What is a ring? -- Abstract Algebra 17 

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.

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!

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'.

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

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

Is there going to be a lecture on group actions?

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

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

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

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.

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 RU-vid that covers them, so could be a good place to break some new ground?

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.

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

If you want it gotta put a ring on it