Sir, your videos are AMAZING. And I love your humour. I have been struggling with all of this stuff but you make it all so simple. Thanks so much for taking the time to have made these videos. Long life abstract algebra
04:00 Whoa! So THAT'S how it works! Now it all clicked! So basically, two-sided ideals are the ring-theoretical analogue of normal subgroups in group theory, right? Because normal subgroups were those for which left cosets = right cosets, which allowed to use them to divide the group evenly with that sub-group as a "template", and form another group from the cosets. I see the similar pattern here: the ideal is used as a "template" to subdivide the set into coset-like subsets, which are basically just "shifted ideal", and then wrap these "cosets" into elements by themselves (the residue classes in this case) to make an algebraic structure out of it. In both cases, the resulting algebraic structure (the quotient group back then, or in this case a finite field) was a "quotient" from that "division" ;> Am I correct?
Yep, that's correct! A slightly broader view is to look at a ring R and suppose you have an equivalence relation ~ on R. You want to find conditions on ~ so that R/~ (the set of equivalence classes of R under ~) is a ring with the natural map R → R/~ being a ring homomorphism (the natural map sends x to [x], the equivalence class of x). Assuming that the natural map must be a ring homomorphism, you get that [x]+[y] = [x+y] and [x][y] = [xy] for any x, y in R. Additionally, you get that [0] is the 0 element of R/~. For [0] to be the 0 element of R/~, if must be the case that for any x and y in [0], x−y is in [x−y]=[x]−[y]=[0]−[0]=[0−0]=[0] as well, so [0] must be an additive subgroup of R. Additionally, for any x in R, you get that [x]+[0] = [x], which shows that all of the equivalence classes must be cosets of [0]: If y is in [0], then x+y is in [x+y]=[x]+[y]=[x]+[0]=[x], so everything of the form x+y (where y is in [0]) is in [x], i.e., everything in the coset of [0] obtained by adding x is in [x]; and if y is in [x], then y−x is in [y−x]=[y]−[x]=[x]−[x]=[x−x]=[0], so everything in [x] is in the coset of [0] obtained by adding x. Now, since R is an Abelian group under addition, we don't have to prove that [0] must be a normal subgroup, but if you wanted to show that, you just need to use that for all x in R, [x]+[0]=[0]+[x]. Lastly, in order for multiplication to work out right, since rings have 0 multiply everything to 0, you need [0][x] = [x][0] = [0] for all x in R. This gives that [0] must be closed under multiplication from R (from both sides). All of this implies that if you want R/~ to be a ring and have the natural map be a ring homomorphism, it must be the case that the equivalence classes are cosets of a two-sided ideal.
@@PunmasterSTP Hello. To be honest this was so long ago I'm not sure what book I was reading at the time. When I was an undergrad we used Gallian's algebra book which might have been what I was talking about. I don't know if it had the best explanation of how you get a field as a quotient of a ring. I remember proving it but not really feeling like I understood why it was true at the time.
Hello, good man. I really thank you for these amazing videos. A question: can you indicate any sequence in your videos about abstract algebra? Guess it can guide us as to what study first....even though there is a quiet natural sequence of topic....Just a suggestion. Greetings from the Caribbean.
Dr. Salomone, do you know about any video that explains (as clear as you do in your videos) what a chain of fields (differential fields indeed) is from the point of view of elementary functions; what I am trying to understand is the role of those sequence of differential fields that explains why some functions dont have elementary function (an elementary primitive: non integrable functions). I have no way of seeing the whole situation as for this. Thanks.
Think of it by analogy with clock arithmetic, let's say, in the modulus 8: Anything multiplied by 0 stays at 0 (as if it were "sucked into a black hole" - it can never escape). But if you multiply, for example, 2·4, it gives 8, which is the same place as 0 too. Similarly, 4·6=24, 24 mod 8 = 0. And also 4·4=16, 16 mod 8 = 0. So we have: 2·4=0, 4·6=0, 4·4 = 0. These are the zero-divisors. What do they have in common? They all share the factor `2` with the modulus. So all numbers that _don't_ share any common factors with the modulus (i.e. are *coprime* with the modulus), will be the units ;) Now how can you extend this idea to other quotient rings? :> Hint: look at 05:30 for some hints ;>
Sir, your videos are AMAZING. And I love your humour. I have been struggling with all of this stuff but you make it all so simple. Thanks so much for taking the time to have made these videos. Long life abstract algebra
Hard to believe this video has only 16 likes after more than 2 years. Many thanks for your videos, these have been so useful for revision!
At least nine years after that it has 143 likes.
I'd say these videos are definitely both prime and maximal!
