Ring Examples (Abstract Algebra)
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- čas přidán 8. 07. 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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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.
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
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
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.
Literally laughed out loud when she said: "This poor ring is having an identity crisis". Think I've been studying too long...
mee too broo mee too
That smirk at the end made my day!!! She was trying so hard not to laugh.
I loved this topic. I didn't know that rings existed in abstract algebra until now. I hope to see much move videos!
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 never can forget the way u helped me.. These videos r really meant a lot to me... Thank u.
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.
Amazing , with these small powerful videos filled with concept I learn everything
I love all this videos. This is the kind of math that I really enjoy and it's explained in an excellent way.
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 #
This was fantastic! Thank you so much!!!! I think you may save me this semester
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.
MASHALLAH.
THE WAY OF TEACHING IS VERY GOOD.
👍👍👍👍
MAY ALLAH BLESS YOU
Muito bom! Continue com essas lições! Obrigado!
Ma'am you are an icon and a legend. Thank you !!
Indeed fantastic series!
Love this series!
Mam your vedios are very helpful
Thanx a lot mam
Lots of well wishes from india
your videos are absolutely fabulous..
just found this channel, really intersting and decent way of teaching
love ur video sm
We're so glad you've found us! 💜🦉
Thank you so much.
Incredible, way of teaching
Thankyou so much
Beautiful videos. One cannot avoid falling in love with math.
Great videos!
Your lecture is so helpful mam!
Love your content !
very helpful. Thank you
Don't worry I have already joined the fellowship of the Ring😆 since childhood, thank you for your wonderful explanation...
Great video!!!
What a brilliant explaining 😊
I like how the background theme song changed when you start introducing the fields :)
Great work
Thank you for best in world classes 😃
Lec are so simple every one can understand easily thank u for making videos
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.
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your lectures are amazing maa'm
it's awsome explanation mam......
I paypaled $20, ♥💕 your content.
Oh my goodness, thank you so much, Joel!! We're so glad you enjoy our videos, and are very humbled by your support. :)
N->regular set
Z->(commutative) ring
Q->field
R->field
C->field
One of my best teacher ..Socratica .
Love from pakistan .. keeping it up ,so that we learn easly ..🇵🇰🇵🇰
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
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.
Really nice
Really it's understandable. Tq mam.
Nice one
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.
The quotient group Z/nZ should be Z/nZ = { [0], [1], [2],..., [n-1] }, where [a] = a + nZ is an equivalence class.
thank u very much mam ...
This video is well done I’m studying for my math teacher’s exam in California that I’m taking in 12 hours
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. 💜🦉
Very useful
Proud to join the fellowship of the ring
Thank you
Can you please do more videos on congruence arithmetic including the euclidean algorithm?
So many questions I had explained in under 8 minutes
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?
Like the shirt like nice color hoping to see some division ring examples too cuz vector spaces right ▶️
I love you too much u just saved me
Wonderful.. ❤️❤️❤️
Good video
identity crisis
fellowship of the rings
P.S.: I love you so much for excavating the fun in math.
reallyy mam.. u r suprb..😄😄..
"...join the fellowship of the ring..." Aaaughhh! Math joke! Math joke! Got a chuckle out of me, though so kudos.
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
Just for the fellowship of the ring, I give 2 thumbs up!!!!
An example of a finite non commutative ring is the set of matrices with elements in Z3
I learned all about algebra and what my Ma'am wants to tell.
Thanks
Thanks
Fantastic
A ring is an abelian group and a monoid such that the monoid operation distributes over the group operation
I just hit rings and then this shows up! I'm good with that! :D
thanx for giving knowledge. from which country you belong kindly tell me i really impress from your lectures
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.
Answer: The Quaternions
Q1.every non zero commmutative ring R cotains maximal ideal
Q2.Show that a ring R is the zero ring i.e R={0} ⇔ 1=0
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 mam...😄 u r suprb..😄😄👌👌👌👌
Mam pls make a video on ideal rings
thank you madam...........
Awesome mam
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?
A non commutative finite ring is set of matriex whoes elements is from Z/nZ ( for every n is element of Z)
The music just adds to the abstraction of this field of math..
Thanks for the video, How can i find the inverse of (1,2) over the ring R = Z5?
Estou de queixo caido com voce Liliana Castro !!!!!!
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)
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.
Unit quaternions with integer coefficients.
Thanks!
Thanks for introducing me to abstract algebra, loving it, its great :D
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. 💜🦉
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?
madam please send a video on ideals
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?