Warning: I made a very simple error at around 4:25. Composition does not satisfy the distributive rule and thus does not make these structure into a ring. Composition does make polynomials with composition into something called a semiring.
You mean near-ring not semiring (although it is easy to forgive this mistake as these algebraic structures are quite obscure). Semirings, also known as "rigs", are what you get when you drop the requirement for additive inverses in a ring, so a semiring is a simultaneous additive semigroup and multiplicative semigroup, as opposed to a ring which is a simultaneous additive group and multiplicative semigroup. the two distributive rules are kept for semirings.
My answers for these warm-up questions: Q2,suppose there exists k in R, s.t. ak=1 Since R is a ring, so there exists r in R s.t. k+b=r(additive closure) then k=r-b ak=a(r-b)=ar-ab=ar-0=ar Since the inverse of a is unique, then k=r, then b=0 It's a contradiction. (bonus)the trivial ring {e}={0}. Thank you professor for your wonderful explanation.
From what I’ve read, the ‘k’ comes from the German word for field, Körper (meaning body). As for the lowercase question, I’m guessing that really only applies to finite fields, because I used capital letters for number fields. I guess it’s just a way to show that the number field K and its integer ring quotient k are related, and of course K is bigger.
Quoting Edward V. Huntington: « Closely connected with the theory of groups is the theory of fields, suggested by GALOIS, and due, in concrete form, to DEDEKIND in 1871. The word field is the English equivalent for DEDEKIND’s term Körper;. KRONECKER’s term Rationalitätsbereich, which is often used as a synonym, had originally a somewhat different meaning. The earliest expositions of the theory from the general or abstract point of view were given independently by WEBER and by Moore, in 1893 […] »
4:22 polynomials with composition do not form a ring because only one of the distributive laws is satisfied - (working with polynomial functions instead of polynomials,) we have (p+q)∘r = p∘r + q∘r because (p+q)∘r(x) = (p+q)(r(x))= p(r(x)) + q(r(x)) = p∘r(x) + q∘r(x) = (p∘r + q∘r)(x). however for the other distributive law we have p∘(q+r)(x) = p(q(x) + r(x)) ≠ p(q(x)) + p(r(x)) = (p∘q + p∘r)(x) in general, so p∘(q+r) ≠ p∘q + p∘r (this distributive law holds only when p is a linear polynomial p(x) = rx for some r, and such polynomials are not closed under multiplication so this subset also isn't a ring)
15:43 Actually, it is not necessary to show they don't have any inverses. The most important condition for a group is that for all a,b€G, a*b€G. But we want to show that Fn-{0} is a group under multiplication, therefore a*b = 0 €/ Fn-{0} shows it can't be a group
I'm trying to relearn linear algebra and I'm loving this approach. Very different, but good explanations. I think while the examples are great, I am actually finding the counter examples even more useful.
So what i was thinking about that warm up is that let there exist an inverse for a, then a^-1(ab)=a^-1(0)=0 so (a^-1*a)b=1*b=b=0 But its not said that b is 0 or it has to be 0, hence there cant be an inverse for a and b.
From someone who learned this material in the opposite order, I like how you front-load the group theoretical preliminaries. In a pure math context, it makes more sense to learn the material this way. The determinant, eigenvalues, and eigenvectors of a matrix with real-valued entries have nice geometric interpretations. I look forward to your discussions on objects like M_{nxn}(Z_5).
The letter K comes from the German word Körper (body). In Portuguese, for example, a similar word is used, corpo, which also stands for body. People rather state results for K just because it is more general.
Presumably the ring where the additive identity is the multiplicative identity would be the trivial ring, where e*e = e = e + e, and the set is the singleton set, {e}.
Love what you are doing here and kudos for wearing the stream of consciousness type errors. I always find I have to proof read my math and do it again and again and ...
In my country, the common notation for fields is F, with the extra line which R, W, C, Z and N also get when talking about the respective groups of numbers
The question about a ring with 0 = 1 was a fun one to ponder! I believe it must be {0} with addition operation 0 + 0 = 0 and multiplication operation 0*0 = 0. Proof: Take any element, a, in a ring, R, where 0 = 1. a + 0 = a = a*1 - since 0 is additive identity and 1 is multiplicative identity a + 0 = a*0 - since 0 = 1 a + 0 = 0 - the result a*0 = 0 for all rings is proved in the lecture a = 0 - once again, 0 is additive identity
okay so this was my attempt at looking at a ring where 0 = 1 suppose we have a ring R, where, the additive inverse = the multiplicative inverse = e then, for any r in R => r+e = r = r*e => (r+e)-r = (r*e)-r => e+e = r*(e-e) => e = r this must be true for any r in R, so R must contain only a singular element, e
for a shorter proof you can see that r * e = r since e is the multiplicative identity but you also have that r * e = e since e is the additive identity, so r = r * e = e -> r = e
Question about finite fields: are there any other finite field that is not isomorphic to F_p or is F_p the complete classification of all finite fields?
