268]

​@@BethKjos In the group ring you don't view the group elements as matrices even if they are usually presented as such. Instead you think of them as abstract elements satisfying the equations of the group law. When forming the group ring, think of the group elements as variables from which you form various polynomial expressions, with the rule that you impose any (multiplicative) equations in the group on the corresponding variables. The multiplicative law will always enable you to simplify these polynomial expressions down into "linear" expressions in which each variable appears with exponent 1, possibly with zero coefficient. The additive structure ignores any relations in the group. This is the reason for the bracket notation: Z[D_3] is meant to parallel notation like Z[x,y], the polynomial ring in two variables, and in that ring there are no additive relations between x and y. In the dihedral group D_3 you have flip s and rotation r. So there are 6 variables in the group ring Z[D_3]: e (=1), r, r^2, s, rs, r^2s. Using integers as coefficients, you can add any combination of these variables together, and you simply add the coefficients. E.g.: (rs + 2r^2s - e) + (4s + 5 rs + e) = 4s + 2r^2s + 6rs. The additive structure is deliberately insulated from any additive structure on any matrix representation of the group. When multiplying you just use distribution and then simplify using the group laws. In D3, you have r^2s = sr and s^2= e, and r^3=e. So (r+s)(e + 3r^2s) =r + s + 3r^3*s + 3sr^2*s = r + 4s + 3rs^2 = r + 4s + 3r = 4r + 4s. (I've only used positive coefficients but there's no restriction against negative coefficients) The confusion with considering the "built-in" additive structure of a group of matrices is that there are potentially more relations among matrices than exist in the group itself. For example, if you have matrices [[1,0],[0,1]],[[1,0],[0,-1]], [[-1,0],[0,1]],[[-1,0],[0,-1]] you have a representation of the "abstract" group with four elements {e,a,b,ab} with a^2=b^2=(ab)^2=e. However, these matrices satisfy linear equations (the sum of the second and third is 0, for example), which would not exist in the group ring.

[@marcin6987]

Can you explain why this exercise became so famus?

[@senaklnc1494]

Thanks! Great exercise

[@ethanbartiromo2888]

Well (2^3)(r^3) cannot technically be combined into (2r)^3 in the way you did it because we don’t know if it’s commutative yet, but either way it gives you the same answer, but technically the logic was circular

[@ethanbartiromo2888]

You actually use the same method I was thinking later in the proof for the a’s and b’s

[@VideoFusco]

Since the ring R is commutative and furthermore each of its elements multiplied by 6 (that is, added to itself six times) is equal to 0, could we say that R is isomorphic to Z_6?

[@LIVIU2003S]

Nope, think about 2r = 0 => 6r = 0. What that means the characteristic of the ring is 6 (by the same reasoning, actually a divisor of 6).

[@Bodyknock]

13:19 I just noticed that strictly working from the definition for a possibly non-commutative ring it’s not immediately obvious that a (-b) = -(ab) (i.e. it hasn’t been shown a times an additive inverse of b is the additive inverse of ab). That seems like you would need this as an additional lemma in the video.

[@schweinmachtbree1013]

you're right that it isn't immediately obvious but it is well-known (it is one of the basic consequences of the ring axioms). you can prove it by using the ring axioms to show that ab + a(-b) = 0 and a(-b) + ab = 0

[@jannesl9128]

I guess the proof is just: ab +a(-b)=a(b+(-b))=a*0=0. So I think it is obvious enough ^^ a(-b)+ab is the same spiel

[@schweinmachtbree1013]

​@@jannesl9128 a*0=0 needs to be proved as a little lemma but yes, not that bad

[@jorex6816]

@@schweinmachtbree1013 wie macht das Schwein? lel

[@juyifan7933]

Its a corolary of distributivity.

