Ring Examples (Abstract Algebra)

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  • Опубліковано 8 лип 2017
  • 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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КОМЕНТАРІ • 251

  • @Socratica
    @Socratica  2 роки тому +27

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

    • @naman4067
      @naman4067 2 роки тому +1

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

    • @lolo6795
      @lolo6795 10 місяців тому +4

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

  • @welovfree
    @welovfree 7 років тому +656

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

    • @readingRoom100
      @readingRoom100 4 роки тому +64

      just do the entire undergrad math curriculum

    • @hy3na739
      @hy3na739 3 роки тому

      nice profile pic mah dud

    • @welovfree
      @welovfree 3 роки тому +3

      @@hy3na739 fellow struggler :)

    • @terrellronin1370
      @terrellronin1370 2 роки тому

      Instablaster

    • @Socratica
      @Socratica  2 роки тому +14

      @@readingRoom100 #Goals

  • @digitsdigitsdigits808
    @digitsdigitsdigits808 7 років тому +316

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

  • @omgopet
    @omgopet 4 роки тому +68

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

  • @fmagarik
    @fmagarik 7 років тому +61

    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

  • @yogitasingh0704
    @yogitasingh0704 6 років тому +87

    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
      @scowell 4 роки тому +3

      Now it's time for Crypto 101! Enjoy.

    • @atulit
      @atulit 3 роки тому

      same here after two years, a night before test

    • @nandy1002
      @nandy1002 3 роки тому +2

      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
      @Yami-bf6je Рік тому

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

  • @swanhtet1
    @swanhtet1 5 років тому +47

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

  • @Imakilla4567
    @Imakilla4567 6 років тому +122

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

  • @__alex.grae__
    @__alex.grae__ 3 роки тому +11

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

    • @toasteduranium
      @toasteduranium Рік тому +2

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

    • @__alex.grae__
      @__alex.grae__ Рік тому +1

      "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.

  • @sheepphic
    @sheepphic 6 років тому +1

    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
    @samcollins2108 7 років тому +7

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

  • @bablidas7236
    @bablidas7236 3 роки тому +7

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

  • @sayy_gaarr
    @sayy_gaarr 4 роки тому +8

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

  • @Omnifarious0
    @Omnifarious0 4 роки тому +2

    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).

  • @escobasingracia962
    @escobasingracia962 7 років тому

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

  • @amansingh-ww2qc
    @amansingh-ww2qc 3 роки тому +2

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

  • @rcarnes3
    @rcarnes3 6 років тому +3

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

  • @oldPrince22
    @oldPrince22 2 роки тому +4

    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 #

  • @tinahayward1604
    @tinahayward1604 2 роки тому +2

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

  • @Fematika
    @Fematika 7 років тому +53

    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
      @Fematika 7 років тому +9

      For a non commutative, finite ring.

    • @hutchisblind
      @hutchisblind 7 років тому +4

      Yes.

    • @greghmn
      @greghmn 5 років тому +1

      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
      @Sam-py9qq 4 роки тому

      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.

  • @Lacerda038
    @Lacerda038 5 років тому +1

    Muito bom! Continue com essas lições! Obrigado!

  • @roadtofitness4208
    @roadtofitness4208 6 років тому +10

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

  • @sujitmohanty1
    @sujitmohanty1 6 років тому

    Indeed fantastic series!

  • @muh.khairulamtsal1635
    @muh.khairulamtsal1635 Рік тому +1

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

    • @Socratica
      @Socratica  Рік тому

      We're so glad you've found us! 💜🦉

  • @zaidnadeem4918
    @zaidnadeem4918 4 роки тому +3

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

  • @Fematika
    @Fematika 7 років тому

    Love this series!

  • @ashishswami7188
    @ashishswami7188 7 років тому

    your videos are absolutely fabulous..

