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Let $R$ be a unique factorization domain and let $a,b\in R$ be distinct irreducible elements. Could anyone tell me which of the following is true?

  1. $\langle 1+a\rangle$ is a prime ideal.

  2. $\langle a+b\rangle$ is a prime ideal.

  3. $\langle 1+ab\rangle$ is a prime ideal

  4. $\langle a\rangle$ is not necessarily a maximal ideal.

I remember the definition of irreducible element: an element $f\in R$ such that there does not exist non-units $g,h$ such that $f=gh$, and in a UFD, prime and irreducible elements coincide. $4$ is prime ideal right?

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5  
It is easy to find counterexamples to the first three by just looking at $\mathbb{Z}$. –  Tobias Kildetoft Aug 2 '13 at 11:52
    
you're right that $4$ will be prime, so think about examples of prime ideals that aren't maximal –  citedcorpse Aug 2 '13 at 11:53
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< and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. –  Zev Chonoles Aug 2 '13 at 11:53
    
@exitingcorpse (and Taxi Driver): $4$ is not prime... –  Zev Chonoles Aug 2 '13 at 12:06
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@ZevChonoles I'm pretty sure that exitingcorpse meant point 4., $\langle a\rangle$ will be a prime ideal (trivially). –  Daniel Fischer Aug 2 '13 at 12:10

1 Answer 1

up vote 4 down vote accepted

Maybe it's me, but I don't get your question. If I understand your question correctly, you're asking that if a and b are irreducible elements in a ring R, which of those ideals must be prime? Is that it? If yes, then the answer is none except the fourth one.

For 1., Take $R=\mathbb{Z}$, then $5$ is irreducible in $\mathbb{Z}$ but $\langle 1+5 \rangle = \langle 6\rangle $ is not a prime ideal.

For 2., set $a=5$ and $b=3$, both are irreducible but $\langle 8\rangle$ is not a prime ideal of $\mathbb{Z}$

For 3., again set $ a=5$ and $b=3$.

For 4., since a is irreducible $\langle a \rangle$ is a prime ideal, is it maximal? well, not necessarily. Take $\langle x^2+y \rangle $ in $\mathbb{R}[x,y]$, it's a prime ideal since $x^2+y$ is irreducible, but it's not maximal since $\langle x^2+y \rangle \subset \langle x,y \rangle$.

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or even more simply, take $\langle x \rangle$. certainly $\mathbb{R}[x, y]/ \langle x \rangle = \mathbb{R}[y]$ is an integral domain, so the ideal is prime, but far from maximal (any of the ideals $\langle x, y^n \rangle$ will contain it) –  citedcorpse Aug 2 '13 at 12:47
    
@exitingcorpse: Yea. That's a better example. –  some1.new4u Aug 2 '13 at 13:10

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