[@azharlatif6228]

Every High School student of elementary Algebra must have asked question arising at the back at their minds, to be dismissed as absurd. This conundrum can only be answered by visiting Abstract Algebra, applying Euclidian Algorithm. Prof. Michael. Penn does it with panache. Stay Blessed Prof. Penn.

[@MyOneFiftiethOfADollar]

Neat that the rule taken for granted in College Algebra, (A^n)(B^n) = (AB)^n would not hold in a ring that is not commutative.

[@schweinmachtbree1013]

yes and no. (AB)^n = (A^n)(B^n) doesn't necessarily hold in a non-commutative ring, but (k · A)^n = k^n · A^n does hold, where m · C is defined as C+C+...+C (m times)

[@Joald]

I thought the natural numbers were just notation shorthands for repeated addition. Yet at 10:45, we're treating them as regular numbers, taking away the cube from the 2. This feels like it needs way more justification.

[@reeeeeplease1178]

(2r)^3 = (r+r)^3 = r^3 + 3r^3 + 3r^3 + r^3 = 8r^3 Using the binomial theorem

[@JM-us3fr]

Not _way_ more justification, but just a simple exercise. Essentially, every ring is a Z module.

[@schweinmachtbree1013]

@@JM-us3fr it is every abelian group that is a *Z* -module. every ring is a *Z* -algebra.

[@schweinmachtbree1013]

@@reeeeeplease1178 It is important to note that the binomial theorem is only valid here because r and r (obviously) commute. The binomial expansion of (a+b)^n holds for all n if and only if a and b commute.

[@JM-us3fr]

@@schweinmachtbree1013 I think you’re mistaken. Z mod n is clearly not a Z-algebra. Edit: We concluded that I was mistaken

[@samosavaglio2141]

Another problem from the Herstein: Let G be a finte group s.t. 3 does not divide it's order, and more we have that for every g and h elements of G (gh)^3=g^3h^3. Prove that G is abelian..

[@khaledchatah3425]

reference? please

[@mrminer071166]

Would have more pedagogic value if you could show A RING with that property that IS commutative, and a ring WITHOUT the property which was NON-commutative!

[@Gretchaninov]

Wow. Super complicated. I usually try to do these problems in my head, but this one was too difficult. I feel vindicated looking at the solution now. I tried (a+b)^3 and a few variants. I also saw that 6 had to divide the characteristic. But never got that far. The question is: why does it work like this? I feel like the solution given doesn't give much insight into the nature of such rings.

[@rmandra]

Thanks!

[@kquat7899]

Great stuff.

[@pawebielinski4903]

Lemma 2 is my favorite.

[@omgopet]

How can non-square matrices form a ring when multiplication is not even well defined? Please explain to me like I'm an engineer, I suck at group theory... edit: I understand all of the steps, but I have no clue where you got the idea to calculate lemmas 1&2. What was the logic, how did you know that you would need those expressions to finish the proof?

[@JohnDoe-ti2np]

I think one doesn't know in advance that lemmas 1 and 2 will finish the proof. The general strategy is to try to derive auxiliary identities by taking the identity r^3 = r and replacing "r" with simple polynomials. Start with polynomials in one variable and tabulate any useful-looking identities that emerge. Then move on to the simplest polynomials in two variables. Try to achieve as much cancellation as possible by looking for common expressions and subtracting them. This trial-and-error approach isn't guaranteed to always work, but luckily it does work here.

[@Khaim.m]

10:10 Can't you also show that (r + r²) = 0 for all nonzero, non-one r? Let r + r² = x r(r + r²) = rx r² + r³ = r² + r = x = rx Therefore x must be 0 or r must be 1.

