What is a ring? -- Abstract Algebra 17

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[@ethanlipson1637]

Note on the commutativity of 1x1 matrices: this is obvious algebraically, but the geometric reasoning is interesting. Matrices with the same eigenvectors commute with each other, and all 1x1 matrices have the same eigenvectors.

[@jamieee472]

michael, your voice sounds really hoarse in this video... if you're not feeling well, please take a break and rest! don't forget to drink more water; and get well soon!

[@spookybrojo]

grateful for these videos that follow Dummit and Foote to help me prepare for my courses

[@MathMan271]

if i remember, in my abstract algebra class (25 years ago), i thought we would say 'ring with unity' for multiplicative identity

[@NotoriousSRG]

“Ring with unity” sounds like what olden people called a wedding ring (they didn’t - but it SOUNDS like it lol)

[@SeeTv.]

Is there going to be a lecture on group actions?

[@JosBergervoet]

Rings are the most interesting! (Groups and fields are fine, of course, but Frodo and Sauron wouldn't fight over those...)

[@schweinmachtbree1013]

the proofs of parts 3 and 4 of the proposition at 14:00 are a bit overcomplicated: 3 follows directly from 2 since (-a)(-b) = -(a(-b)) = -(-ab) = ab, and for 4 one can prove that the multiplicative identity in a ring is unique exactly the same way one proves that the identity in a group is unique: 1·1' = 1 and 1·1' = 1' so 1 = 1'.

[@lukehibbs6723]

Exactly... 1=1*1'=1'......... and we are done. First = is bc 1' is an identity and second = is bc 1 is an identity.

[@robshaw2639]

Will the course return to group theory for the sylow theorems and conjugacy classes?

[@michaelwynne2310]

Love your videos, Michael! Do you think you'll cover modules at all? I know they're not typically covered in a first course on algebra, but I'm yet to find a good lecture series on KZbin that covers them, so could be a good place to break some new ground?

[@mathmajor]

Maybe as a stand-alone series -- a course on more advanced ring theory

[@maxrang7303]

I don't understand the first "observation", isn't addition commutative by definition? since (R,+) is an Abelian group

[@schweinmachtbree1013]

Yes, addition being commutative is part of the definition of a ring. The observation says that for rings _with identity_ , commutativity of addition follows from the rest of the axioms, so in this case one can get away with saying just "(R,+) is a group" which is then necessarily abelian.

[@scottmiller2591]

Note that not every author uses a ring to mean not having multiplicative identity. Wikipedia, for instance, defines a ring to be what Prof. Penn calls ring with unity. It's a little frustrating to pick up a paper and not know which definition of ring they are using. I suspect this is discipline-dependent, although it's different for different authors even in the same discipline.

[@homerthompson416]

Is this course finished with group theory before getting to the Sylow theorems?

[@mathmajor]

My current plan is to make a "mini-series" on Sylow theorems after this course is finished.

[@homerthompson416]

@@mathmajor Nice, can't wait to see it!

[@scottmiller2591]

I think your proof of uniqueness of 1 only proves that 1 itself has a unique identity, not that in general _all_ elements a have the _same_ unique identity. It's easy to prove that 1 is unique for all a, though, starting from for all a in A, a 1 = a, then assume a second identity exists 1': a 1' = a. Since this is true for all a, it must be true for a=1, resulting in 1 1' = 1. Then since the identity works from either side (a 1 = 1a, a 1' = 1' a), then 1 1' = 1' 1 = 1, but as previously noted, 1 1' = 1', therefore 1 = 1', contradicting the assumption that there was another identity for a, therefore 1 is unique for all a in A. I think this is the general proof, but I could be wrong - feel free to point out any mistakes, chat.

[@NotoriousSRG]

If you want it gotta put a ring on it

[@NotoriousSRG]

I wrote this before reading the other comments and now I’m ugly laughing and scaring my dogs