# When does $\mathbb{Z}[\zeta_m]$ contain divisors of $2$ (besides units)?

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.

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 $$\phi(p)=p-1$$.

• 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. – Christian Woll Mar 17 at 0:24
• 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 $p$. – Lubin Mar 17 at 5:02