[@arnedecadt8975]

No it doesn't follow directly because of be zero divisors, in Z_10 for example r=6 and x=6+6^2=42=2 gives x=r x and (r-1)x=10=0

[@teeweezeven]

In the proof of lemma 2, it is stated that 2³r³ = (2r)³. This is true, but may need more consideration as it isn't true in a general ring that a³b³=(ab)³. (Higher mathematics would maybe say that rings are modules over the integers but you can also work out (r+r)³ by cross multiplying)

[@schweinmachtbree1013]

a³b³ ≠ (ab)³ doesn't have anything to do with 2³r³ = (2r)³ because the 2 is not an element of the ring. If the ring has a multiplicative identity 1_R (which in this video is not assumed) _then_ 2 can be thought of as an element of the ring, and we have 2³r³ = (2r)³ because 2 := 1_R + 1_R commutes with r (2, and more generally n := 1_R + ... + 1_R, is in the center of any unital ring). In the non-unital case (this video) it is good to denote integer multiples of elements by n · r instead of nr, and then we have 2³ · r³ = (2 · r)³ and in general n^k · r^k = (n·r)^k. As you say, integer multiples fit into the context of *Z* -modules - specifically, an abelian group is a *Z* -module in a natural way, while a (not necessarily unital) ring is a *Z* -algebra in a natural way. Some other natural module structures are that an abelian group of exponent n (that is, an abelian group in which every element has order at most n) is naturally a *Z* /n *Z* -module (so in particular an abelian group of prime exponent p is naturally a *Z* /p *Z* -vectorspace), and a ring of characteristic n is naturally a *Z* /n *Z* -algebra.

[@iabervon]

(2r)^3 means (r+r)^3 (since 2 isn't being used here as a ring element), and (r+r)^3 is 8 terms that are all r^3 because associativity implies that powers of r commute.

[@jeffreyblack666]

Doesn't your proof of lemma 2 require commutivity, as you go from 2*2*2*r*r*r to 2*r*2*r*2*r?

[@MuffinsAPlenty]

There is a little bit more work to be done for lemma 2, but it's not a problem. The idea is that "integers" in a ring are defined as n = 1_R+1_R+...+1_R (n terms) [at least, if n is nonnegative]. From this definition, the commutativity of 1_R, and the distributive properties, you can prove that "integers" commute with everything in the ring under multiplication.

[@MuffinsAPlenty]

Oops! I should be a bit more careful here, since we're not assuming R has a 1 element in this problem. In that case, the work is slightly more difficult, but not terrible, in that, given an element r of R and a nonnegative integer n, we would define nr as r+r+...+r (n terms). From this, you can prove that nr*mr = (nm)r^2, similarly using the distributive properties and the fact that r commutes with itself.

[@philstubblefield]

Wow, déjà vu! I used the Herstein "Topics in Algebra" textbook nearly 40 years ago, so it's very strange to hear it still cited!

[@ConManAU]

Is there an example of a non-trivial ring where the r=r^3 property holds?

[@ichtusvis]

Z/3Z: the ring of integers modulo 3

[@mguzjebesku2591]

Integers modulo 3.

[@reeeeeplease1178]

As the others pointed out, Z/3. As to why: a^n == a (mod n) So we even know that in Z/n we have r^n = r (for all n)

[@kristianwichmann9996]

@@ichtusvis I would say that's a trivial example. Are there others?

[@schweinmachtbree1013]

@@reeeeeplease1178 This is not true; that only holds when n is prime, which is Fermat's little theorem. For non-prime n the generalization is Euler's theorem, which implies that a^{ϕ(n)+1} == a (mod n) for all integers a that are coprime to n. but this still doesn't say that a^m = a for all a in *Z* /n where m = ϕ(n)+1; it only says that a^m = a when a is coprime to n, i.e. when a is a unit, that is a ϵ ( *Z* /n)^× ⊂ *Z* /n. In *Z* /n it is possible that r^k ≠ r for all k > 1 - for example in *Z* /4 we have 2^2 = 4 = 0 so 2^k = 0 ≠ 2 for all k > 1.

