Ring Examples (Abstract Algebra)

Rings are one of the key structures in Abstract Algebra. In this video we give lots of examples of rings: infinite rings, finite rings, commutative rings, noncommutative rings and more!
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We recommend the following textbooks:
Dummit & Foote, Abstract Algebra 3rd Edition
amzn.to/2oOBd5S
Milne, Algebra Course Notes (available free online)
www.jmilne.org/math/CourseNote...
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Teaching​ ​Assistant:​ ​​ ​Liliana​ ​de​ ​Castro
Written​ ​&​ ​Directed​ ​by​ ​Michael​ ​Harrison
Produced​ ​by​ ​Kimberly​ ​Hatch​ ​Harrison
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[@Socratica]

Sign up to our email list to be notified when we release more Abstract Algebra content: snu.socratica.com/abstract-algebra

[@naman4067]

I wanted to dislike due to bad joke but video is soo good I can't

[@lolo6795]

@@naman4067 : clever jokes are for clever people, sorry for u.

[@welovfree]

Thumb up if you want Socractica to do a playlist on: Number Theory, Topology, Linear Algebra ...etc

[@readingRoom100]

just do the entire undergrad math curriculum

[@hy3na739]

nice profile pic mah dud

[@welovfree]

@@hy3na739 fellow struggler :)

[@terrellronin1370]

Instablaster

[@Socratica]

@@readingRoom100 #Goals

[@digitsdigitsdigits808]

"This poor ring is having an identity crisis." You and me both, even-numbered matrix. You and me both.

[@bonbonpony]

What are the odds?...

[@AbhishekThakur-wl1pl]

@@bonbonpony Monoids

[@omgopet]

Come for the algebra lesson, stay for the puns. The delivery is amazing on both.

[@yogitasingh0704]

An example of finite non-commutative ring is a finite MATRIX. And the way of teaching is really very wonderful, I have learnt Group Theory from your videos in my previous college semester and now in this semester, you are again making it very easy to learn Ring Theory. 🙏🙏 Thanks a lot SOCRATICA🙏 for giving us an excellent teacher🙏.... Best wishes from INDIA....🙏

[@scowell]

Now it's time for Crypto 101! Enjoy.

[@atulit]

same here after two years, a night before test

[@nandy1002]

well if we say a finite ring with no identity and non-commutative then we can say finite even integer matrix is a ring for that

[@Yami-bf6je]

Hey i see you r an indian may i ask which college r u in

[@Imakilla4567]

Literally laughed out loud when she said: "This poor ring is having an identity crisis". Think I've been studying too long...

[@souvikbiswas284]

mee too broo mee too

[@fmagarik]

If you liked it then you should have put a group on it, such that it is abellian under addition, a monoid under multiplication and the distributive property holds

[@maknimariem3979]

😂😂

[@hybmnzz2658]

Comment of the decade

[@__alex.grae__]

Love the video. One note from a German speaker: “Zahl” is number (singular), “Zahlen” is numbers (plural), “zahlen” is pay/paying (verb).

[@toasteduranium]

How do the latter two differ? Capitalization only? Or pronunciation as well?

[@__alex.grae__]

"Zahlen" (numbers) and "zahlen" (to pay) are pronounced the same but keep in mind that German language will heavily conjugate verbs - English does not so much. Ich zahle, du zahlst, er/sie/es zahlt, wir zahlen, ihr zahlt, sie zahlen.

[@swanhtet1]

In this "Fellowship of the Ring" you are my lady Gandalf.

[@sheepphic]

These are some of my favourite math videos! I've always wanted to learn abstract algebra, but it was always just a jumble of notation. Thanks for making these great videos to help people learn.

[@samcollins2108]

I loved this topic. I didn't know that rings existed in abstract algebra until now. I hope to see much move videos!

[@bablidas7236]

I never can forget the way u helped me.. These videos r really meant a lot to me... Thank u.

[@escobasingracia962]

I love all this videos. This is the kind of math that I really enjoy and it's explained in an excellent way.

[@Omnifarious0]

Your bad puns, so carefully and thoughtfully delivered are amazing. I couldn't do better myself, and that's saying something (specifically, that I couldn't do better myself).

[@amansingh-ww2qc]

Amazing , with these small powerful videos filled with concept I learn everything

[@sayy_gaarr]

That smirk at the end made my day!!! She was trying so hard not to laugh.

[@rcarnes3]

Yep. I'm now a Patreon contributor. Excellent presentation.

