5

Where $a, b, c \in \mathbb Z$?

I know that if in an UFD, $\langle c, a + b \sqrt{d} \rangle$ would boil down to a principal ideal. But it seems to me that in $\mathbb Z[\sqrt{-5}]$, for any purely real integer $c$ that is irreducible, the ideal $\langle c \rangle$ can be properly contained in some suitable choice of $\langle c, a + b \sqrt{-5} \rangle$ that is itself properly contained within the ring.

quid
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John-Luke
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3 Answers3

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Yes. Let $I$ be an ideal of $\mathbb{Z}[\sqrt{-5}]$. Then $I \cap \mathbb{Z}$ is an ideal of $\mathbb{Z}$ and hence principal, take $I \cap \mathbb{Z} = (c)$. Then, as abelian groups, $I/c \mathbb{Z}$ injects into $\mathbb{Z}[\sqrt{-5}]/\mathbb{Z} \cong \mathbb{Z}$. Let the image be generated by the coset $\mathbb{Z}+b \sqrt{-5}$ and lift this coset to $a+b \sqrt{-5}$ in $I$. Then $a+b \sqrt{-5}$ and $c$ generate $I$ as an abelian group and, therefore, as an ideal.

David E Speyer
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    I feel this does not answer the question, which is in my understadning rather if it is always necessary to have two elements. See the body of the question and the comment. – quid Apr 14 '15 at 13:50
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    @quid It definitely answers the question in the quote box. The OP's comments after that leave me a little confused, though. Possibly, he or she doesn't understand that an ideal can have many different generating sets? EG $\langle 1+\sqrt{-5} \rangle = \langle 2, 1+\sqrt{-5} \rangle$, so this ideal is both of the form $\langle a+b \sqrt{-5}, c \rangle$ and $\langle d+e \sqrt{-5} \rangle$. – David E Speyer Apr 14 '15 at 13:58
  • The quoted box is there since 5 minutes only and not due to OP. The question is in my reading if there exist (nonzero) pinicipal prime ideals (the OP thinking they do not exist). – quid Apr 14 '15 at 14:00
  • If John-Luke knows about "disguised" principal ideals in PIDs, surely he knows about different generators, e.g., $\langle 12, 20 \rangle$ for $\langle 4 \rangle$ in $\mathbb{Z}$, and that principal ideals can be similarly disguised in non-PIDs. – Robert Soupe Apr 15 '15 at 01:03
  • I'm not disagreeing with you about different generating sets in general, but I'm confused by your example of $\langle 1 + \sqrt{-5} \rangle = \langle 2, 1 + \sqrt{-5} \rangle$. Since $N(1 + \sqrt{-5}) = 6$, how can $\langle 1 + \sqrt{-5} \rangle$ contain numbers of norm 4? – Robert Soupe Apr 15 '15 at 01:06
  • @RobertSoupe You're absolutely right, oops! $\langle 1+\sqrt{-5} \rangle = \langle 6, 1+\sqrt{-5} \rangle$, not what I wrote. – David E Speyer Apr 15 '15 at 02:58
5

A Dedekind domain $\,D\,$ satisfies what's known as the $\,1\ 1/2\,$ (one and a half) generator property: given an ideal $\,I\subset D\,$ and $\,0\neq i\in I\,$ then there is some $\,j\in I\,$ such that $\,I = (i,j).\,$

For example, if $\, (n) = I\cap \Bbb Z\,$ then $\, I = (n,w)\,$ for some $\,w,\, $ as you seek.

Bill Dubuque
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5

No, it is not the case that for any irreducible integer $c$ the principal ideal $(c)$ is properly contained in an $(c, a + b \sqrt{-5})$ that is properly contained in the full ring.

If $p$ is a prime number number that is inert in $\mathbb Z[\sqrt{-5}]$ then essentially by definition $(p)$ is a prime ideal in $\mathbb Z[\sqrt{-5}]$.

Now, as a Dedekind domain $\mathbb Z[\sqrt{-5}]$ is one-dimensional and thus each non-zero prime ideal is maximal. Thus for an inert prime $p$ the ideal $(p)$ is not properly contained in any ideal but the full ring.

As explained in other answers it is possible to write every ideal in the form $(c, a + b \sqrt{-5})$, and in particular $(p)$, in the form $(p, a + b \sqrt{-5})$, but in the case I mention there is no proper containment.

It is also true that every prime ideal of $\mathbb Z[\sqrt{-5}]$ contains some prime number and thus the principal ideal $(p)$ for some prime number $p$. Thus, if you drop the condition that $(p)$ is properly contained in $(p, a + b \sqrt{-5})$ then (but only then) what you say seems to be true is in fact true.

Community
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quid
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  • I think this is a good answer, I think it's the one the OP will accept. But, for the sake of less algebraically sophisticated people like me, could you give a concrete example? Such as take $13$, which is definitely irreducible in this ring, and maybe also prime. If $\langle 13 \rangle$ is a prime ideal, then any $\langle 13, a + b \sqrt{-5} \rangle$ turns out to be the whole ring, right? – Mr. Brooks Apr 14 '15 at 20:39
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    Thanks for the feedback. Indeed, $13$ is inert as it does not divide the disrimianant, i.e. $-20$, and $-5$ is not a quadratic residue modulo $13$, that is $-5$ cannot be written as $a^2$ modulo $13$. However, it is not exactly true that $(13, a + b \sqrt{-5})$ is the full ring for every choice of $a,b$. If $a=b=0$ or $a=b=13$ or $a= 13, b=-26$ and so on, then this ideal is just $(13)$. It is however true for any $a + b \sqrt{-5}$ outside $(13)$ that is not divisible by $13$. – quid Apr 14 '15 at 21:07
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    So any choice of $a \equiv 0 \pmod {13}$ and likewise for $b$ makes $\langle 13, a + b \sqrt{-5} \rangle = \langle 13 \rangle$ and otherwise $\langle 13, a + b \sqrt{-5} \rangle = \mathbb{Z}[\sqrt{-5}]$? – Mr. Brooks Apr 14 '15 at 21:23
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    Yes. $a+b\sqrt{-15}$ is a mulitple of $13$ if and only if both $a$ and $b$ are multiples of $13$. In this case the ideal is unchanged. Otherwise it is the full ring. – quid Apr 14 '15 at 21:26