Question

Studying for a test for tomorrow. Just hoping to throw out some things/questions to make sure I'm on the right track. The test is in Algebraic Structures and the focus is on Rings, Fields, and Integral Domains. I'd also welcome some questions to be thrown at me of material people would expect to be on a test of those topics.

I know this isn't the normal format of a question on this site, but it seems to be fair game and within the spirit of what this site is about :)

Answer 1

I'll use this question to get things started...

Decide whether the indicated operations are defined (closed) on the set, and give a ring structure. If a ring is not formed, tell why this is the case. If a ring is formed state whether the ring is commutative, whether it has unity, and whether it is a field.

\[n \mathbb{Z}\]with the usual addition and multiplication
This is a ring
It is commutative because for all a in Z, na=an
It does not have unity because for all even n, there is no identity.
It is not a field because a field is a division ring with unity and per the previous statement, this ring does not have unity.

Answer 2

prove that if R is a ring and P is a prime ideal then \(R/P\)is a domain
and if M is maximal ideal then \(R/M\) is a field

Answer 3

I found a proof of this in the book. It's a chapter past what tomorrow's test will cover. The only specific type of ideal that we conceivably have to be prepared for is a principal ideal which I do need to take a decent look at tonight. But aside from that the only thing I think I really have to know about ideals for tomorrow is that for a subring I of a ring R for i in I and r in R that ir is in I and ri is in I.

Answer 4

Also any thought on my answers from the first question?

Answer 5

oh sorry i didn't look carefully

Answer 6

One thing is that this class isn't quite as in depth as an Abstract Algebra course is. We do "some" proofs but not as much as a full blown Abstract Algebra. I think the principal ideal is as far as she plans to cover on ideals. She said she is going to skip factor rings all together and I believe she implied the other ideals were more geared towards that.

Answer 7

\(n\mathbb Z\) does not have 1 is not if n is even or odd. it just isn't in there. \(n\mathbb Z\) is an ideal in \(\mathbb Z\) and if any ideal contains 1 it is the whole ring

Answer 8

At this point I'm just looking at nZ as a ring. This question is from the earlier part before we get to ideals. So all I need to find here is if it's commutative, if it has unity, and if it is a field.

I can see why nZ doesn't have a 1 for even n's and even for all odd n's besides one. But doesn't it exist for n=1? I know that's kind of trivial because once I've identified an n for which one does not exist, I've excluded nZ from having unity, but just trying to understand what you are saying.

Answer 9

I just used n is even as one instance of a counterexample, but I do realize that most odd cases will not have an identity either.

Answer 10

for n = 1 you get all of \(\mathbb Z\) back