Abstract Algebra | Principal Ideals of a Ring

[@stanfordjb]

This is a masterpiece.

[@batmanrobinson7882]

Seems like in your proof at 10:00 you need a strictly positive number. If you take 0, you get a trivial ideal instead of something useful.

[@kangablue7372]

I was wondering about that, thanks for clarifying that (as all ideals must contain the zero element).

[@ericvosselmans5657]

if you keep on watching, Michael says exactly that around 14:00

[@darrenpeck156]

Beautiful lecture series, thank you!

[@ibrahimyahmadi3897]

Why did you use the division algorithm and assume that r = 0?

[@harrisonbennett7122]

In the statement of the division algorithm, the condition for the existence of r is that 0

[@MathStudent_Moon_Star]

Super helpful🙏

[@shweta.k6471]

Very helpful

[@arghyachakraborty1151]

I want to ask you, does x-c, c in C belongs to ideal generated by x^2+1 in C[x] that is I=? I think it doesn't belong to, because degree of x-c is 1

[@giannisniper96]

the ideal is made out of polynomials of the form (x^2+1)*p(x) for a generic polynomial p(x), so x-c is not in the ideal

[@انتصارحسينعبابنه]

Thank you Have you an email?

[@sssingh5217]

Thank u

[@janerush1193]

this was helpful thank you

[@Rafau85]

In the first proof you didn't show 0∈(a) and x+y∈(a) for all x,y∈(a). An ideal is a subgroup of R with respect to +!