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.
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'.
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!
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.
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 RUclips that covers them, so could be a good place to break some new ground?
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.
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.
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.
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'.
Exactly... 1=1*1'=1'......... and we are done. First = is bc 1' is an identity and second = is bc 1 is an identity.
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!
grateful for these videos that follow Dummit and Foote to help me prepare for my courses
if i remember, in my abstract algebra class (25 years ago), i thought we would say 'ring with unity' for multiplicative identity
“Ring with unity” sounds like what olden people called a wedding ring (they didn’t - but it SOUNDS like it lol)
Rings are the most interesting! (Groups and fields are fine, of course, but Frodo and Sauron wouldn't fight over those...)
Is there going to be a lecture on group actions?
I don't understand the first "observation", isn't addition commutative by definition? since (R,+) is an Abelian group
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.
Will the course return to group theory for the sylow theorems and conjugacy classes?
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 RUclips that covers them, so could be a good place to break some new ground?
Maybe as a stand-alone series -- a course on more advanced ring theory
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.
which brand chalk are you using?
Is this course finished with group theory before getting to the Sylow theorems?
My current plan is to make a "mini-series" on Sylow theorems after this course is finished.
@@mathmajor Nice, can't wait to see it!
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.
If you want it gotta put a ring on it
I wrote this before reading the other comments and now I’m ugly laughing and scaring my dogs