  • @cuntyclown
    @cuntyclown Рік тому +1

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

  • @MatematicasNuevoLeon
    @MatematicasNuevoLeon 7 років тому

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

  • @theultimatereductionist7592
    @theultimatereductionist7592 5 років тому +4

    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
      @theultimatereductionist7592 5 років тому +2

      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
    @chandrakalachauhan470 2 роки тому

    Incredible, way of teaching
    Thankyou so much

  • @zahidrafiq2943
    @zahidrafiq2943 3 роки тому

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

  • @MdShahid-fx2iw
    @MdShahid-fx2iw 4 роки тому

    Your lecture is so helpful mam!

  • @Headon2580
    @Headon2580 5 місяців тому

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

  • @Belmogaming6002
    @Belmogaming6002 4 роки тому

    Thank you for best in world classes 😃

  • @Khazam1992
    @Khazam1992 5 років тому +7

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

  • @hectorblandin1027
    @hectorblandin1027 Рік тому

    Love your content !

  • @kunslipper
    @kunslipper 6 років тому +1

    Thank you so much.

  • @victoralejadromc
    @victoralejadromc 2 роки тому

    Great videos!

  • @humletnobel7792
    @humletnobel7792 9 місяців тому

    What a brilliant explaining 😊

  • @jeannymath6349
    @jeannymath6349 6 років тому

    very helpful. Thank you

  • @eringreene9482
    @eringreene9482 4 роки тому +11

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

    • @javiervera6318
      @javiervera6318 4 роки тому

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

    • @johnb1391
      @johnb1391 4 роки тому

      ​ 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
      @dkprasad100 4 роки тому +3

      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
      @eringreene9482 4 роки тому +1

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

  • @muzafarhussain6878
    @muzafarhussain6878 4 роки тому

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

  • @kingprogramming
    @kingprogramming 5 років тому

    Great video!!!

  • @IjazKhan-fm4si
    @IjazKhan-fm4si 3 роки тому

    Great work

  • @radhaballavnandi3155
    @radhaballavnandi3155 4 роки тому

    your lectures are amazing maa'm

  • @LocNguyenCrypto
    @LocNguyenCrypto 6 років тому +1

    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?

  • @tauamatuatabuanaba3125
    @tauamatuatabuanaba3125 9 місяців тому

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

  • @theflaggeddragon9472
    @theflaggeddragon9472 7 років тому

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

  • @karthikkrishnan6717
    @karthikkrishnan6717 2 роки тому

    Nice one

  • @valor36az
    @valor36az 4 роки тому

    So many questions I had explained in under 8 minutes

  • @ACZ29
    @ACZ29 7 років тому

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

  • @macmos1
    @macmos1 6 років тому +3

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

  • @abdullahtrabulsiah3603
    @abdullahtrabulsiah3603 2 роки тому

    Thank you

  • @objective_truth
    @objective_truth 4 роки тому +1

    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.

  • @geogeo14000
    @geogeo14000 3 роки тому

    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
    @ujjalboro5127 3 роки тому

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

    • @Socratica
      @Socratica  3 роки тому

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

  • @marklusala8397
    @marklusala8397 3 роки тому

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

  • @kavithamicheal9821
    @kavithamicheal9821 5 років тому

    Really it's understandable. Tq mam.

  • @desalewtefera1128
    @desalewtefera1128 6 років тому

    Really nice

  • @ATD909
    @ATD909 4 роки тому

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

  • @nolanhauck9390
    @nolanhauck9390 2 роки тому

    Good video

  • @noellundstrom7447
    @noellundstrom7447 6 років тому

    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?

  • @cameronspalding9792
    @cameronspalding9792 3 роки тому

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

  • @JoelBondurant
    @JoelBondurant 7 років тому +44

    I paypaled $20, ♥💕 your content.

    • @Socratica
      @Socratica  7 років тому +14

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

  • @oneandonly66
    @oneandonly66 6 років тому

    thank u very much mam ...

  • @brouqtv6388
    @brouqtv6388 5 років тому

    Fantastic

  • @naveenbabu3689
    @naveenbabu3689 6 років тому

    Very useful

  • @elizabethcornell5745
    @elizabethcornell5745 6 років тому +1

    I love you too much u just saved me

  • @kavitha.s3863
    @kavitha.s3863 3 роки тому

    Awesome mam

  • @ronycb7168
    @ronycb7168 Рік тому

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

  • @sartajmuzafer9636
    @sartajmuzafer9636 2 роки тому +1

    Wonderful.. ❤️❤️❤️

  • @Stafford674
    @Stafford674 4 роки тому

    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?

