Ring Definition (expanded) - Abstract Algebra

A ring is a commutative group under addition that has a second operation: multiplication. These generalize a wide variety of mathematical objects like the integers, polynomials, matrices, modular arithmetic, and more. In this video we will take an in depth look at the definition of a ring.

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We recommend the following textbooks:
Dummit & Foote, Abstract Algebra 3rd Edition

Milne, Algebra Course Notes (available free online)

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Teaching​ ​Assistant:​ ​​ ​Liliana​ ​de​ ​Castro
Written​ ​&​ ​Directed​ ​by​ ​Michael​ ​Harrison
Produced​ ​by​ ​Kimberly​ ​Hatch​ ​Harrison

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[Uuuuhhhhhh]

"Consider these four sets"
Me: "okay"
"The integers.."
Me: "Mhm"
"The real numbers.."
Me: "Yeah"
"2-by-3 matricies with real entries and
Me: ఠ_ఠ

[doodelay]

For those learning this in the future and reading this comment, remember that these definitions are almost entirely based on a single concept and that is the notion of "closure." If you understand closure and see why you need it to even sensibly discuss operations between elements in a set, then you'll see where these definitions of rings, fields, and other abstract algebra definitions arise. Without a notion of closure, every time you perform an operation in a set A that you prefer, you may find that the given operation, O, yields an element not in your set. So, at that point you say that A is not closed under O. So, you don't do O when studying A and this will allow u to strictly remain in A and study it freely.

[TinyMaths]

Three hours of head-scratching while staring blankly at my lecture handout. Four and a half minutes into this video, complete clarity.. Thank you.

[oppproto]

I’m studying immunology so I don’t know what I’m doing on an abstract algebra video but it’s great

[escobasingracia]

You can't imagine how much I was waiting this series to continue. Thanks Socratica

[sterlingveil]

How am I only now discovering this channel in 2019? Oh how the YouTube algorithm scorns me! Instantly subbed.

[starbooi]

Please do a series on complex analysis (final year, undergraduate).

[mahimakalani]

Thank you Mam and you entire team for helping us understand the intuition of these concepts of ABSTRACT Algebra :)

[drewtmacha]

Please make a linear algebra series! I really enjoy the abstract algebra playlist from Socratica. I recently took an online class in abstract algebra, and I didn't fully understand the concepts until I discovered this playlist. The teaching and animations make it much more accessible than reading from a book with no illustrations (as was my case). I also took a graduate-level linear algebra class online, and, well--same thing. We read from a textbook without any illustrations, and I didn't understand much. I remember the class was mostly structured to work proofs here and there building up concepts until we developed the Jordan normal form of a matrix, and then after that I was especially lost, to where I can't even remember what topics were taught. I would be extremely interested in a playlist of linear algebra videos covering those sorts of topics.

[irynado]

Omg... 7 yrs ago I was looking for this everywhere. This and partial differential equations.

[ishanmaheshwari]

Lebesgue integration...
Pls pls pls pls pls ....mam upload these...🙏🙏🙏🙏🙏

[nickhodgskin]

Thank you much!! My head has been swimming with all these different words and structures, but you've helped to make it succint for me, and easy to visualise!! :D

[MuffinsAPlenty]

There are uses for rings without a multiplicative identity, but I'm in the camp that believes the official definition of a ring should include a multiplicative identity.

This comes from the convention of the empty product - a product with no factors.

Technically, multiplication in a ring is a binary operation, so it has exactly two inputs: no more and no less. Now, there is a very natural way to extend the definition of multiplication in a ring to take on more than two inputs, and that is to do the multiplication recursively. In other words, a·b·c is defined to be (a·b)·c, and a·b·c·d is defined to be ((a·b)·c)·d. But there are, a priori, many ways one could define multiplication for more than two inputs. For example, we could have chosen to define a·b·c to be a·(b·c) or to define a·b·c·d to be a·(b·(c·d)) or (a·b)·(c·d) or (a·(b·c))·d or a·((b·c)·d). Luckily, rings require multiplication to be associative, so the associative property of multiplication allows us to extend from two factors to more than two factors without worry about making a choice in these definitions (since all possible choices end up being equal).

So this is all nice for products of two or more elements, but it doesn't allow us to say that we can take a product of any finite number of elements. In order to do that, we have to be able to take a product with one factor or even zero factors. These situations can sometimes be useful! For example, if you want to take the product of all elements in a ring satisfying a certain property, then you need these sorts of notions if your ring only has one element satisfying the property or if your ring has no elements satisfying the property at all. Based on using the associative property to extend multiplication to more than two factors, we can again turn to the associative property for extending multiplication to fewer than two factors. If we demand that a product with one factor is consistent with the general associative property, we must conclude that the product of one element is that very element. Similarly, if we demand that a product with zero factors is consistent with the general associative property, we must conclude that the product of zero elements is a multiplicative identity.

So if we want to be able to take a product of any finite amount of elements in a ring, we must have a multiplicative identity in the ring.

This mindset also explains why many texts which require rings to have a multiplicative identity then go on to require that a subring must have the same multiplicative identity as the ring itself and why ring homomorphisms must send the multiplicative identity to the multiplicative identity. All computations in a subring should be consistent with the same computations in the ring itself, so the empty products should yield the same results (giving that we should have the same multiplicative identity). Moreover, homomorphisms are multiplicative functions and as such should send the empty product to the empty product (giving that the multiplicative identity should be mapped to the multiplicative identity).

Again, this is not to say that non-unital rings or non-unital ring homomorphisms are unimportant. But philosophically, it seems to me as though they should be viewed as lacking a little bit of the structure that a ring "should" have (empty products).

[highermathematics-bxmi]

Assalam-o-alaikum jazak Allah mam Allah shower his countless blessings upon you Allah succeeded us In every field of life

[sunilrampuria]

A simple example of a ring without identity is the set of even integers.
Division rings are also called skew field.
A algebraic structure which has all the properties of a group except that of an inverse is called a monoid. So a ring (with identity) is a group under addition and a monoid under multiplication.

[Grassmpl]

Talk about projective space and Grassmannians. I'm confused with the exterior products of a vector space.

[timhourigan]

I can't thank you enough, Socratica! I can now make a concise taxonomy/categorization of all of those objects we find in abstract algebra. An added bonus: I can now use and "translate" the textbooks I bought all those years ago that my professors never used during the course. Huzzah!!

[benjaminmcgahee]

What happens in the ring, stays in the ring. True for algebra and in boxing!

[MrWandalen]

If I understand correctly, mistake in the video at 2:19. Polynomial with complex coefficients has division which always keep it in the set, so should be check ( not cross ) at 2:19. Fraction complex coefficients is convertable to polynomial with complex coefficients by multiplication by denominator. Zero in denominator is only exception.

[bayupatten]

doing a level further maths right now and it felt like i finally have learned enough to comprehend videos like this
then you said "quirky four dimensional abstract"
....