[@f5673-t1h]

The first lemma can be done like so (rings are defined to have a multiplicative identity 1): r+1 = (r+1)^3 =r^3 + 3(r^2+r) + 1 = (r+1) + 3(r^2+r). Compare both sides and 3(r^2+r) = 0.

[@schweinmachtbree1013]

yes but the lemma in the video is more general because it doesn't assume a multiplicative identity, and the proof of it is just an adaptation of the proof in the unital case (that you've just given) to the non-unital case; instead of examining 1 + r = (1 + r)^3 one examines r + r^2 = (r + r^2)^3

[@skallos_]

Some books define rings as not having identity by default. Others use the terminology Rngs (no i since no identity). The problem is more general in that it is true for rings without identity.

[@kwichmann2777]

Rings aren't always defined as having a multiplicative identity. Specifically, this was not done here.

[@MichaelPennMath]

I take the general definition of a ring to not have a multiplicative identity. I would call a ring with multiplicative identity a "ring with 1" or a "ring with unity".

[@ConManAU]

I like the notation where a ring with a multiplicative identity is a r1ng and one without is a rng, but that’s probably just me.

[@tcmxiyw]

It sounds like you have a microphone behind the board.

[@林進生-k5l]

nice exercise if we want to calculate sin[(10/13)°] we will try to find some n and define r satisfy r^(n) = r

[@Sup3rdud4]

It seems like if you had any odd power of r being equal to r you will always have a commutative ring. Is this a correct intuition?

[@schweinmachtbree1013]

more is true: for any n ≥ 2, if r^n = r for all r in _R_ then _R_ is commutative (and in fact even more is true, as other commenters have mentioned)

[@desco8121]

Very nice video, but i'm not quite sure if the passage at 10:54 is correctly justified how you did it, because you defined nr=r+...+r n times, so i'm not sure if it is straightforward to claim then that 2³r³=(2r)³, because the first one is adding r to itself 8 times while the second is taking the cube of (2r). Anyway, the equality is true and it can be shown by expanding (2r)³=(r+r)³=...=r³+...+r³=8r³. It could also just be that i didn't get how you thought of that passage, or maybe you did the same thing i did just not going deeply in what it means, but anyway it's true and the rest of the video is amazing as always.

[@sbares]

It works out because r commutes with itself

[@ChristopherEvenstar]

Fantastic!

[@willemesterhuyse2547]

He didn't replace all instances of r with r cubed i.e. RS = f(r^3) was not used (timestamp 11:10). Isn't this invalid?

[@willemesterhuyse2547]

It's invalid, notice: 8r - 2r != 2r - 2r for r an element of integers.

[@theartisticactuary]

Set is as good a name as anything for a set. Group is as good a name as anything for a group. Although it's easy for me to envisage a world where groups are called sets and sets are called groups. Vector space is a pretty good name for a vector space: I can see where they got that from. But why are rings called rings? Is it just because integers mod n fit the definition and look like rings?

[@russellsharpe288]

wikipedia says "The term "Zahlring" (number ring) was coined by David Hilbert in 1892 and published in 1897. In 19th century German, the word "Ring" could mean "association", which is still used today in English in a limited sense (for example, spy ring), so if that were the etymology then it would be similar to the way "group" entered mathematics by being a non-technical word for "collection of related things". According to Harvey Cohn, Hilbert used the term for a ring that had the property of "circling directly back" to an element of itself (in the sense of an equivalence). Specifically, in a ring of algebraic integers, all high powers of an algebraic integer can be written as an integral combination of a fixed set of lower powers, and thus the powers "cycle back". For instance, if a3 − 4a + 1 = 0 then a3 = 4a − 1, a4 = 4a2 − a, a5 = −a2 + 16a − 4, a6 = 16a2 − 8a + 1, a7 = −8a2 + 65a − 16, and so on; in general, an is going to be an integral linear combination of 1, a, and a2." You might ask the same question about fields. Again wikipedia has the answer: "In 1871 Richard Dedekind introduced, for a set of real or complex numbers that is closed under the four arithmetic operations, the German word Körper, which means "body" or "corpus" (to suggest an organically closed entity). The English term "field" was introduced by Moore (1893)." The fact that it was German mathematicians who developed the abstract theory of fields and used the word Körper for them is doubtless why in field theory nowadays field are still generally called K,L,M... (rather than F, G, H...)