  • @seroujghazarian6343
    @seroujghazarian6343 Рік тому

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

  • @rngwrldngnr
    @rngwrldngnr 5 років тому +1

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

    • @MuffinsAPlenty
      @MuffinsAPlenty 5 років тому

      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.

  • @naman.sharma1
    @naman.sharma1 4 роки тому

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

  • @sreejaps2428
    @sreejaps2428 6 років тому

    Mam pls make a video on ideal rings

  • @aoungorayaa7459
    @aoungorayaa7459 5 років тому

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

  • @ilguerrierodragone129
    @ilguerrierodragone129 7 місяців тому

    Proud to join the fellowship of the ring

  • @elnurazhalieva1262
    @elnurazhalieva1262 5 років тому +15

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

    • @ZiggyNorton
      @ZiggyNorton 5 років тому +5

      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
      @elnurazhalieva1262 5 років тому +2

      @@ZiggyNorton Yeah, absolutely

    • @chetanpatidar3900
      @chetanpatidar3900 3 роки тому

      Yes that's right

    • @llhammer3075
      @llhammer3075 3 роки тому

      you've blown my mind

  • @riyaagrawal269
    @riyaagrawal269 6 років тому

    reallyy mam.. u r suprb..😄😄..

  • @muhammadshafqat1935
    @muhammadshafqat1935 2 роки тому

    In integer mod n ring how can we check inverse w.r.t addition property?

  • @theultimatereductionist7592
    @theultimatereductionist7592 5 років тому +1

    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
      @AhmedIsam 5 років тому

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

  • @nowornever5598
    @nowornever5598 5 років тому

    Thanks

  • @saurabhsingh-ow7ue
    @saurabhsingh-ow7ue 3 роки тому

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

  • @jean-francoistremblay7744
    @jean-francoistremblay7744 3 роки тому

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

  • @bonbonpony
    @bonbonpony 4 роки тому

    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?

  • @BareClause
    @BareClause 3 роки тому

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

  • @MrBroybros
    @MrBroybros 7 років тому

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

  • @whatupnosy6994
    @whatupnosy6994 3 роки тому

    A matrix describing vectors on a spherical surface is a ring of finite mod n elements. There, I am now a member of the Fellowship Of The Ring

  • @ajayganta4778
    @ajayganta4778 6 років тому +1

    madam please send a video on ideals

  • @MrityunjaySinghVictor
    @MrityunjaySinghVictor 4 роки тому

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

  • @ericdew2021
    @ericdew2021 4 роки тому

    Any chance of getting to algebraic topology?

  • @SuperStingray
    @SuperStingray 6 років тому

    Unit quaternions with integer coefficients.

  • @nymphaea96
    @nymphaea96 3 роки тому

    The music just adds to the abstraction of this field of math..

  • @riyaagrawal269
    @riyaagrawal269 6 років тому

    really mam...😄 u r suprb..😄😄👌👌👌👌

  • @asishkhamrui8685
    @asishkhamrui8685 6 років тому

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

  • @worldboy9684
    @worldboy9684 2 роки тому

    Thanks!

    • @worldboy9684
      @worldboy9684 2 роки тому

      Thanks for introducing me to abstract algebra, loving it, its great :D

    • @Socratica
      @Socratica  2 роки тому

      Oh my goodness, THANK YOU so much for your kind support! We're so glad you're enjoying learning about AA - it's really our fondest hope to help people enjoy learning more. 💜🦉

  • @antoniovieira388
    @antoniovieira388 6 місяців тому

    Estou de queixo caido com voce Liliana Castro !!!!!!

  • @marcushellstrom1157
    @marcushellstrom1157 7 років тому

    Quaternions, and matrices steps on the way to triangulating any square matrice?