Abstract Algebra | Units and zero divisors of a ring.

[@oqardZ]

7:30 if 1 (linear comb. of m and n) is multiple of gcd, gcd divides 1 and therefore gcd equals 1.

[@Pika250]

Proof that a unit cannot be a zero-divisor. Let TU = 1. If T were a zero-divisor, there would be an S nonzero such that ST = 0. However, we would have 0 = 0∙U = (ST)U = S(TU) = S∙1 = S, and this is our desired contradiction

[@harsh25186]

@7.06. mx-ny=1.

[@運慶-w3s]

22:34 ±2,±3,±4…is neither a zero divisor nor a unit in Z.

[@prettymuchanobody6562]

I started my sophomore year in college last week Abstract algebra is really confusing to me but your videos are doing wonders Keep it up

[@Maria-yx4se]

I really like the aesthetics of your vids, using chalks and blackboards somehow gives a more formal approach to mathematics.

[@eamon_concannon]

Thanks. I've learned a lot from your number theory and abstract algebra videos, subjects of which I had very little knowledge of till now. However, I have a problem at 14:54. I could not show that ax≠0 (mod n) for all m.n and d satisfying conditions here after spending an entire evening on it. Actually, I think that there is a counterexample to ax≠0 (mod n) in the reverse direction of this proof. Take m = 10, n = 22, then mx + ny = d is 10(-2)+22(1) = 2 so ax = (n/d)x= 11(-2) = -22 = 0(mod 22). m=10 and a=11 is a pair of zero divisors as 10.11 = 0 mod 22 suggesting that a and not ax is the comrade zero divisor for m. Since amx=0 mod n, gcd(x,n)=1 and d

[@mathematicalexpert208]

hi

[@ingevangastel7435]

20:05 I get that this proofs that A is a left zero divisor, but how does it also proof that A is a right zero divisor and therefore a zero divisor?

[@GOATvoldemort]

Can you please solve Putnam 1999 Question no. A2. Please

[@MichaelPennMath]

I'll put it together today!

[@iabervon]

In a finite ring, it can't be neither: Consider all non-0 x. If ax=0, a is a zero divisor. If ax=1, a is a unit. Otherwise, we have |R|-1 non-0 values for x, and |R|-2 non-0, non-1 values for ax, so, by the pigeonhole principle, there are x and y, such that ax=ay and x-y is not 0. But then a(x-y)=ax-ay=0, so a is a zero divisor.

[@mathematicalexpert208]

ples i sport

[@amuthaganesang8353]

Really I easily understand your way of explanation. Really love from India

[@Ali-bp8co]

At 11:11, why can we commute (a and m), (a and n) when we multiply the LHS of mx + ny = 1 by a?

[@franciscoazevedo1476]

Tem uma forma de determinar os divisores de zero sem precisar fazer a tabela de multiplicação? Quero determinar os divisores de zero do anel Z20.

[@dr.shakirmajid9644]

beautifully explained

[@calculus8399]

Nice

[@mathmusingcenter7732]

Very helpful video, totally understand and your approaches very good to accquire the knowledge. From India🇮🇳.

[@stefanandyjapan7293]

Never been so confused in my life

[@mathematicalexpert208]

nice

[@yaxinqi6660]

0 is neither a unit nor a zero divisor, right? Since we have restricted zero divisors to non-zero elements, to begin with?

[@MuffinsAPlenty]

Whether or not 0 is classified as a zero-divisor depends on definition. If most of the rings you care about are all going to be integral domains (for example, if you're going to study algebraic number theory), usually 0 is not considered a zero-divisor. However, if many of the rings you care about will have zero-divisors (for example, if you're going to study algebraic geometry), then usually 0 is classified as a zero-divisor (in nontrivial rings). This is because the theory of zero-divisors is _much_ easier to state when 0 is considered a zero-divisor. Using the definitions given in this video, for sure, 0 is neither a unit nor a zero-divisor.