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Let $K$ be an algebraic extension field (not necessarily finite) of $\mathbb{Q}$. Let $\mathscr{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$.

Then, is $\mathscr{O}_K$ Noetherian?

If $K$ is a finite extension of $\mathbb{Q}$, then $\mathscr{O}_K$ can be shown to be Noetherian. However, if we do not assume that finiteness condition, is it still true?

(It can be shown that $\mathscr{O}_K$ is an integrally closed domain of Krull-dimension $1$. Hence, if it is Noetherian, it is a Dedekind domain.)

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4 Answers 4

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We can take $K = \bar{\mathbb Q}$, the algebraic closure of $\mathbb Q$. Then it is easy to see that every element of $\mathscr O_K$ admits a square root, hence $\mathscr O_K$ contains no irreducible elements. But any element in a Noetherian domain is a product of irreducibles.

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Other answers have shown examples of infinite-degree number fields $K$ (that is, $K$ is algebraic over $\mathbf Q$ with $[K:\mathbf Q] = \infty$) where $\mathcal O_K$ is provably not Noetherian. They show an answer to your question is "sometimes no." After thinking these over you might get the idea that the ring of integers in an infinite-degree number field is never Noetherian. I will show how to recursively create an infinite-degree number field $F$ such that its ring of integers $\mathcal O_F$ is Noetherian, and thus forms a Dedekind domain. So another answer to your question is "sometimes yes." I found the construction below on p. 111 of Schilling's book The Theory of Valuations, which is old enough (1950) that it says fields containing $\mathbf Q$ have characteristic $\infty$ instead of characteristic $0$.

Before starting I'll describe briefly the idea behind the construction. We will build the big field as a tower of (ordinary finite-degree) number fields in which we arrange that in the $k$th layer of the tower the prime ideals lying over the first $k$ primes remain prime in all further stages. This will imply that in the field $F$ at the top of the tower (an infinite-degree number field) every prime ideal of $\mathcal O_F$ is finitely generated, and that turns out to be enough to know all ideals of $\mathcal O_F$ are finitely generated.

Let $p_1, p_2, p_3, \ldots$ be the sequence of all prime numbers and $n_1, n_2, n_3, \ldots$ be a sequence of integers all greater than $1$ (e.g., $n_j = 2$ for all $j$). Set $F_1 = \mathbf Q$. Suppose we have constructed an increasing tower of (finite-degree) number fields $F_1 \subset \cdots \subset F_k$ for some $k\geq 1$ such that $[F_{i+1}:F_i] = n_i$ for $1\leq i \leq k-1$ if $k\geq 2$. (Ignore the degree condition if $k=1$.) To build $F_{k+1}$, let $\{\mathfrak p_{\alpha}\}$ be the finite set of prime ideals in $\mathcal O_{F_k}$ lying over $p_1,\ldots,p_k$. The residue fields $\mathcal O_{F_k}/\mathfrak p_\alpha$ are finite, and over every finite field there is an irreducible of each degree. So for each $\alpha$ there is a monic irreducible $\pi_{\alpha}(x)$ in $(\mathcal O_{F_k}/\mathfrak p_{\alpha})[x]$ of degree $n_k$. Using the Chinese remainder theorem coefficientwise in degrees $0, 1, \ldots, n_k-1$, there is a monic polynomial $\pi(x) \in \mathcal O_{F_k}[x]$ of degree $n_k$ such that $\pi(x) \equiv \pi_\alpha(x) \bmod \mathfrak p_\alpha$ for all $\alpha$. Set $F_{k+1} = F_k(r_k)$ where $r_k$ is a root of $\pi(x)$.

