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I've been thinking a lot about ideals and factor rings, and I came upon the following. First, consider $\mathbb{Z}[\zeta_8]$, which is the cyclotomic ring of integers such that all $z\in\mathbb{Z}[\zeta_8]$ can be written as $\Sigma_{n=0}^3 a_n\omega^n,\ a_n\in\mathbb{Z},\ \omega=\frac{\sqrt{2}+i\sqrt{2}}{2}$.

With our definition of $\omega$, we can make the following claims:

  • $\omega=\frac{\sqrt{2}+i\sqrt{2}}{2}$
  • $\omega^2=i$
  • $\omega^3=\frac{-\sqrt{2}+i\sqrt{2}}{2}$
  • $\omega^4=-1$

From the above, we can also find the values of $\omega^5,\omega^6$, and $\omega^7$, though we will not necessarily need them.

Now, consider the ideal $\langle 1+3\omega\rangle$. What are we left with when we evaluate $\mathbb{Z}[\zeta_8]/\langle 1+3\omega\rangle$?

My logic was this: owing to the fact that $1+3\omega+\langle 1+3\omega\rangle=\langle 1+3\omega\rangle$, we can say that in this subring, $1+3\omega=0$. From this, we can also make the following claims by first subtracting $1$ from both sides and then repeatedly multiplying by $\omega$:

  • $3\omega = -1$
  • $3\omega^2 = -\omega$
  • $3\omega^3= -\omega^2$
  • $3\omega^4 = -\omega^3$

We now pause and remember that, per our definition of $\omega$, $\omega^4=-1$. We can therefore make a series of substitutions to say the following:

  • $\omega^3=3$
  • $\omega^2=-9$
  • $\omega=27$
  • $81=-1$

Then, simply by adding one on either side, we see that $82=0$.

Therefore, since we have a process such that any $z=a_0+a_1\omega+a_2\omega^2+a_3\omega^3$ can be turned into an integer, and we have made the claim that $z+82=z$, we come to the conclusion that $\mathbb{Z}[\zeta_8]/\langle 1+3\omega\rangle\cong\mathbb{Z}_{82}$.

Is this a sound proof? I can't find a flaw in my own logic, but then again, I do not attest to be a master of ring theory. It just seems odd and somewhat eerie to me that the factor ring would turn out to be so relatively simple; I'd have expected it to be the external direct product of several other groups. Furthermore, something feels off to me with regards to the process that any integer in this ring can be turned into a strictly real integer. It feels especially uncanny because there are real integers in $\mathbb{Z}[\zeta_8]$ that aren't in $\mathbb{Z}$ (for example, $\sqrt{2}=\omega-\omega^3\in\mathbb{Z}[\zeta_8]$), but all the elements of the factor ring seem to be exactly elements of $\mathbb{Z}$. Did I make a flaw in my reasoning somewhere, or is it really this simple?

