Prime and Maximal Ideals -- Abstract Algebra 20 

31:44 what you call the “nilradical” of an ideal is usually just called the “radical”, while the word nilradical usually only refers to the radical of the zero ideal.

Michael, drink some tea, eat some chicken soup and go to bed and nurse that upper respiratory tract ailment! You’re too important to risk your voicebox. 👍🏻☕️🍜

Are you planning on covering modules inside this series?

8:20 seems like in the left diagram there should be an arrow from (4) down to (12) instead of the arrow from (4) to (6)

@40:00 this argument doesn’t seem to work when n or m equal to 1. Then it’s not true that either nm-k or k is greater than equal to max{m,n} But that’s okay because if n or m equal to 1, then that means either r or s are in the original ideal I. Therefore, (r-s)^nm is in the ideal I since every term in the binomial expansion is in I from I being an ideal and closure in I.

Several important subtle points were overlooked in this video. For the proofs of the backwards directions of the results "P is prime ⇔ R/P is an integral domain" and "M is maximal ⇔ R/M is a field" you need to prove that P and M are proper ideals, as this is part of the definition of being a prime/maximal ideal - this is true because the trivial ring 0 is neither an integral domain nor a field and R/I ≅ 0 ⇒ I = R. Next I think it would have been beneficial to write down the proof that every maximal ideal is prime rather than just saying it - we have: I is maximal ⇔ R/I is a field ⇒ R/I is an integral domain ⇔ I is prime. For the very first result ("claim") the proof was incomplete; to say "the prime ideals of R are those of the form ..." is an if and only if statement: any prime ideal is of that form, and conversely any ideal of that form is prime. So the claim means "I ⊆ *Z* is prime ⇔ I = {0} or I = p *Z* where p is prime". you proved the forwards direction and half of the backwards direction (I = {0} ⇒ I is prime) - the other half of the backwards direction amounts to what is in the red box on the left side of the board ("compare"): this can be phrased in terms of the principal ideal p *Z* as "when ab ∈ p *Z* we know a ∈ p *Z* or b ∈ p *Z* ". Lastly in the example at 30:47 you need to show that both factors are nonzero, i.e. 1+i + I ≠ 0 + I and 1-2i + I ≠ 0 + I which is equivalent to 1+i, 1-2i ∉ I = (3-i). that is, 1+i and 1-2i are not Gaussian integer multiples of 3-i - this is easy to show but important not to overlook. For the two examples on this board, the former benefits from the classification of prime and maximal ideals since the isomorphism *Z* [x]/(x) ≅ *Z* is very easy, but the later does not benefit; phrasing it in terms of the quotient ring *Z* [i]/(3-i) only obscures things: working directly in *Z* [i] the proof that (3-i) is not prime becomes just 3-i = (1+i)(1-2i) ∈ (3-i) but 1+i ∉ (3-i) and 1-2i ∉ (3-i). And finally at 32:52 {0, 6} = (6) is typo'd as "{0, 6} = " ( is sometimes used as an alternative notation for (a), as you have said, but based on the rest of the videos it's clear that in this course you're using "" and "" for cyclic subgroups and generated subgroups and "(a)" and "(a_1, a_2, ...)" for principal ideals and generated ideals)

40:32 someone already pointed out that it should be (r-s)^(m+n). There’s a second mistake though that (-1) should not be assumed to exist because we are given a commutative ring and unity may not exist. So the binomial expansion should instead be (-s)^k. It shouldn’t be hard to show that (-s)^n belongs to I and use the same argument to conclude that radical of I is a sub ring.

0:56 the statement is false: 1 is not prime, but 1 divides any integer.

15:43 don't we need to show that if I is not equal to R[x] that I = (x^2 + 1)? If not, why not?