[@HuntBobo]

What’s an integral domaine ? An integratable domain?

[@byronwatkins2565]

There seems to be an order of algebraic operations preference between the two operators that is not well-defined. Does the non-commutative operator ALWAYS take precedence over the commutative operator?

[@masonhart2908]

Notation for ring operations follows PEMDAS.

[@gcewing]

Precedence is just a matter of notation, but the usual convention is that we use multiplication-like notation for the operation that distributes over the other one, for which we use addition-like notation.

[@stighemmer]

As long as you use + and •, it is assumed that they follow the ordinary rules like "ab" means "a • b" and the normal order of operation applies. Once an author starts using other symbols for their operations, they had better make clear how they work together.

[@tomholroyd7519]

Quite nice! a^3 is interesting. In Z2/(x^2+x+1) = GF(4) (which you did in an earlier video) a^3 is the validity function, the modal operator "possible". Of course there are 4 elements so a^4 = a. r^3 = r probably implies r

[@drdca8263]

Huh? The “possible” operator acts on propositions. What would it mean to apply it to an element of a field?

[@charlesgantz5865]

The originator of Ring theory, if you can say that anyone originated a branch of math, was Emmy Noether, most famous nowadays for Noether's theorem in physics. Also, David Hilbert, who I'm sure everyone on your channel is familiar with, said Rings are not like a wedding ring. Instead he likened it to a ring of thieves.

[@MuffinsAPlenty]

I have heard people suggest "ring of thieves" explanation before, but I had been under the impression that it was not explicitly stated, but merely a hypothesis people came up with after the fact. Do you have a reference that Hilbert actually named a ring in the same way as "ring of thieves"? I would love to see that reference if you have one!

[@JohnVKaravitis]

5:46 "coefficients" of a matrix? You meant "elements", no?

[@MuffinsAPlenty]

I would say entries.

[@abraxas4947]

Awesome proof

[@ronflypotato4242]

Where can i find more videos about academic math subjects either just learning or solving standard questions or even harder questions about those subjects

[@chritophergaafele8922]

Maybe at 2:37 you should have said (R, •) instead

[@VictorAgababov]

So, given the definition, that was given in the beginning, ie without the rule that multiplication by 0 yields 0 - how do we assume that (in previous to last step) multiplying left by b (or a) yields 0 on the right?

[@schweinmachtbree1013]

It follows from the definition: we have 0x = (0+0)x = 0x + 0x so cancelling (subtracting) 0x on both sides gives 0 = 0x (and the same argument works when 0 is on the right). Interestingly we need subtraction to prove 0x = 0; if one considers "rings without subtraction", which are called semirings, then 0x = 0 is not provable so it needs to be added to the axioms.

[@SM321_]

Make some video discussiong some really deep stuff :))

[@inf0phreak]

Another nice one: If every commutator of a ring is idempotent, then the ring is commutative.

[@ow7398]

Great video, but what actually is a 'ring' in abstract algebra and how does it differ from a field?

[@russellsharpe288]

He actually defines a ring at the very beginning.

[@ConManAU]

Fields are rings where the multiplication operation is invertible except when zero is involved. So Z_5 is a field since every value from 1 to 4 has a multiplicative inverse, but Z_6 is not a field since 2x = 1 has no solution mod 6.

[@ow7398]

So he does! Thanks, I somehow interpreted that bit as properties of rings rather than defining features (despite it saying defn)

[@thomashamilton564]

Great!

[@stipendi2511]

I don't understand lemma 2. How do you show that 8r = 2r * 2r * 2r ?