[@roadtofitness4208]

Mam your vedios are very helpful Thanx a lot mam Lots of well wishes from india

[@tinahayward1604]

This was fantastic! Thank you so much!!!! I think you may save me this semester

[@oldPrince22]

How to construct a finite non-comm ring. If one uses the trick introduced in the video, one can take all 2 by 2 matrices whose entries only be 1 or 0. And addition/multiplication all usual matrix operations but under mod 2. Then (01,00)(01,10)=(10,00) but (01,10)(01,00)=(00,01) hence non-comm. Finite is obvious because we have 4 entries and each entry can be either 0 or 1 thus #

[@Fematika]

Do the n x n matrices mod(n), meaning ((a mod(n), b mod(n)), (c mod(n), d mod(n))), with all of the usual operations, though each element is now mod(n).

[@Fematika]

For a non commutative, finite ring.

[@hutchisblind]

Yes.

[@greghmn]

By that token, you can also come up with a non-commutative finite rng (my way of notating the lack of mult id), like nxn matrices with entries that are elements of xZ/yZ, where x divides y, x

[@Sam-py9qq]

If anyone finds it unclear, this ring is finite because it contains (only) the matrices with elements ∊ ℤ (mod n), and closed because the elements of any product or sum thereof reduce to ℤ (mod n). Specifically, the order of this ring (in the "size of set" sense) is n^(n·n) since there are n variants for every n·n position ⇒ n^(n·n) total variants.

[@muh.khairulamtsal1635]

just found this channel, really intersting and decent way of teaching love ur video sm

[@Socratica]

We're so glad you've found us! 💜🦉

[@Lacerda038]

Muito bom! Continue com essas lições! Obrigado!

[@zaidnadeem4918]

MASHALLAH. THE WAY OF TEACHING IS VERY GOOD. 👍👍👍👍 MAY ALLAH BLESS YOU

[@MatematicasNuevoLeon]

Beautiful videos. One cannot avoid falling in love with math.

[@sujitmohanty1]

Indeed fantastic series!

[@ashishswami7188]

your videos are absolutely fabulous..

[@Fematika]

Love this series!

[@MdShahid-fx2iw]

Your lecture is so helpful mam!

[@zahidrafiq2943]

Lec are so simple every one can understand easily thank u for making videos

[@theultimatereductionist7592]

6:27 Wedderburn's Theorem: there are no finite noncommutative division rings (rings all of whose nonzero elements have multiplicative inverses). But finite noncommutative non-division rings: matrices over a Z/n with n composite might work.

[@theultimatereductionist7592]

Don't even need n to be composite. The 16-member ring of all 2-by-2 matrices over Z/2 is noncommutative: M = 1 in all entries except 0 in (1,2) N = 1 in all entries except 0 in (2,1) MN = 1 in all entries except 0 in (2,2) NM =1 in all entries except 0 in (1,1) The 4 matrices with 0s in all entries except 1 in one entry have no inverse.

[@chandrakalachauhan470]

Incredible, way of teaching Thankyou so much

[@muzafarhussain6878]

One of my best teacher ..Socratica . Love from pakistan .. keeping it up ,so that we learn easly ..🇵🇰🇵🇰

[@Khazam1992]

I like how the background theme song changed when you start introducing the fields :)

[@Belmogaming6002]

Thank you for best in world classes 😃

[@jeannymath6349]

very helpful. Thank you

[@Headon2580]

your teaching technique is so good i like it .thanks❤❤❤❤👍👍

[@kunslipper]

Thank you so much.

[@cuntyclown]

Ma'am you are an icon and a legend. Thank you !!

[@eringreene9482]

Example of a finite noncommutative ring, maybe The set of 2x2 Matrices where the entries are from The integers mod n (Z/nZ)

[@javiervera6318]

That has identity Since 1 belongs to Z/nZ. So te matrix with 1 in the diagonal belongs to that set

[@johnb1391]

​ Javier Vera What about the zero matrix? It's determinant is zero so it does not have an inverse matrix (so no identity since A^-1 does not exist).

[@dkprasad100]

that answer is correct. That ring is denoted by M[Zn] which has finite number of elements and non-commutative under matrix multiplication. It is Abelian under matrix addition and thus a ring.

[@eringreene9482]

John B remember that in a ring, there doesn’t neccesarily need to be multiplicative inverses.

[@humletnobel7792]

What a brilliant explaining 😊

[@victoralejadromc]

Great videos!

[@hectorblandin1027]

Love your content !

[@kingprogramming]

Great video!!!

[@theflaggeddragon9472]

Can you please do more videos on congruence arithmetic including the euclidean algorithm?

