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Or equivalently, in which $\mathbb{Z}[\zeta_m]$ is $2$ reducible? And how does one construct any such divisors?

$\bullet\ \textbf{My attempt}$

The smallest example seems to be $m=4$ with $2=(1+i)(1-i)$. I thought at first I'd found other examples like $(2\zeta_3+2)(-\zeta_3)=2$ in $\mathbb{Z}[\zeta_3]$. But this is is really just unit migration since $(\zeta_3+1)(-\zeta_3)=1$. I looked at $\mathbb{Z}[\zeta_5]$ and $\mathbb{Z}[\zeta_7]$ and found only the same false positives. So in general I'd like to exclude factorizations of the form $(2u^{-1})(u)=2$.

The only attempt I've managed so far is to note that, given $\alpha \in \mathbb{Z}[\zeta_m]$ and an automorphism on $\mathbb{Q}(\zeta_m)$, $$\phi_k:\zeta_m \rightsquigarrow \zeta_m^k\quad\text{where}\quad \gcd(k,m)=1,$$ if $\alpha|2$ then $\phi_k(\alpha)|2$ also. Thus we could make use of the homomophism $$F(\alpha)=\prod_{(k,m)=1}\phi_k(\alpha)\in\mathbb{Z}$$ noting that $F(\alpha)|2^{\varphi(m)}$. For example, if $m$ is divisible by a prime $p$ of the form $4k+3$, then $F(\alpha)$ will be of the form $a^2+pb^2$ with $a,b\in\mathbb{Z}$. And there are only a finite number of choices for $a,b$ to check. For example, following this method for $p=7$ yields $$1+\zeta_7 + \zeta_7^2+\zeta_7^3|2^5.$$ But I didn't know how to carry on from there.

Some SageMath reveals that $m=7,15,17$ yield nontrivial factorizations of $2$.

$\bullet\ \textbf{Motivation}$

I was trying to categorize Pythagorean triples in general rings of integers (focusing on $\mathbb{Z}[\zeta_m]$ in particular). It seems that for $\alpha,\beta,\gamma\in K $ satisifying $\alpha^2+\beta^2=\gamma^2$, and supposing that the triple is not a multiple of a smaller example, it follows that $$\alpha=\sigma\mu\nu,\quad 2\beta=\tau^+\mu^2-\tau^-\nu^2,\quad\text{and}\quad 2\gamma=\tau^+\mu^2+\tau^-\nu^2$$ where $\mu,\nu,\tau^\pm,\sigma \in K$, $\tau^-\tau^+=\sigma^2$, and $\tau^+,\tau^-|2$. Thus a categorization of the divisors of $2$ also yields a categorization of Pythagorean triples. Choosing $\tau^+=\tau^-=\sigma=2$ recovers the full parametrization of reduced Pythagorean triples over the integers.

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I’m sorry that no one else has jumped in to answer your question, since I’m definitely not the best person to do that.

For the integers $\Bbb Z$, a nonzero number is irreducible/indecomposable/prime if and only if it generates a prime ideal. So, $(5)$ is prime because if $mn\in(5)$, it must be that either $m$ or $n$ is divisible by $5$, i.e. in $(5)$. For algebraic integer rings, the story is quite different: for example, in $\Bbb Z[\sqrt{-6}\,]$, $2$ is still irreducible (not the product of two nonunits) but its ideal is not prime, since $(\sqrt{-6}\,)^2\in(2)$ without either factor $\sqrt{-6}$ being in the ideal.

As a result, I want to answer an easier question than the one you asked, saying, “For which $m$ is $(2)$ still a prime ideal in $\Bbb Z[\zeta_m]$?” When you deal with a ring of the form $\Bbb Z[\alpha]$, things will be a lot harder for you if it is not the full ring of algebraic integers in its fraction field $\Bbb Q(\alpha)$. Fortunately, a root of unity will always generate the ring of integers in its field, so that issue does not come up.

In general, if you have the ring of integers $I$ of an algebraic number field $K$ and a prime $p$ of $\Bbb Z$, the corresponding ideal $(p)$ will not be prime in $I$: you may have seen what can happen: $(p)=\mathfrak P_1^{e_1}\mathfrak P_2^{e_2}\cdots\mathfrak P_g^{e_g}$: there will be $g\ge1$ prime ideals upstairs dividing $p$, appearing with the multiplicities $e_1,\cdots e_g$. These last numbers are called the ramification indices associated to the various primes. And worse: in each case, $\mathfrak P_j$ is a maximal ideal of $I$, so that $I/\mathfrak P_j$ is a field naturally containing $\Bbb F_p$, so of form $\Bbb F_{p^{f_j}}$. (The exponent there shows up too tiny: it’s $f_j$.) This number $f_j$ is called the residue field extension degree of $\mathfrak P_j$. And magically, $[K:\Bbb Q]=\sum_0^ge_jf_j$. That’s the arithmetic background we need to use here.

In our special case, a number of simplifications make things much easier: first, as long as $m$ is odd, $2$ is unramified, that is, all the indices $e_j$ are equal to $1$. Further, when an extension is normal (Galois), all the ramification indices are equal, similarly for the residue extension degrees, so we get $[K:\Bbb Q]=fg$. And of course that is the case for $K=\Bbb Q(\zeta_m)$.

What does that give us? If $g=1$, then $(2)$ does not split as an ideal, in other words, remains a prime ideal of $I=\Bbb Z[\zeta_m]$. But $g=1$ means that $f=[\Bbb Q(\zeta_m):\Bbb Q]=\phi(m)$, and for reasons I don’t want to go into here, the Galois group $\text{Gal}^{\Bbb Q(\zeta_m)}_{\Bbb Q}$ is isomorphic to the corresponding Galois group of $\Bbb Z[\zeta_m]/\mathfrak P$ over $\Bbb F_2$, which is always cyclic.

Now I can say: if $m$ is divisible by two or more odd primes, then the Galois group is not cyclic, and $(2)$ must split. Furthermore, if adjunction of the $p$-th roots of unity to $\Bbb F_2$ gives an extension of degree less than $\phi(p)=p-1$, again $(2)$ must split.

I think I’ve said enough. I’ll leave it to you to answer the question of which primes $p$ have the property that adjoining the $p$-th roots of unity to $\Bbb F_2$ gives you an extension of degree $\phi(p)=p-1$.

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  • $\begingroup$ I spent a few hours chewing on $[\mathbb{F}_2[\zeta_p]:\mathbb{F}_2]=p-1$ with no success. Looks like in general I've got a lot more to learn about field extensions! Thanks for the answer though. The new vocabulary terms are helpful. $\endgroup$ – Christian Woll Mar 17 at 0:24
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    $\begingroup$ Well, @ChristianWoll, you’re looking for the smallest extension of $\Bbb F_2$ that contains the $p$-th roots of unity, in other words the smallest $2^n-1$ that’s divisible by $p$. For $p=5$, for instance that’s $n=4$, so $(5)$ doesn’t split. On the other hand, for $p=7$, you already get $7|(2^3-1)$, so that $(7)$ does split. So you can answer the question easily enough for any given prime $p$. $\endgroup$ – Lubin Mar 17 at 5:02

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