$\mathbb{Z}[\sqrt d]$ and gcd(which point do I have a wrong?)

Question

For the ring $\mathbb{Z}[\sqrt -5]$ and say its ideal $P = \langle 3, 1+ \sqrt -5 \rangle$

Since $\langle a, b \rangle = \langle d \rangle$ where $d= (a,b)$ if the gcd, d exists

Then $\langle 3, 1+ \sqrt -5 \rangle = \langle d \rangle$ where $d= (3, 1+ \sqrt -5)$

Put the norm $N(a) = (r+s \sqrt t)(r-s \sqrt t) = r^2 + s^2 t$ for $a (= r+s \sqrt t) \in \mathbb{Z[\sqrt t]}$

Hence $N(d) \vert N(3) = 9$ and $N(d) \vert N(1+ \sqrt -5) = 6$

$N(d) = 1$ or $3$.

But there are not $d$ $s.t.$ $N(d) = 3$ for $d \in \mathbb{Z}[\sqrt -5]$

So, $N(d) = 1$ Therefore $d$ is the unit of the $P $

Since $\mathbb{Z}[\sqrt -5]$ is commutative ring with unity and unit $d \in P$

Threfore, $P$ = $\mathbb{Z}[\sqrt -5]$

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Question) It is constadict with my proof's conclusion and the fact I knew that $P \neq \mathbb{Z}[\sqrt -5]$.

Then which point do I have a wrong? I can't figure out what point is error.

Answer 1

You've got your causality wrong between greatest common divisors and generators of ideals.

Let $R$ be a commutative ring with $1$, and let $a,b\in R$. A greatest common divisor of $a$ and $b$ is an element $d\in R$ such that:

1. $d|a$ and $d|b$; and
2. If $c|a$ and $c|b$, then $c|d$.

Greatest common divisors need not exist; but when they exist, they are unique up to the equivalence relation of "associates" ($x$ and $y$ are associates if and only if they divide each other). In domains, they are unique up to units.

Now, if the ideal $(a,b)$ is principal, $(a,b)=(d)$, then $d$ will be a greatest common divisor of $a$ and $b$.

Indeed, $a,b\in(d)$, hence there exist $x,y\in R$ with $dx=a$ and $dy=b$, so $d|a$ and $d|b$. Moreover, since $d\in (a,b)$, there exist $r,s\in R$ such that $d=ra+sb$. Thus, if $c|a$ and $c|b$, then $c|(ra+sb)=d$. Hence, $d$ is a greatest common divisor of $a$ and $b$.

However, it is possible for $a$ and $b$ to have a greatest common divisor $d$, but for the ideal $(a,b)$ to not be equal to the ideal $(d)$. For example, in a UFD, any pair of elements have greatest common divisors. So in $\mathbb{Z}[x]$, $x$ and $2$ have greatest common divisors: in fact, since they are both prime but not associates, their greatest common divisors are $1$ and $-1$. But the ideal $(x,2)$ is not equal to the ideal $(1)$. This is because in arbitrary domains, it is not the case that the greatest common divisor can be written as a linear combination of $a$ and $b$. (Rings in which this can be done for any pair of elements are called Bezout domains, which is a stronger property than just any pair of elements having a greatest common divisor).

So, you write:

Since $\langle a,b\rangle = \langle d\rangle$ where $d=(a,b)$ if the gcd, $d exists

but this is wrong. You are saying that if the gcd exists, then it must generate the ideal $(a,b)$ which is therefore principal.

But the implication goes the other way: if the ideal $(a,b)$ is principal, then its generator is a gcd of $a$ and $b$.

So, you show that in $\mathbb{Z}[\sqrt{-5}]$, any common divisor of $3$ and $1+\sqrt{-5}$ must have norm $1$, and therefore be a unit. That means that the greatest common divisor of $3$ and $1+\sqrt{-5}$ is $1$ (or $-1$). There are no problems here.

But what does this mean? It means that if the ideal $(3,1+\sqrt{-5})$ is principal, then it would have to be equal to the ideal $(1)$, because the generator would have to be the gcd of $3$ and $1+\sqrt{-5}$, which is $1$. But we know that $(3,1+\sqrt{-5})$ is not the ideal $(1)$. That means that this ideal cannot be a principal ideal. So $\mathbb{Z}[\sqrt{-5}]$ is not a Principal Ideal Domain.

(And since it is a Dedekind domain; or since it is the ring of integers of a finite extension of $\mathbb{Q}$, this also implies that it is not a UFD)