user26857
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AKemats
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    I agree with your proof that $82 = 0$ in the quotient, and that everything is equivalent to an integer, but it might be the case that $41 = 0$ (or it might not be), so something remains to be checked. – hunter Mar 21 '22 at 18:50
  • Another approach is to use polynomial long division to find the remainder of $z$ modulo $1 + 3 \zeta$ and then use relations on $\zeta$ to simplify. This might give you some clarity as to why the structure collapses into something so small. – CyclotomicField Mar 21 '22 at 19:04
  • @hunter Very salient point! I just did the work and showed that $82=(1+3\omega)(1-3\omega+9\omega^2-27\omega^3)$, and thus $82\in\langle 1+3\omega\rangle$. Admittedly, I have yet to check whether $41$ is in the set yet, but if I know this and it looks like its factorization cannot be divided by $2$, do I even have to check? – AKemats Mar 21 '22 at 19:32
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    @AKemats One way to formalize the "factorization cannot be divided by 2" idea: The only way for an integer $n$ to be in $\langle 1+3\omega \rangle$ is if $n = (1+3\omega) x + (\omega^4+1) y$ for some $x$, $y \in \mathbb{Z}[\omega]$. But working mod 2, we notice that $1+3\omega \equiv 1+\omega$ and $\omega^4+1 \equiv (1+\omega)^4$, so both terms are divisible by $1+\omega$. The only integer divisible by $1+\omega$ is 0, so $n \equiv 0 \pmod{2}$. Hence $41 \not\in \langle 1+3\omega \rangle$. – Ant Mar 21 '22 at 20:04
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    (And if you were worried that $2$ could be in $\langle 1+3\omega \rangle$, you can repeat the same argument mod 41 and show that both $1+3\omega$ and $\omega^4+1$ are divisible by $\omega-27$.) – Ant Mar 21 '22 at 20:05
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    Well, if you’re willing to use a Theorem from Algebraic Number Theory, the fact drops out. The Theorem is that if $\xi\in R\subset K$ where $R$ is the ring of integers of the number field $K$, then $|R/\langle\xi\rangle|$ is equal to the absolute value of the Norm of $\xi$, that’s the Galois-theoretic Norm. In your case, you can get the Norm down to $\Bbb Q(i)$ as $(1+3\sqrt i,)(1-3\sqrt i,)=1-9i$. The Norm of this down to $\Bbb Q$ is $82$, and there you are.(You have an abelian group of order $82$, so it’s definitely cyclic.) – Lubin Mar 21 '22 at 21:19
  • I'd second @Lubin's suggested approach, since it is essentially algorithmic, and independent of specifics. Re-usable. – paul garrett Mar 21 '22 at 21:31
  • Putting the calculation through a program I wrote for myself to do Groebner basis calculations in $\mathbb{Z}[x]$, I find that $\langle \Phi_8(x), 3x+1 \rangle = \langle x^4+1, 3x+1 \rangle = \langle x+55, 82 \rangle$. – Daniel Schepler Mar 21 '22 at 22:53
  • In fact, your arguments are pretty easy to adapt to show that $x-27$ and 82 are both in $\langle x^4+1, 3x+1 \rangle$, so $\langle x-27, 82 \rangle \subseteq \langle x^4+1, 3x+1 \rangle$. On the other hand, it should be trivial to show that $x^4+1$ and $3x+1$ are both in $\langle x-27, 82 \rangle$ as well, giving the opposite inclusion. – Daniel Schepler Mar 21 '22 at 23:08

1 Answers1

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So, first, as in comments, yes, your computations are correct, as far as I (and others) can see. And I can corroborate them, though by other means...

The "other means", partly suggested by @Lubin's comment, involve some (very standard, very useful) basic algebraic number theory.

First, again, the cardinality of a ring of algebraic integers modulo a principal ideal $\langle \alpha\rangle$ is the absolute value of the Galois norm of $\alpha$. "For sure!" :) This much is essentially algorithmic, though the computations can be a bit burdensome. But no lucky guessing is required.

Second, more specific to the situation, we find that the norm is $82=2\cdot 41$. Thus, ("by algebraic number theory"), since the prime factors occur to the first power (not higher), that quotient ring is exactly a sum $\mathbb F_2\oplus \mathbb F_{41}$. Some explanation:

Since the prime $2$ is ramified in both quadratic subextensions $\mathbb Q(i)$ and $\mathbb Q(\sqrt{2})$, it is totally ramified in the whole extension. And, yes, the prime $41$ splits in both quadratic subextensions, because there is a $\sqrt{-1}$ in $\mathbb F_{41}$ (since $4|(41-1)$) and there is a $\sqrt{2}$ there, as well, because (quadratic reciprocity, if one likes) $41=1\mod 8$.

The total ramification on one hand, and complete splitting, on the other, are situations in which residue class fields of extensions are isomorphic to residue class fields of the base ring. So, yes, $1+3\omega$ generates an ideal that's a product of the unique prime $\tau$ lying over $2$ and some one of the four primes $\mathfrak p$ lying over $41$. Yes, $\mathfrak o/\tau\approx \mathbb Z/2$ by the total ramification, and $\mathfrak o/\mathfrak p\approx \mathbb Z/41$ by the complete splitting.

So, beyond cardinalities, $\mathfrak o/\langle 1+3\omega\rangle$ is isomorphic as a ring to $\mathbb Z/2\oplus \mathbb Z/41\approx \mathbb Z/82$ (thinking of Sun-Ze's theorem).

In fact, there's no way that the norm of an element can be a product of two distinct rational primes without something of this sort occurring. So, in fact, as soon as we see that the Galois norm is a product of two distinct rational primes, we know that the residue class field is the product of the two rings $\mathbb Z_2$ and $\mathbb Z_{41}$. The further explication of total ramification versus complete splitting is partly implied by this, but/and does give more details.

(As was deeply impressed on me by interactions with faculty in grad school, notably Bernard Dwork, Nick Katz, ... "alg no th" is not just a thing that we do to kill time, but includes a collection of very useful ideas and methods, that clarify seemingly-mundane situations.)

paul garrett
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