Abstract Algebra | Maximal and prime ideals.

We prove some classic results involving maximal and prime ideals. Specifically we prove the an ideal P is prime iff R/P is an integral domain. Further, we prove that an ideal M is maximal iff R/M is a field.
www.michael-penn.net
www.researchgate.net/profile/...
www.randolphcollege.edu/mathem...

[@werner0prinz]

Sie sind ein ausgezeichneter Mathematiker, der einfach super erklären kann. Genau dieser Fall, wenn R ein Integritätsring oder ein kommutativer Ring mit 1 sei bzw. ein Hauptidealring ist, habe ich das mit den maximalen und primen Idealen nie verstanden, Sie konnten es leicht verständlich und ausführlich erklären, trotz meinen bescheidenen Kenntnissen in Englisch. In Deutschland fand ich hierzu weder in den Fachbüchern noch im deutschen Internet vergeblich eine so einleuchtende Erklärung.

[@YitzharVered]

Always great to have some Michael Penn before an abstract Algebra exam! Very helpful for Group theory, lets hope this helps me with rings!

[@darrenpeck156]

Thank you. Wonderful standard of lecturing.

[@darrenpeck156]

Are we using the correspondence theorem for the maximal ideal and field proof? must correspond to I between M and R.

[@madhavestark3173]

For the prime ideal, shouldn't the ideal be a proper subring of R

[@phuocbui8182]

What is define of and and why (a +M)(b+M) is in ?, thank you very much

[@darrenpeck156]

How do the generators a and M work for the ideal?

[@asimislam6194]

Hi Fantastic thanks, please can you tell me (at 11.30) why 1+M must be in the ideal (a,M) ? You chose b in R to find the inverse of (a+M) but will this mean 1+M has to be in the ideal (a,M) ....please explain, thank you

[@dibeos]

I do not see the difference between the definition of a prime ideal and an ideal (in general). In both cases, whenever ab belongs to an ideal I, either a is in I and b is any element of the ring R, or vice-versa. So what’s the difference between the two definitions?

[@natepolidoro4565]

No not necessarily

[@dibeos]

@@natepolidoro4565 thank you for the detailed explanation.

[@henk7747]

@@dibeos A prime ideal is just an ideal with the extra condition that ab in P => a in P or b in P Here is an ideal which is not a prime ideal: Consider 4Z = {4x | x is in Z} This is an ideal of Z because it is a subgroup of Z (under addition), and it absorbs elements under multiplication, in the sense that any integer times a multiple of 4 is still a multiple of 4. It is not a prime ideal because 2*2 is in 4Z but 2 is not in 4Z.

[@gregnisbet]

An ideal I is a subset of a ring R that is closed under addition (a in I and b in I implies a + b in I) and under arbitrary multiplication (r in R and a in I implies ra in I). Let R \ I denote the elements of R that are not elements of I. This is different from the notation R/I which denotes the quotient ring. A prime ideal I is one where R \ I is closed under multiplication. For example, in the integers, (2), the ideal generated by two contains all the even numbers. Its complement is the set of odd numbers and odd numbers are closed under multiplication. As others have pointed out, not every ideal is a prime ideal, for example (4) or (10). As another example, consider the zero ideal (0). It consists only of 0 and nothing else. It's always an ideal, but whether it's a prime ideal or not depends on the ring. In the integers, this ideal is prime (since the product of two nonzero integers is nonzero). In Z/7Z (the integers mod 7), (0) is also prime. However, in Z/10Z, (0) is not a prime ideal because 2 * 5 = 0 (mod 10), so the nonzero elements of Z/10Z are not closed under multiplication.

[@natepolidoro4565]

4:38 It seems like this theorem doesn't use the fact that R is commutative and has an identity, so it is true for all rings R. Am I wrong?

[@serkanbasatlk3322]

I think you're wrong

[@astriiix]

integral domains require the ring to be commutative and provided with identity before the condition of not having zero divisors, so it is a needed but not useful condition for the proof

[@hybmnzz2658]

Way too late to say this but the thumbnail has a mistake