What a cute love story about zero and p. You have a knack for writing those with abstract algebra. ;)
04:00 Whoa! So THAT'S how it works! Now it all clicked! So basically, two-sided ideals are the ring-theoretical analogue of normal subgroups in group theory, right? Because normal subgroups were those for which left cosets = right cosets, which allowed to use them to divide the group evenly with that sub-group as a "template", and form another group from the cosets. I see the similar pattern here: the ideal is used as a "template" to subdivide the set into coset-like subsets, which are basically just "shifted ideal", and then wrap these "cosets" into elements by themselves (the residue classes in this case) to make an algebraic structure out of it. In both cases, the resulting algebraic structure (the quotient group back then, or in this case a finite field) was a "quotient" from that "division" ;> Am I correct?
Yep, that's correct!
A slightly broader view is to look at a ring R and suppose you have an equivalence relation ~ on R. You want to find conditions on ~ so that R/~ (the set of equivalence classes of R under ~) is a ring with the natural map R → R/~ being a ring homomorphism (the natural map sends x to [x], the equivalence class of x).
Assuming that the natural map must be a ring homomorphism, you get that [x]+[y] = [x+y] and [x][y] = [xy] for any x, y in R. Additionally, you get that [0] is the 0 element of R/~.
For [0] to be the 0 element of R/~, if must be the case that for any x and y in [0], x−y is in [x−y]=[x]−[y]=[0]−[0]=[0−0]=[0] as well, so [0] must be an additive subgroup of R. Additionally, for any x in R, you get that [x]+[0] = [x], which shows that all of the equivalence classes must be cosets of [0]:
If y is in [0], then x+y is in [x+y]=[x]+[y]=[x]+[0]=[x], so everything of the form x+y (where y is in [0]) is in [x], i.e., everything in the coset of [0] obtained by adding x is in [x];
and if y is in [x], then y−x is in [y−x]=[y]−[x]=[x]−[x]=[x−x]=[0], so everything in [x] is in the coset of [0] obtained by adding x.
Now, since R is an Abelian group under addition, we don't have to prove that [0] must be a normal subgroup, but if you wanted to show that, you just need to use that for all x in R, [x]+[0]=[0]+[x].
Lastly, in order for multiplication to work out right, since rings have 0 multiply everything to 0, you need [0][x] = [x][0] = [0] for all x in R. This gives that [0] must be closed under multiplication from R (from both sides).
All of this implies that if you want R/~ to be a ring and have the natural map be a ring homomorphism, it must be the case that the equivalence classes are cosets of a two-sided ideal.
Thank you for this. I understand everything so much better than I did just reading my book.
Out of curiosity, what book were you using?
@@PunmasterSTP Hello. To be honest this was so long ago I'm not sure what book I was reading at the time. When I was an undergrad we used Gallian's algebra book which might have been what I was talking about. I don't know if it had the best explanation of how you get a field as a quotient of a ring. I remember proving it but not really feeling like I understood why it was true at the time.
@@CallMeIshmael999 Ah gotcha, thanks for letting me know!
Maybe I'm confused here, but 6Z/2Z is just [0] right?
9:40 I thought he said that (Z/6Z)/(Z/2Z) was equivalent to Z/3Z.
Hello, good man. I really thank you for these amazing videos. A question: can you indicate any sequence in your videos about abstract algebra? Guess it can guide us as to what study first....even though there is a quiet natural sequence of topic....Just a suggestion. Greetings from the Caribbean.
Exploring Abstract Algebra II: Contents
Dr. Salomone, do you know about any video that explains (as clear as you do in your videos) what a chain of fields (differential fields indeed) is from the point of view of elementary functions; what I am trying to understand is the role of those sequence of differential fields that explains why some functions dont have elementary function (an elementary primitive: non integrable functions). I have no way of seeing the whole situation as for this. Thanks.
Thanks for the explanation. But I didn't really get u after 12:30 onwards..
Great visual aids!
Sir how to find zero divisor and unit in Quotient ring
Think of it by analogy with clock arithmetic, let's say, in the modulus 8:
Anything multiplied by 0 stays at 0 (as if it were "sucked into a black hole" - it can never escape). But if you multiply, for example, 2·4, it gives 8, which is the same place as 0 too. Similarly, 4·6=24, 24 mod 8 = 0. And also 4·4=16, 16 mod 8 = 0. So we have: 2·4=0, 4·6=0, 4·4 = 0. These are the zero-divisors. What do they have in common? They all share the factor `2` with the modulus. So all numbers that _don't_ share any common factors with the modulus (i.e. are *coprime* with the modulus), will be the units ;)
Now how can you extend this idea to other quotient rings? :>
Hint: look at 05:30 for some hints ;>
thanks sir thanks alot
Good video but you are always so fast in your explanation and demonstrations
Have you tried 0.75x speed?
Sir plz ans my Question