17:23 Suppose ab∈M and a ∉ M where M is a maximal ideal of the integral domain D. The ideal = + M = D since is larger than M. This means that 1 ∈ so 1=ax+m for some x ∈ D and m ∈ M. Therefore b= bax + bm ∈ M as ab ∈ M, so a maximal ideal of an integral domain is a prime ideal. I have learnt a lot of new and interesting concepts from this abstract algebra series. Great work!
After proving point 1 of the Lemma, the other two points actually follow by observing that saying that a and b are associates is equivalent to saying that a divides b and b divides a (hence (a) is included in (b) and (b) is included in (a)), and that the units of D are precisely the elements associated to the identity 1 (which of course generates the whole D as an ideal). I wanted to thank you for your excellent videos, and ask you what is your favourite book (or books) to learn these topics. I'm currently studying them again on my own on the book Basic Algebra I by N. Jacobson, and although I find it an amazing book (I like its very terse style and its didactic approach, which leaves many important results as exercises), very often I found myself frustrated by some of its most difficult exercises, for which the author provides no hints.
Please try A Book of Abstract Algebra by C. Pinter, Dover Publication. It has solutions for some exercises. Or try Basic Abstract Algebra by R. Ash, also from Dover Publication. Ash's book is for graduate students, it is a wonderful book because it provides solution for all problems.
Slight issue, you claim that in an arbitrary ring "prime < irreducible" but this is only true for domains. You need the cancellation property to prove that every prime is irreducible. p=ab => p|a => a=pr => p=prb *=>* 1=rb.
Hey heads up Michael, there is playlist on your channel called 'The Floor is Lava'. Even though you've made the video inside of it private, we can still see that such a playlist exists so I think there is a seperate option to private the playlist.
That was amazing😊😊😊
17:23 Suppose ab∈M and a ∉ M where M is a maximal ideal of the integral domain D.
The ideal = + M = D since is larger than M. This means that 1 ∈ so 1=ax+m for some x ∈ D and m ∈ M.
Therefore b= bax + bm ∈ M as ab ∈ M, so a maximal ideal of an integral domain is a prime ideal.
I have learnt a lot of new and interesting concepts from this abstract algebra series. Great work!
After proving point 1 of the Lemma, the other two points actually follow by observing that saying that a and b are associates is equivalent to saying that a divides b and b divides a (hence (a) is included in (b) and (b) is included in (a)), and that the units of D are precisely the elements associated to the identity 1 (which of course generates the whole D as an ideal).
I wanted to thank you for your excellent videos, and ask you what is your favourite book (or books) to learn these topics. I'm currently studying them again on my own on the book Basic Algebra I by N. Jacobson, and although I find it an amazing book (I like its very terse style and its didactic approach, which leaves many important results as exercises), very often I found myself frustrated by some of its most difficult exercises, for which the author provides no hints.
Please try A Book of Abstract Algebra by C. Pinter, Dover Publication. It has solutions for some exercises. Or try Basic Abstract Algebra by R. Ash, also from Dover Publication. Ash's book is for graduate students, it is a wonderful book because it provides solution for all problems.
@@DrPeterkuah thank you very much for your recommendations! I'll definitely check them out.
Thanks this help me in my 8th grade math studies
In the proof for 3 you can just use the fact that as a is a unit the ideal generated contains 1
6:50 It appears that b must be non-zero here to ensure that b(1- xy)= 0 implies 1 - xy = 0 by cancellation rule in an integral domain.
Slight issue, you claim that in an arbitrary ring "prime < irreducible" but this is only true for domains.
You need the cancellation property to prove that every prime is irreducible.
p=ab => p|a => a=pr => p=prb *=>* 1=rb.
link for this playlist please?
Hey heads up Michael, there is playlist on your channel called 'The Floor is Lava'. Even though you've made the video inside of it private, we can still see that such a playlist exists so I think there is a seperate option to private the playlist.
The video becomes public later this morning. I am going to put all of my videos involving the floor function in this playlist.
To contain is to divide!