Every finite field is has p^n elements for a prime p and a natural number n. On the other hand there exists for all p and n such a field (and its unique). For n=1 this field is just the integers mod p with the ordinary addition and multiplication. For n>1 the field is NOT (!) the integers mod p^n. Because the only m, such that the ring of integers mod m is a field are the primes (otherwise the divisors of m are zero divisors, which do not exist in fields). These fields with p^n elements for n>1 are harder to describe: One way is to take an irreducible polynomial f over F_p and look at the quotient F_p[x]/(f). This is (isomorphic to) F_{p^n} where n is the degree of f.
The operation that he defined combines two objects of the given set by either using addition, multiplication or some other operation that combines the objects. It seems like the operating is addition in the example so 7+7=14 and 14 mod 12 = 2
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)
Warning: I made a very simple error at around 4:25. Composition does not satisfy the distributive rule and thus does not make these structure into a ring. Composition does make polynomials with composition into something called a semiring.
22:47 an even smaller mistake is you forgot a minus sign in the (rs) on the left side of the bottommost equation there
You mean near-ring not semiring (although it is easy to forgive this mistake as these algebraic structures are quite obscure). Semirings, also known as "rigs", are what you get when you drop the requirement for additive inverses in a ring, so a semiring is a simultaneous additive semigroup and multiplicative semigroup, as opposed to a ring which is a simultaneous additive group and multiplicative semigroup. the two distributive rules are kept for semirings.
7*7=49 mod 12= 1
Yor forgven Mike.
Polynomials do NOT form a ring with addition/composition because the distributive property does not hold: a∘(b+c) ≠ (a∘b)+(a∘c)
embarrassing slip.... i'll cut it out...
At 6:50, It should be 7^2 = 49 = 1 mod 12.
My answers for these warm-up questions:
Q2,suppose there exists k in R, s.t. ak=1
Since R is a ring, so there exists r in R s.t. k+b=r(additive closure)
then k=r-b
ak=a(r-b)=ar-ab=ar-0=ar
Since the inverse of a is unique, then k=r, then b=0
It's a contradiction.
(bonus)the trivial ring {e}={0}.
Thank you professor for your wonderful explanation.
From what I’ve read, the ‘k’ comes from the German word for field, Körper (meaning body). As for the lowercase question, I’m guessing that really only applies to finite fields, because I used capital letters for number fields. I guess it’s just a way to show that the number field K and its integer ring quotient k are related, and of course K is bigger.
Quoting Edward V. Huntington: « Closely connected with the theory of groups is the theory of fields, suggested by GALOIS, and due, in concrete form, to DEDEKIND in 1871. The word field is the English equivalent for DEDEKIND’s term Körper;. KRONECKER’s term Rationalitätsbereich, which is often used as a synonym, had originally a somewhat different meaning. The earliest expositions of the theory from the general or abstract point of view were given independently by WEBER and by Moore, in 1893 […] »
Yep, field as in Abstract Algebra is in Spanish "Cuerpo", aka body, and not the literal translation of field, "campo".
LMAO, I literally said "If it doesn't make any sense, check the german term" and found it on Wikipedia
@@joelklein3501 we forgiiiiive you Germanyyyy
In swedish it is called Kropp (body)
25:25 {0}
25:12 That homework is a classic
26:15 Good Place To Stop
4:22 polynomials with composition do not form a ring because only one of the distributive laws is satisfied - (working with polynomial functions instead of polynomials,) we have (p+q)∘r = p∘r + q∘r because (p+q)∘r(x) = (p+q)(r(x))= p(r(x)) + q(r(x)) = p∘r(x) + q∘r(x) = (p∘r + q∘r)(x). however for the other distributive law we have p∘(q+r)(x) = p(q(x) + r(x)) ≠ p(q(x)) + p(r(x)) = (p∘q + p∘r)(x) in general, so p∘(q+r) ≠ p∘q + p∘r
(this distributive law holds only when p is a linear polynomial p(x) = rx for some r, and such polynomials are not closed under multiplication so this subset also isn't a ring)
I’m surprised I didn’t know that, thanks! Is there a name for a structure weaker than a ring that polynomials with composition is an instance of?
