6

  1. Given an algebraic number field $K$ and its ring of integers $\mathcal{O}_K$, is $\mathcal{O}_K$ always isomorphic to $\mathbb{Z}[X]/(f(x))$, for some irreducible polynomial $f(x)$?

  2. Since $\mathcal{O}_K/\mathcal{m}$ is a finite field for any non-zero prime ideal $m \subset \mathcal{O}_K$, it is isomorphic to some $\mathbb{F}_p[X]/(\widetilde{f}(x))$. If the statment 1. is true, is it also true that the $\widetilde{f}(x) = f(x) \bmod{p}$, and $\mathcal{O}_K/\mathcal{m} \cong \mathbb{Z}[X]/(p,f(x)) \cong \mathbb{F}_p[X]/(\widetilde{f}(x))$?

If these are not true for any algebraic number field could you point me in the right direction? I have been reading some notes, that are apparently incomplete, and I would appreciate a reference to where I could find the proofs of these (or the most similar general) statements.

Watson
  • 24,404
baltazar
  • 1,631
  • 1
    I don't have the tools to easily carry out this construction, but find any cubic number field that has three degree $1$ primes lying over $2$. Every algebraic integer $\alpha$ will have the same image in two of the residue fields $\mathbf{F}_2$, and therefore the image of $\mathbf{Z}[\alpha]$ is the diagonal (proper) subring of the corresponding residue ring $\mathbf{F}_2 \times \mathbf{F}_2$. Since $\mathcal{O}$ surjects onto the residue ring, we see $\mathbf{Z}[\alpha] \subset \mathcal{O}$. –  Dec 12 '14 at 22:03
  • @Hurkyl I understand. Thank you. – baltazar Dec 12 '14 at 22:45

2 Answers2

9

Let me elaborate on the "strange" behavior of ring of integers. Let $K$ be an algebraic number field with ring of integers $\mathcal O_K$. Given an algebraic integer $\alpha \in \mathcal O_K$, let us denote by $\mathbf Z[\alpha]$ the $\mathbf Z$-algebra generated by $\alpha$. (This is isomorphic to $\mathbf Z[X]/(f)$, where $f$ is the minimal polynomial of $\alpha$).

Now the basic question is: Does there exist $\alpha \in \mathcal O_K$ such that $\mathcal O_K = \mathbf Z[\alpha]$? (In this case some people say $K$ is monogenic or $K$ admits a power integral basis).

Let $i(\alpha) = [ \mathcal O_K : \mathbf Z[\alpha]]$ be the order of $\mathcal O_K/\mathbf Z[\alpha]$. Does there always exists $\alpha \in \mathcal O_K$ such that $i(\alpha) = 1$ (which is equivalent to $\mathcal O_K = \mathbf Z[\alpha]$)? In general the answer is no. Consider for example the number field $K$ definied by a root of $X^3 + X^2 - 2X - 8 \in \mathbf Z[X]$. Then it is not hard to show that $2$ divides $i(\alpha)$ for all $\alpha \in \mathcal O_K$. In particular we can never have $\mathcal O_K = \mathbf Z[\alpha]$. Already in 1871, Dedekind gave this example and knew about these so called common non-essential discriminant divisor (gemeinsame außerwesentliche Diskriminantenteiler).

These common non-essential discriminant divisors are not the only obstruction. For a deeper investigation of this topic, one can use the so called index form.

I hope this gives you enough information/keywords to find more in the literature. Tell me if you need explicit references.

Hans Giebenrath
  • 848
4

The $\mathbb{Z}$-module $M=\mathbb{Z}[X]/(f)$ is generated as an algebra by a single element in $M$, since $M = \mathbb{Z}[ \overline{X} ]$ where $\overline{X} = X + (f)$. So if it was true, what you were asking in question 1, it will follow that the ring of integers are generated by a single algebraic integer. This is not true in general. Look for an example of a ring of integers that fails to be generated by a single element.

Nicolas Bourbaki
  • 5,553
  • I understand. On the other hand I know that always $\mathcal{O}_K/\mathcal{m} \cong \mathbb{F}_p[X]/(\widetilde{f}(x))$ for some irreducible $\widetilde{f}(x)$, so it seems strange to me that the ring of integers itself is not isomorphic to any quotient of $\mathbb{Z}[X]$. – baltazar Dec 12 '14 at 21:36
  • @baltazar the quotient rings $\mathcal O_K/(p)$ need not be $\mathbf F_p[x]/(f(x))$ for an irreducible $f(x) \bmod p$ since $\mathcal O_K/(p)$ need not be a field. In any case, the rings $\mathcal O_K/\mathfrak a$ are much simpler than $\mathcal O_K$. For example, $\mathcal O_K$ can have nonprincipal ideals, but for nonzero ideals $\mathfrak a$ in $\mathcal O_K$, all ideals in $\mathcal O_K/\mathfrak a$ are principal. That's because nonzero ideals in $\mathcal O_K$ have the form $(\alpha,\beta)$ where one generator can be chosen arbitrarily among the nonzero elements of the ideal. – KCd May 05 '23 at 16:45