[@IjazKhan-fm4si]

Great work

[@ACZ29]

it's awsome explanation mam......

[@LocNguyenCrypto]

So, we need a finite set of elements and matrix. We can limited a set by using { module, char, int, etc in computer science, other set } Is there a way for not using matrix?

[@radhaballavnandi3155]

your lectures are amazing maa'm

[@noellundstrom7447]

My answer for the final question would be a ring consisting of the 2x2 matrices where all the elements of the matrix are the integers mod n. The ring would be commutative under addition from the definition of a matrix and because the integers mod n also being commutative. And of course matrix multiplication is non-commutative. Am I right?

[@ATD909]

This video is well done I’m studying for my math teacher’s exam in California that I’m taking in 12 hours

[@objective_truth]

In fact, every ring is a group, and every field is a ring. A ring is a group with an additional operation, where the second operation is associative and the distributive properties make the two operations "compatible". A field is a ring such that the second operation also satisfies all the group properties (after throwing out the additive identity); i.e. it has multiplicative inverses, multiplicative identity, and is commutative. In mathematics, a ring is one of the fundamental algebraic structures used in abstract algebra. It consists of a set equipped with two binary operations that generalize the arithmetic operations of addition and multiplication. Through this generalization, theorems from arithmetic are extended to non-numerical objects such as polynomials, series, matrices and functions. A ring is an abelian group with a second binary operation that is associative, is distributive over the abelian group operation, and has an identity element (this last property is not required by some authors, see § Notes on the definition). By extension from the integers, the abelian group operation is called addition and the second binary operation is called multiplication.

[@aoungorayaa7459]

thanx for giving knowledge. from which country you belong kindly tell me i really impress from your lectures

[@macmos1]

The quotient group Z/nZ should be Z/nZ = { [0], [1], [2],..., [n-1] }, where [a] = a + nZ is an equivalence class.

[@kavithamicheal9821]

Really it's understandable. Tq mam.

[@ronycb7168]

Like the shirt like nice color hoping to see some division ring examples too cuz vector spaces right ▶️

[@tauamatuatabuanaba3125]

Don't worry I have already joined the fellowship of the Ring😆 since childhood, thank you for your wonderful explanation...

[@desalewtefera1128]

Really nice

[@oneandonly66]

thank u very much mam ...

[@marklusala8397]

Thanks for the video, How can i find the inverse of (1,2) over the ring R = Z5?

[@JoelBondurant]

I paypaled $20, ♥💕 your content.

[@Socratica]

Oh my goodness, thank you so much, Joel!! We're so glad you enjoy our videos, and are very humbled by your support. :)

[@geogeo14000]

Great video as always, but a ring A can existe without identity element "1_A" ? because when I read the definitions given on french website and in my french course, the present of 1_A an identity element is required, same for sub-rings

[@ujjalboro5127]

I CAN LEARN ABSTRACT ALGEBRA ONLY FROM SOCRATICA. THANK YOU SO MUCH SOCRATICA.

[@Socratica]

We're so glad you're watching with us!! It really inspires us to make more videos when we hear that we're helping. 💜🦉

[@karthikkrishnan6717]

Nice one

[@valor36az]

So many questions I had explained in under 8 minutes

[@naveenbabu3689]

Very useful

[@abdullahtrabulsiah3603]

Thank you

[@elizabethcornell5745]

I love you too much u just saved me

[@sartajmuzafer9636]

Wonderful.. ❤️❤️❤️

[@theultimatereductionist7592]

5:59 The ADDITIVE structure of rings is a group: an abelian group, specifically. But, don't say rings, in general, are a subset of all groups. In general the multiplicative structure on rings is not a group.

[@AhmedIsam]

Rings by definition come with elements that form a group. So, yes, any ring is a group under addition.

[@riyaagrawal269]

reallyy mam.. u r suprb..😄😄..

[@MrBroybros]

I just hit rings and then this shows up! I'm good with that! :D

[@naman.sharma1]

I learned all about algebra and what my Ma'am wants to tell. Thanks

[@bonbonpony]

05:12 Can you talk some more about those ideals? I don't see them being introduced anywhere on this playlist. 06:46 Dying inside a little bit when reading that from the prompter there, eh? :) OK, I guess that the 2×2 matrices with coefficients being integers mod n is the non-commutative finite ring we're looking for?

[@sreejaps2428]

Mam pls make a video on ideal rings

[@cameronspalding9792]

An example of a finite non commutative ring is the set of matrices with elements in Z3

[@Stafford674]

Once we have established the definitions of various types of ring, is there anything else that can be said about them. Do all commutative finite rings have some property in common. If so, what is it? If not, what is the point of all this?

