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?
$\langle 1+a\rangle$ is a prime ideal.
$\langle a+b\rangle$ is a prime ideal.
$\langle 1+ab\rangle$ is a prime ideal
$\langle a\rangle$ 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?
Best Answer
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 $\langle x^2+y \rangle \subset \langle x,y \rangle$.