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I have the following question:

Answer the following questions for each part:

  1. Is it an integral domain?
  2. Is it a principal ideal domain?
  3. Is it a field?

a. $\mathbb{Z}/13\mathbb{Z}$;

b. $\mathbb{Z}/20\mathbb{Z}$;

c. $\mathbb{Q}[X]/(f)$ with $f=X^2+X+1$;

d. $R[X]$ with $R := \mathbb{Z}/4\mathbb{Z}$;

e. $R[X]/I$ with $R:=\mathbb{Z}/2\mathbb{Z}$ and $I=(X^2+X+1)$.

For a. I know that it is an integral domain because 13 is prime, also that this is then a field however I don't know how to show/disprove that it is a PID.

Any help with this would be great, thank you.

In the question $\mathbb{Z}$ are the integers, $\mathbb{Q}$ are the rational numbers. Thanks.

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    $\begingroup$ Rather than asking 24 questions of varying difficulty in a single post, you should rather focus on the one (or two) you are currently stuck on. $\endgroup$ – rschwieb May 12 '14 at 18:01
  • $\begingroup$ Sorry, didn't realise that I asked 24 questions, I thought I asked 8. I put a lot up to get peoples opinions and extra help, if you don't want to answer it don't, no need to post sarcastic comments. $\endgroup$ – Justin May 12 '14 at 18:03
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    $\begingroup$ Dear Justin: Eight is still way more than one or two... and I didn't mean to sound sarcastic at all. I'd like you to get the best help possible, and asking a focused question is a sure way to do that :) $\endgroup$ – rschwieb May 12 '14 at 18:03
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    $\begingroup$ do not bite that delicious hand feeding you, as my grandmother always told... $\endgroup$ – Max May 12 '14 at 18:10
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    $\begingroup$ Sorry @rschwieb I didn't see your post above, I misunderstood what you were saying $\endgroup$ – Justin May 12 '14 at 18:14
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For 1. I know that it is an integral domain because 13 is prime, also that this is then a field however I don't know how to show/disprove that it is a PID.

Hint #1:

You should try to get your head around this conceptual chain:

fields $\subseteq $ principal ideal domains $\subseteq$ domains

Then you can go through the list and try to determine which ones are fields and which aren't. Once you identify fields, you don't have to revisit them again. Then ask about PIDs...

Hint #2:

Prove for yourself that the quotient of a principal ideal ring is still a principal ideal ring. That means, of course, any quotients of principal ideal domains are principal ideal rings. A quotient of a PID by a prime ideal would again be a PID.

Hint #3:

It should definitely have been proven in your text or class that $F[x]$ is a PID for any field $F$. The proof usually uses the division algorithm to demonstrate that there is a unique monic polynomial of minimal degree that generates each nonzero ideal.

Hint #4:

In a couple cases above, it's easy to notice by observation that the ring contains two nonzero elements that multiply to zero. Those can't be domains of course (or PIDs or fields...). One of the rings even contains nonzero elements which square to zero...

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    $\begingroup$ Thank you, I know what if something is an integral domain then it implies that it is a field, because obviously it contains all units, I also know that a PID means that each ideal is a principal ideal, i.e for I in R we have that I:=(n) but I just don't know how to show when this is the case $\endgroup$ – Justin May 12 '14 at 18:07
  • $\begingroup$ @Justin I have a feeling your asking, basically, "How do you tell when a domain is principal or not?" $\endgroup$ – rschwieb May 12 '14 at 18:12
  • $\begingroup$ yes, that's exactly it @rschwieb. I have been trying to work it out for weeks. For example in Z/13Z is it something to do with each element being applied to itself over and over again giving an ideal? i.e. 3^2=9, 3^3=27=1, then that means that it is an ideal? This could be completely wrong $\endgroup$ – Justin May 12 '14 at 18:17
  • $\begingroup$ $\Bbb Z/13\Bbb Z$ is a quotient ring of $\Bbb Z$. If you are a bit fuzzy on what quotient rings look like, then you'll have some luck if you search around on this site for questions about quotient rings. Many people have asked for explanations about what quotient rings look like in various cases, including ones you mentioned. $\endgroup$ – rschwieb May 12 '14 at 18:18
  • $\begingroup$ Thanks for your hints, they were helpful I knew 1 and 3, however I didn't know 2, so thank you! So for example ℤ[X] is a principle ideal domain because ℤ is? $\endgroup$ – Justin May 12 '14 at 18:22

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