[@bethhentges]

Because r commutes with itself w.r.t. multiplication (even if two distinct element do not).

[@stipendi2511]

@@bethhentges Wow, I can't believe I didn't think of that! Thank you so much

[@thomasoa]

Easier to prove lemma 1 with (1+r)^3=1+r=1+r^3

[@jimiwills]

Nice

[@Macieks300]

how would you come up with these two lemmas or the whole proof for that matter?

[@bethhentges]

By playing with the defining property of R and what is means to be commutative.

[@timelsen2236]

Super Like!

[@freddyfozzyfilms2688]

8:54 doesn't using the binomial theorem assume commutivity?

[@freddyfozzyfilms2688]

nvm didnt know that addition is always commutative

[@JosBergervoet]

Why is 3-2 equal to 1? (used at kzbin.info/www/bejne/eYesf6dubpKIias ) As long as the only things we know are the ring properties, how does that follow?

[@schweinmachtbree1013]

it's not saying that the number 3 minus the number 2 equals the number 1; it's saying that 3 copies of an element minus 2 copies of that element is 1 copy of that element

[@reeeeeplease1178]

He's just using n*r as a shorthand for r + r + ... + r (n times), which is why the 8r^3 = (2r)^3 step *could* lead to problems bit it works out fine

[@JosBergervoet]

So it is using the natural number properties. (Solving problems often requires using more knowledge than stated in the problem! Like using complex number properties when contour integrals are used to solve innocent-looking real number integrals.)

[@TheAzwxecrv]

In a ring R with elements q, s and r, we have (qs)r = q(sr), and call this "rule" P. Then, can't we proceed as follows: r^3 = r•r•r =r. Then by rule P, we get (r•r)•r = r•(r•r). Call r•r as s, and we get s•r = r•s, which proves commutativity.

[@lutingchen8958]

This only proves s.r = r.s when s = r.r, not general commutativity for any s and r

[@BrollyyLSSJ]

You just proved that for all r in R there exists s in R such that s•r = r•s, while for commutativity you need arbitrary s in R. As an aside - this s is not even necessarily different from r - consider trivial ring R = {0} for example.

[@thejaunecnaguy5113]

I tried to do it on paper on my own before watching the solution and I naturally looked at (a +- b)^3, and if I hang on for a little longer I probably would've added these two quantities However, I don't see how the two lemmas come up, they seem to appear out of nowhere and the way we use them isn't natural to me (I'm like "yes it works, but... Whah...? Why would you think about doing that...?") Are those a consequence of some theory or just gigabrain algebraic manipulations/results of multiple tries ? Do you know a way to explain how these two results come to one's mind ?

[@juyifan7933]

The order of solution was changed, if you were to solve this problem you would first get to 2(ab-ba)=0 and in trying to conclude, probably after a few hours trying different expressions and uses of the r^3=r hypothesis you would come up with something like those lemmas. In general you can only come up with that stuff after hours trying everything else. Anyway, thats how real math problems are, they require exploration, stuff that you can just straightforwardly solve are not problems, they are exercises.

[@khaledchatah3425]

@@juyifan7933 this exactly what happened when i tried to solve it. i took me so much time. I should point out u can also get different results or come up with different lemmas based on what u ended up with form ur starting point, his starting point was (a+b)^3 and (a-b)^3 but yes its always the same thing. Time will be spent very fast when trying to start from nowhere

[@dmytrolevin738]

This proof gives me flashbacks from my abstract algebra and functional analysis courses - we did lots of such exercises, where the proof technically works but is completely artificial. It looks like it was written back to front. How did those lemmas appear? How did all those "let's take..." steps appear? The proof doesn't show you that. And I hated such proofs, because they are kinda useless and does not teach you anything. It often looked to me that my professor just remembers these proofs and reproduces them mechanically, cause he teaches the same thing every year.