[@neuronclasses1415]

Plz....explain mam The set of all continuous real-valued functions of a real variable whose graphs pass through the point (1,0) is a commutative ring without unity without unity under the operations of pointwise addition and multiplication, i.e., the operations (f+g)(a) = f(a)+g(a) and (fg)(a)=f(a)g(a)

[@brouqtv6388]

Fantastic

[@seroujghazarian6343]

N->regular set Z->(commutative) ring Q->field R->field C->field

[@nolanhauck9390]

Good video

[@MrityunjaySinghVictor]

A non commutative finite ring is set of matriex whoes elements is from Z/nZ ( for every n is element of Z)

[@jean-francoistremblay7744]

Just for the fellowship of the ring, I give 2 thumbs up!!!!

[@amyp7067]

Im taking abstract alg. as a summer class right now and strangely enough we are starting out by learning Rings first, this mainly has to do with the text we use (Hungerford) so it's hard to find material that mirrors this when I'm trying to find out more info on Rings and some theorems of the sort most online materials reference groups but I don't learn about groups until next week. Anyone else have a similar experience? Any other text you would recommend?

[@thewalkingcrow8946]

My education path covered Abstract Algebra first, where we went over rings, but a summer course was offered called Group Theory as an elective. So we didn't define a Ring using the concept of groups, but those of us that had taken that Group Theory knew how to shorten the definition.

[@MrCmon113]

That's very odd. Groups (or Halfgroups or Monoids) should be the first Algebraic structure anyone learns about.

[@ronnies.3440]

11 months later and I am in your position. Currently taking abstract this summer and we began with rings. Any suggestions on helpful textbooks/resources?

[@elnurazhalieva1262]

Hmm, finite noncommutative ring? What about ring of matrices whose elements are from set Z/nZ?

[@ZiggyNorton]

That's what I believe as well. Since matrices are non-commutative, regardless of the entries, they will be non-commutative. Since the integers mod n is finite, there is a finite number of matrices with entries from this set.

[@elnurazhalieva1262]

@@ZiggyNorton Yeah, absolutely

[@chetanpatidar3900]

Yes that's right

[@llhammer3075]

you've blown my mind

[@ajayganta4778]

madam please send a video on ideals

[@kavitha.s3863]

Awesome mam

[@nowornever5598]

Thanks

[@riyaagrawal269]

really mam...😄 u r suprb..😄😄👌👌👌👌

[@rngwrldngnr]

5:35 if the integers mod (some prime) is a field, wouldn't that require there to be a multiplicative inverse for 0?

[@MuffinsAPlenty]

No. A field is a ring in which every _nonzero_ element has a multiplicative inverse. The axioms of a ring (namely, an Abelian group under addition and distribution of multiplication over addition) _forces_ 0 to multiply every ring element to 0. As such, it is _impossible_ for zero to have a multiplicative inverse (except in the fairly stupid case where 0 = 1). Therefore, the best you can do for multiplicative inverses is to have every _nonzero_ element have a multiplicative inverse. So that is the requirement to have a field/division ring.

[@saurabhsingh-ow7ue]

thank you madam...........

[@BareClause]

A ring is an abelian group and a monoid such that the monoid operation distributes over the group operation

[@claytonbenignus4688]

You are the Abstract Algebra I wish I had in 1974. The “man” I had could and did make anything boring while kindling resentment for himself and his subject.

[@claytonbenignus4688]

Answer: The Quaternions

[@tommaybe7854]

identity crisis fellowship of the rings P.S.: I love you so much for excavating the fun in math.

[@sotosmath6284]

in the integers mod 3 consider the matrix A= ( 1 2 and B=(1 1 then A times B is not the same as B times A 0 2) 1 1)

[@vinca43]

Nice, digestible videos overall. I disagree with the Venn diagram however. It makes no sense to say that the set of rings is contained in the set of groups. Given an element R from the set of rings, R is not in the set of groups since R has 2 binary operations. You can say that, given a ring, (R,+,x), (R,+) is a group. To extend the argument, you would not say that the set of groups is contained in the set of sets, because a group has a binary operation, and sets do not.

[@jayaprakash3056]

How to prove that (N,+,*) is not field in linear algebra? Can u plz tell me solution.

[@asishkhamrui8685]

Mam you help to solve some problem in ring theory?????

[@SuperStingray]

Unit quaternions with integer coefficients.

[@ericdew2021]

"...join the fellowship of the ring..." Aaaughhh! Math joke! Math joke! Got a chuckle out of me, though so kudos.