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I know the following fact.

An integral domain $R$ is UFD $\iff$ any nonzero prime ideal of $R$ possess a nonzero prime element.

In other words, if an integral domain is not a UFD, there must exist at least one prime ideal that does not contain any prime element. However, I’m struggling to come up with an example of such a prime ideal of a domain.

For instance, $\mathbb Z[\sqrt{-5}]$ is not UFD but which prime ideal in this ring not containing a prime element?

Fuat Ray
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2 Answers2

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Idea: you could start with irreducible elements, which are not prime, e.g., $2,3,1\pm \sqrt{-5}$. However, the ideals $\mathfrak{p}=\left<2,1+\sqrt{-5}\right>, \mathfrak{q}=\left<3,1+\sqrt{-5}\right>$ are prime in $\mathbb{Z}[\sqrt{-5}]$.

Edit: For a proof that $\mathfrak{p}$ doesn't contain a prime element $\pi$, I had in mind that the norm of the ideal $\mathfrak{p}$ is $N(\mathfrak{p})=[\Bbb Z[\sqrt{-5}]:\mathfrak{p}]=2$, and that $\pi \mid N(\mathfrak{p})=2$. So, for $\pi=a+b\sqrt{-5}$, we have $N(\pi)\mid N(2)$, i.e., $$ a^2+5b^2\mid 4 $$ in $\Bbb Z$, which yields $(a,b)=(\pm 2,0),(\pm 1,0)$. But $\pi=2$ is not prime in $\Bbb Z[\sqrt{-5}]$, see below:

Proving whether ideals are prime in $\mathbb{Z}[\sqrt{-5}]$

Irreducible but not prime in $\mathbb{Z}[\sqrt{-5}] $

Dietrich Burde
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Let me expand on Dietrich Burde's answer.

Let $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$.

  • Direct computations shows that we have $I=\{a+b\sqrt{-5}\mid a\equiv b \ \mod 2\}$. It is not difficult to deduce that $I$ is maximal, hence prime.

First $, I\neq R$. Moreover, let $J$ an ideal of $R$ containing $I$.

Assume $J\neq I$. If $z=a+b\sqrt{-5}\in J\setminus I$, then $a-b$ is odd . Hence $a=b+2m+1$ for some $m\in \mathbb{Z}$. But $z=a-b+b(1+\sqrt{-5})=1+2m+b(1+\sqrt{-5})$. Since $J$ contains $I$, it contains $2$ and $(1+\sqrt{-5})$. But is also contains $z$, and since $J$ is an ideal, we deduce that $J$ contains $1$. Hence $J=R$ and $I$ is maximal.

  • If $\pi=a+b\sqrt{-5} \in I$ is prime, $(\pi)$ is a prime ideal of $R$ , so $(\pi)\cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. This intersection is not zero since it contains $\pi \bar{\pi}=\vert \pi\vert^2$. Hence $(\pi)\cap \mathbb{Z}=p\mathbb{Z}$, where $p$ is a prime number.

Now $(\pi)\cap \mathbb{Z}=p\mathbb{Z}\subset I\cap\mathbb{Z}= 2\mathbb{Z}$. Hence $2\mid p$ is $\mathbb{Z}$, and since $p$ is a prime number, $p=2$.

  • We finally get that $(\pi)\cap \mathbb{Z}=2\mathbb{Z}$. Hence $2 \mid \pi $ in $R$ and $\vert \pi\vert^2\mid 4$ in $\mathbb{Z}$. Hence $a^2+5b^2=1,2,$ or $4$, and simple then computations then show that $\pi=\pm 1,\pm 2$.

Since $\pm 1$ are units, we are left to consider $\pi=\pm 2$.

These two elements are not prime: we have $(1+\sqrt{5})(1-\sqrt{5})=6\in 2R$, but $1\pm \sqrt{-5}\notin 2R$.

GreginGre
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  • We have $N(\pi)=a^2+5b^2$, dividing $4$. How do you obtain $a=b=\pm 1$ then? – Dietrich Burde May 02 '25 at 10:01
  • Sorry, I made a stupid mistake. I modified my answer, everything should be correct now. – GreginGre May 02 '25 at 22:47
  • If it's not too hard to explain, what method would you use to show that $I$ is maximal? The primary way I'm familiar with is to show that $R/I$ is a field, but I'm not immediately seeing it. (I'm still learning algebra, so this might be a "me" problem.) – apnorton May 03 '25 at 00:31
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    Done. My original proof of $\pi\mid 2$ was flawed, so added details. Basically, This is the same idea as Dietrich's answer,but with avoiding the norm of an ideal (which is the right way to think about it though) – GreginGre May 03 '25 at 07:56
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    @apnorton I have included two links in my answer about this, the first one shows that the quotient is a field (where you say you are not immediately seeing it, which is understandable). – Dietrich Burde May 03 '25 at 08:09