Abstract Algebra | Some basic exercises involving rings.

We present some basic results involving rings.
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[@muckchorris9745]

First lemma was a very nice short proof, havent seen b4 thanks!

[@darrenpeck156]

Clever way of constructing an inverse for 1 + nullpotent.

[@kalebyirgalem6688]

at 8:47 the factiorization of 1+x^m only work if m is odd ??

[@ethanbottomley-mason8447]

In a boolean ring, wouldn't you have (-1)^2 = -1 but also have (-1)^2 = 1 meaning -1 = 1. Does this mean that the multiplicative identity is the same as the additive identity? If 1 = -1, then 0 = 1 + 1 which means that 0 = 1 right?

[@nickyin7781]

maybe -1 isn‘t inside the Boolean ring

[@ethanbottomley-mason8447]

@@nickyin7781 Z/2Z is a boolean ring which makes it clear that this makes sense. In Z/2Z, 1 + 1 = 0 and -1 = 1.

[@valeriobertoncello1809]

Everything you said except 0 = 1, because 0 = 1 + 1 =/= 1 in a boolean ring

[@dehnsurgeon]

at 9:00; isn't that factorisation only correct if (-1)^(m-1) is 1

[@Jkfgjfgjfkjg]

It doesn't matter. Even if (-1)^(m - 1) is -1, when you multiply the two factors you get 1 - x^m which is still 1, because x^m = 0.

[@carl3260]

16:05: how is multiplicative commutativity assumed?

[@余淼-e8b]

Z[i] is a polynomial ring whose coefficients are in Z. Since Z is commutative, Z[i] is commutative.

[@cykkm]

Hi Michael, I'm going through ring theory part only in the Abstract Algebra playlist order (kzbin.info/aero/PL22w63XsKjqxaZ-v5N4AprggFkQXgkNoP), thank you, it's an excellent refresher! Just wanted to let you know, this vid comes as #58 in the list, but I think it should be after #62: units and zero divisors are introduced in #59, and homomorphisms in #62.