[@pcaraoulanis]

You have to know this Here's the def it starts with...

[@zazinjozaza6193]

*sol ring thumbnail* *sol ring thumbnail*

[@scowell]

Excellent explanation! I have a math minor, and never really understood the purpose of rings... it's like, let's create a crazy set of numbers with some condition and see what happens. The condition you choose forms the ring. What do the members of this set look like? Does the set of integers mod 3 fulfill the conditions?

[@adamnevraumont4027]

Like most of abstract algebra, rings are a paring away of the properties of numbers; what do we know if we know less? Here, we have sensible commuting + and distributing * and typically + and * identities. You can start with any ring and "mod out" ideals, like Z mod multiples of 3. You can take sums and formal polynomials of Rings and get rings. You can do something analogous to a matrix over a ring and get a ring. Then mod them out. Lots of fun to be had.

[@MuffinsAPlenty]

Don't know if this is a helpful way to think about things. I don't think it's historically accurate, but I think it could be used as motivation. Suppose you want to study permutations of something. You can study the properties of permutations and come up with the structure of a group to abstractify permutations. In the course of studying groups, you may consider the collection of endomorphisms of an Abelian group (call its operation addition). What's interesting about the collection of endomorphisms of an Abelian group is that it has two meaningful operations: addition of endomorphisms (defined by pointwise addition of the outputs of the endomorphisms) and function composition. You can study the structure of this collection of endomorphisms of this Abelian group and come up with the structure of a ring.

[@scottmiller2591]

Sheesh. I followed everything, but I'll be damned if I know what motivated some of these equations in the first place.

[@orthoplex64]

nice

[@dandjr1546]

So every element of R satisfies r^3=r ? Sounds like a boring ring.

[@lifesahobby]

This is like some nightmare I was forced to endure as a child . The problem with studying physics , organic chemistry and applied maths is that once you understand these subjects you become a knob end with problem solving . I studied these subjects , you can know too much . I should have been paid for being taught these subjects as they messed up my life .

[@cycklist]

Here's an idea for you. Make some videos about interesting unsolved problems in maths.

[@goodplacetostop2973]

And the homework is to solve those problems 👍

[@vinvic1578]

I feel like that might be difficult since so many specific problems that haven't been solved require a very graduate level understanding ? I guess he can explain very general open questions ?

[@hugohugo37]

And solve them

[@thejelambar82]

@@vinvic1578 Collatz Conjecture, also many open problem in labelling graph theory is easy to understand

[@aweebthatlovesmath4220]

@@hugohugo37 yeah I'm sure that's easy.

[@dipandeb2005]

I wonder if Sauron knew this

[@bethhentges]

9:42 Dr. Penn says, “… because this guy turns into r.” He could have said “… because this turns into r.” He could have said, “… because r to the 3rd turns into r.”

[@gregy1570]

weird "q" symbol. took me right out of the video.

[@MuffinsAPlenty]

Over the years of being a mathematician, I have developed quirks of writing certain letters and numbers to avoid confusing them with other letters and numbers. I write my q's like that to avoid confusing them with 9's. (It also put "extra" lines in z and 7.)

[@StrangeWorld11191]

something starting with "you should know this" could hardly be usefull ...but ok, excess energy that i have today gives me ability to overly extend my curiosity ,so ill watch it after all.

[@lifesahobby]

Love the way this guy makes mind abuse look cool though . I can imagine he has a surf board and a multi coloured dog that speaks . But for me ... It's the stuff that filled my nightmares and made my childhood a scary place . I wish they had ADHD when I was a kid .

[@edoardopurini7062]

soft once you've gone beyond the soft length accidentally. i'm tempted to do everytNice tutorialng in the sa step sequencer rather than

[@mrminer071166]

Very pretty, at the level of nonsensical symbol-fiddling.

[@shameemkhan8176]

It forms an invisible probable loop in real and prime numbers.