15:43
Actually, it is not necessary to show they don't have any inverses. The most important condition for a group is that for all a,b€G, a*b€G.
But we want to show that Fn-{0} is a group under multiplication, therefore
a*b = 0 €/ Fn-{0} shows it can't be a group
I'm trying to relearn linear algebra and I'm loving this approach. Very different, but good explanations. I think while the examples are great, I am actually finding the counter examples even more useful.
So what i was thinking about that warm up is that
let there exist an inverse for a, then a^-1(ab)=a^-1(0)=0
so (a^-1*a)b=1*b=b=0
But its not said that b is 0 or it has to be 0, hence there cant be an inverse for a and b.
In Z12, 7 x 7 = 49 = 1 mod 12
From someone who learned this material in the opposite order, I like how you front-load the group theoretical preliminaries. In a pure math context, it makes more sense to learn the material this way.
The determinant, eigenvalues, and eigenvectors of a matrix with real-valued entries have nice geometric interpretations. I look forward to your discussions on objects like M_{nxn}(Z_5).
The letter K
comes from the German word Körper (body). In Portuguese, for example, a similar word is used, corpo, which also stands for body. People rather state results for K
just because it is more general.
Presumably the ring where the additive identity is the multiplicative identity would be the trivial ring, where e*e = e = e + e, and the set is the singleton set, {e}.
Love what you are doing here and kudos for wearing the stream of consciousness type errors. I always find I have to proof read my math and do it again and again and ...
In my country, the common notation for fields is F, with the extra line which R, W, C, Z and N also get when talking about the respective groups of numbers
The question about a ring with 0 = 1 was a fun one to ponder! I believe it must be {0} with addition operation 0 + 0 = 0 and multiplication operation 0*0 = 0.
Proof:
Take any element, a, in a ring, R, where 0 = 1.
a + 0 = a = a*1 - since 0 is additive identity and 1 is multiplicative identity
a + 0 = a*0 - since 0 = 1
a + 0 = 0 - the result a*0 = 0 for all rings is proved in the lecture
a = 0 - once again, 0 is additive identity
okay so this was my attempt at looking at a ring where 0 = 1
suppose we have a ring R,
where, the additive inverse = the multiplicative inverse = e
then, for any r in R
=> r+e = r = r*e
=> (r+e)-r = (r*e)-r
=> e+e = r*(e-e)
=> e = r
this must be true for any r in R, so R must contain only a singular element, e
for a shorter proof you can see that r * e = r since e is the multiplicative identity
but you also have that r * e = e since e is the additive identity, so r = r * e = e -> r = e
k is for Körper, the german term for field in algebra.
22:49 should be -(rs) = r(-s)
Question about finite fields: are there any other finite field that is not isomorphic to F_p or is F_p the complete classification of all finite fields?
Every finite field is has p^n elements for a prime p and a natural number n. On the other hand there exists for all p and n such a field (and its unique).
For n=1 this field is just the integers mod p with the ordinary addition and multiplication. For n>1 the field is NOT (!) the integers mod p^n. Because the only m, such that the ring of integers mod m is a field are the primes (otherwise the divisors of m are zero divisors, which do not exist in fields). These fields with p^n elements for n>1 are harder to describe: One way is to take an irreducible polynomial f over F_p and look at the quotient F_p[x]/(f). This is (isomorphic to) F_{p^n} where n is the degree of f.
At 6:50 7 times 7 = 14!! Is that new math? Not sure if that was supposed to be addition, or an error in the multiplication.
It's also not 14 double factorial 🤪
The operation that he defined combines two objects of the given set by either using addition, multiplication or some other operation that combines the objects. It seems like the operating is addition in the example so 7+7=14 and 14 mod 12 = 2
@@ibrahimn628 Right after that he said anything times 1 is 1, so we have 0, 1, 2, 3. He wasn’t having a great day!
what is a number ring? mentioned at ~20:40
I'm confused. Is this series supposed to precede or follow the Abstract Algebra course you just finished?
(-a+bi)/(ai^2+bi^2) is another complex conjugate. Has anyone noticed this?
directly equal to a-bi/(a^2+b^2) by simplifying the i^2 and cancelling the negatives
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html
P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)