Ring Examples (Abstract Algebra)
ฝัง
- เผยแพร่เมื่อ 8 ก.ค. 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
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Milne, Algebra Course Notes (available free online)
www.jmilne.org/math/CourseNote...
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I wanted to dislike due to bad joke but video is soo good I can't
@@naman4067 : clever jokes are for clever people, sorry for u.
Thumb up if you want Socractica to do a playlist on: Number Theory, Topology, Linear Algebra ...etc
just do the entire undergrad math curriculum
nice profile pic mah dud
@@hy3na739 fellow struggler :)
Instablaster
@@readingRoom100 #Goals
"This poor ring is having an identity crisis."
You and me both, even-numbered matrix. You and me both.
What are the odds?...
@@bonbonpony Monoids
Come for the algebra lesson, stay for the puns. The delivery is amazing on both.
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
😂😂
Comment of the decade
In this "Fellowship of the Ring" you are my lady Gandalf.
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....🙏
Now it's time for Crypto 101! Enjoy.
same here after two years, a night before test
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
Hey i see you r an indian may i ask which college r u in
Literally laughed out loud when she said: "This poor ring is having an identity crisis". Think I've been studying too long...
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Love the video. One note from a German speaker: “Zahl” is number (singular), “Zahlen” is numbers (plural), “zahlen” is pay/paying (verb).
How do the latter two differ? Capitalization only? Or pronunciation as well?
"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.
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.
I loved this topic. I didn't know that rings existed in abstract algebra until now. I hope to see much move videos!
That smirk at the end made my day!!! She was trying so hard not to laugh.
I never can forget the way u helped me.. These videos r really meant a lot to me... Thank u.
I love all this videos. This is the kind of math that I really enjoy and it's explained in an excellent way.
Amazing , with these small powerful videos filled with concept I learn everything
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).
Yep. I'm now a Patreon contributor. Excellent presentation.
This was fantastic! Thank you so much!!!! I think you may save me this semester
Mam your vedios are very helpful
Thanx a lot mam
Lots of well wishes from india
Muito bom! Continue com essas lições! Obrigado!
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).
For a non commutative, finite ring.
Yes.
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
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.
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We're so glad you've found us! 💜🦉
MASHALLAH.
THE WAY OF TEACHING IS VERY GOOD.
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MAY ALLAH BLESS YOU
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 #
Indeed fantastic series!
Love this series!
Beautiful videos. One cannot avoid falling in love with math.
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Lec are so simple every one can understand easily thank u for making videos
Incredible, way of teaching
Thankyou so much
Ma'am you are an icon and a legend. Thank you !!
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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.
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.
very helpful. Thank you
Thank you so much.
Example of a finite noncommutative ring, maybe The set of 2x2 Matrices where the entries are from The integers mod n (Z/nZ)
That has identity Since 1 belongs to Z/nZ. So te matrix with 1 in the diagonal belongs to that set
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).
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.
John B remember that in a ring, there doesn’t neccesarily need to be multiplicative inverses.
Can you please do more videos on congruence arithmetic including the euclidean algorithm?
I like how the background theme song changed when you start introducing the fields :)
Love your content !
Great videos!
What a brilliant explaining 😊
Great video!!!
One of my best teacher ..Socratica .
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Great work
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?
it's awsome explanation mam......
your lectures are amazing maa'm
The quotient group Z/nZ should be Z/nZ = { [0], [1], [2],..., [n-1] }, where [a] = a + nZ is an equivalence class.
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Oh my goodness, thank you so much, Joel!! We're so glad you enjoy our videos, and are very humbled by your support. :)
Really it's understandable. Tq mam.
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.
Thanks for the video, How can i find the inverse of (1,2) over the ring R = Z5?
Don't worry I have already joined the fellowship of the Ring😆 since childhood, thank you for your wonderful explanation...
thank u very much mam ...
Nice one
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
Really nice
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?
I love you too much u just saved me
An example of a finite non commutative ring is the set of matrices with elements in Z3
Like the shirt like nice color hoping to see some division ring examples too cuz vector spaces right ▶️
Very useful
This video is well done I’m studying for my math teacher’s exam in California that I’m taking in 12 hours
Wonderful.. ❤️❤️❤️
thanx for giving knowledge. from which country you belong kindly tell me i really impress from your lectures
I CAN LEARN ABSTRACT ALGEBRA ONLY FROM SOCRATICA. THANK YOU SO MUCH 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. 💜🦉
Thank you
reallyy mam.. u r suprb..😄😄..
So many questions I had explained in under 8 minutes
Mam pls make a video on ideal rings
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?
Hmm, finite noncommutative ring? What about ring of matrices whose elements are from set Z/nZ?
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.
@@ZiggyNorton Yeah, absolutely
Yes that's right
you've blown my mind
madam please send a video on ideals
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?
Good video
I just hit rings and then this shows up! I'm good with that! :D
really mam...😄 u r suprb..😄😄👌👌👌👌
A ring is an abelian group and a monoid such that the monoid operation distributes over the group operation
Fantastic
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.
Rings by definition come with elements that form a group. So, yes, any ring is a group under addition.
In integer mod n ring how can we check inverse w.r.t addition property?
5:35 if the integers mod (some prime) is a field, wouldn't that require there to be a multiplicative inverse for 0?
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.
Unit quaternions with integer coefficients.
thank you madam...........
Please describe division ring
Mam you help to solve some problem in ring theory?????
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)
A non commutative finite ring is set of matriex whoes elements is from Z/nZ ( for every n is element of Z)
I learned all about algebra and what my Ma'am wants to tell.
Thanks
Thanks
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?
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.
That's very odd. Groups (or Halfgroups or Monoids) should be the first Algebraic structure anyone learns about.
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?
how to find cayley table is associative
Just a guess: matrices with elements from the integers mod n.
Any chance of getting to algebraic topology?
Quaternions, and matrices steps on the way to triangulating any square matrice?
All the way triangulated matrices?
Please give me video of product of two rings
Awesome mam
Don't fields also need multiplicative inverses?