I'm trying to show that $\mathbb{Z}[\sqrt{-5}]$ is not a principal ideal domain without using the fact that all PIDs are UFDs and $\mathbb{Z}[\sqrt{-5}]$ isn't a UFD, as was done in this answer.
I'm curious whether this proof attempt works and secondarily curious whether there's a faster way to do it that's consistent with the general spirit, assuming that the proof is valid.
I'm trying to show that $\mathbb{Z}[\sqrt{-5}]$ is not a principal ideal domain.
Let $h$ be $\sqrt{-5}$.
Proof that the ring $R = \mathbb{Z}[h]$ contains a non-principal ideal. The ring $R$ is a subring of $\mathbb{C}$. $R$ is equipped with a norm $|{\bigcirc}|$. I will treat the norm map as having type $R \to \mathbb{R}_{\ge 0}$. Note that this norm respects multiplication $|ab| = |a||b|$.
Let $I = (2, 1 + h)$. I will now show that $I$ is not a principal ideal.
A unit is an element with a multiplicative inverse.
An irreducible is an irreducible element. An element $a$ is irreducible if and only if $a$ is not 0, a is not a unit, and for all $c$ and $d$ such that $cd = a$, $c$ is a unit or $d$ is a unit.
Lemma 41: $|a+bh|$ is greater than or equal to maximum of $|a|$ and $|b|$.
The absolute value of $h$ in the complex numbers is $\sqrt{5}$, between $2$ and $3$, which is strictly greater than $1$.
It holds that $|a+bi| = \sqrt{|a||a| + |b||b|}$ among complex numbers.
Therefore, the value of $a + bh$ in $R$ is bounded below by the maximum of $a$ and $b$ and bounded above by twice the maximum of $a$ and $b$.
End of proof of Lemma 41.
Lemma 51: An ideal generated by two irreducibles $J = (p,q)$ where $p\nmid q$ is principal if and only if it contains $1$.
Let $J$ be $(p, q)$ where $p$ and $q$ are two irreducibles and $p \nmid q$. Suppose $J$ is principal. Let's name the generator; suppose $J = (i)$.
Since $p$ is in $J$, it holds that $i|p$. Since $q$ is in $J$, it holds that $i|q$. Since $p \nmid q$, any common factor of $p$ and $q$ must be a unit. Thus $i$ is a unit. Thus $J$ contains $ii^{-1} = 1$.
For the converse, if $J$ contains $1$, then it contains everything, including $p$ and $q$.
End of proof of Lemma 51.
Lemma 102: $I$ does not contain $1$.
The formula for an arbitrary element of $I$ is:
$$ (a+b)(2) + (c + dh)(1+h) $$
i.e. the sum of
$$ \begin{bmatrix} 2a & 2bh \\ c & ch \\ -5d & dh \end{bmatrix} $$
The units term is thus $2a + c — 5d$ and the $h$ term is thus $2b + c + d$.
Consider the system of equations:
$$ 2a + c – 5d = 1 \\ 2b + c + d = 0 $$
Consider those equations mod $2$.
$$ c + d \equiv_2 1 \\ c + d \equiv_2 0 $$
This system of equations clearly has no solution, therefore $1$ is not in $I$.
End of proof of Lemma 102.
Lemma 103: $2$ is irreducible in $R$.
The only numbers with an absolute value less than $2$ by Lemma 41, up to multiplication by a unit, are $0$, $1$, $2$, $h$, $1+h$, and $1-h$.
Of these, $0$ is not a factor of anything. $1$ and $2$ are both allowed by the definition of a prime, and each of $h$, $1+h$, and $1-h$ have a modulus that is larger than $2$.
Therefore $2$ is prime in $R$.
Lemma 104: $1+h$ is prime in $R$
The true modulus of $1+h$ is $\sqrt{6}$, which is less than three.
It suffices to consider, up to mulitplication by a unit:
- $0, 1, 2$
- $h, 1+h, 2+h$
- $-h, 1-h, 2-h$
- $2h, 1+2h, 2+2h$
- $-2h, 1-2h, 2-2h$
$0$ is forbidden.
$1$ works and is permitted.
$2$ does not divide $1+h$.
$h$ does not divide $1+h$. Multiples of $h$ have the form $h(a+bh) = -5b+ah$. Therefore the closest nonzero multiples of $h$ to zero are $h(5-h) = 5+5h = 5(1+h)$ and $h(h-5) = -5-5h = -5(1+h)$, neither of which are $1+h$ and both of which are strictly further away from $0$ than $1+h$.
$1+h$ works and is permitted.
$2+h$ has a modulus that is too large.
$-h$ is a multiple of $h$.
$1-h$ has products of the form $(a+5b)+(b-a)h$ where $a, b \in \mathbb{Z}$. Those products are too large unless $b$ is zero, and when $b$ is zero we never get a multiple of $1+h$.
$2-h$ has a modulus that's too large.
$2h$ is a multiple of $h$.
$1+2h$ has a modulus that's too large.
$2+2h$ is a multiple of $1+h$.
$-2h$ is a multiple of $h$.
$1-2h$ has a modulus that's too large.
$2-2h$ is a multiple of $1-h$.
Theorem: $(2, 1+h)$ is a non-principal ideal of $R$.
By 103 and 104, $2$ and $1+h$ are both irreducible. Therefore $I$ is an ideal generated by two irreducibles.
By 102, $I$ does not contain $1$.
Therefore, by 51, $I$ is not principal.
End of proof of theorem.
Best Answer
The basic idea is fine, but there’s a lot of unnecessary material. For checking a specific example, you don’t usually need to prove general statements, and as we see, we only really need that 2 is irreducible and doesn’t divide $1+\sqrt5$.
Set $I=(2,1+\sqrt5)$. Then $I$ is not the unit ideal:
If $1=2(a+b\sqrt5)+(1+\sqrt5)(c+d\sqrt5)$, then $$ 1=2a+c+5d \quad\textrm{and}\quad 0=2b+c+d $$ which as you say cannot be solved (reduce mod 2).
If $I=(x)$ is principal, then $2=xy$, so $4=|x|^2|y|^2$, so $|x|^2$ is either 1, and $x$ is a unit, or 4 and $y$ is a unit, or else $|x|^2=2$. For $x=a+b\sqrt5$ this gives $$ 2 = a^2+5b^2 $$ which again cannot be solved (reduce mod 5).
Thus the only possibility is $I=(2)$, in which case $1+\sqrt5=2z$, so taking norms again gives $6=4|z|^2$, a contradiction. Thus $I$ is not principal.