Since $\pi(x)$ is monic and is irreducible when reduced modulo a (nonzero) prime ideal of $\mathcal O_{F_k}$ (use any $\mathfrak p_\alpha$), $\pi(x)$ is irreducible over $F_k$ and thus $[F_{k+1}:F_k] = \deg \pi(x) = n_k \geq 2$. For each $\alpha$ the polynomial $\pi(x)$ reduced modulo $\mathfrak p_\alpha$ is irreducible, so ${\rm disc}(\pi(x)) \not\equiv 0 \bmod \mathfrak p_\alpha$ (irreducibles over finite fields are separable) and thus the prime ideal $\mathfrak p_\alpha$ in $\mathcal O_{F_k}$ remains prime when extended to an ideal in the integers of $F_k(r_k) = F_{k+1}$. (This is an analogue over $F_k$ of the fact that if $f(x) \in \mathbf Z[x]$ is monic and $p$ is a prime number such that $f(x) \bmod p$ is irreducible over $\mathbf Z/(p)$, and we set $K = \mathbf Q(r)$ where $f(r) = 0$, then the natural embedding $\mathbf Z[x]/(f(x)) \cong \mathbf Z[r] \hookrightarrow \mathcal O_K$ becomes an isomorphism when we reduce mod $p$: $(\mathbf Z/p)[x]/(\overline{f}(x)) \cong \mathcal O_K/p\mathcal O_K$ because ${\rm disc}(f(x)) \not\equiv 0 \bmod p$, so $p\mathcal O_K$ is prime in $\mathcal O_K$.) So all prime ideals in $\mathcal O_{F_k}$ lying over $p_1,\ldots,p_k$ remain prime when extended to ideals in $\mathcal O_{F_{k+1}}$.

This recursive construction of number fields $F_k$ can be continued indefinitely and $[F_k:\mathbf Q] \geq 2^{k-1}$, so the union (or direct limit) $F$ of all $F_k$ is a field, algebraic and of infinite degree over $\mathbf Q$. By construction, the prime ideals over $p_k$ in $F_k$ are inert in $F_m$ for all $m \geq k$. Let $\mathcal O_F$ be the algebraic integers in $F$ (union of all $\mathcal O_{F_k}$).

Claim: every prime ideal in $\mathcal O_F$ is finitely generated.

Proof of claim: Let $\mathfrak p$ be a prime ideal in $\mathcal O_F$. Without loss of generality it is not $(0)$, so $\mathfrak p$ has a nonzero element of some $\mathcal O_{F_j}$, so $\mathfrak p \cap \mathcal O_{F_j}$ is a nonzero prime ideal in $\mathcal O_{F_j}$. Let $p$ be the prime number lying below $\mathfrak p \cap \mathcal O_{F_j}$, so $p = p_k$ for some $k$. Then $\mathfrak p \cap \mathcal O_{F_k}$ is a prime ideal over $p_k$ in $\mathcal O_{F_k}$, so by design this prime ideal is inert in $F_m$ for all $m\geq k$. Since $\mathcal O_{F_k}$ is Noetherian there is a finite generating set $S$ for $\mathfrak p\cap \mathcal O_{F_k}$ as an ideal in $\mathcal O_{F_k}$, and in each $\mathcal O_{F_m}$ with $m\geq k$ the extension of $\mathfrak p \cap \mathcal O_{F_k}$ to an ideal in $\mathcal O_{F_m}$ is prime, so $S$ generates $\mathfrak p \cap \mathcal O_{F_m}$ as an ideal in $\mathcal O_{F_m}$. Each element of $\mathfrak p$ is in $\mathcal O_{F_m}$ for some $m\geq k$, so $S$ generates $\mathfrak p$ as an ideal in $\mathcal O_F$.

If every prime ideal in a commutative ring (with identity...) is finitely generated then a theorem of I.S. Cohen (which uses Zorn's lemma) says every ideal is finitely generated, so $\mathcal O_F$ is Noetherian.

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If $K=\mathbb Q\left[2^{1/2^m}\mid \forall m\in\mathbb Z^+\right]$ then $\mathcal O_{K}$ has ideals $I_j=\left\langle 2^{1/2^j}\right\rangle$ with $I_j\subsetneq I_{j+1}$.

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Can't we take $K$ to be the field of all algebraic numbers? Then the integral closure of $\mathbb Z$ in $K$ is the ring of all algebraic integers. I believe this is not noetherian - for example, the principal ideals generated by $2$, $2^{1/2}$, $2^{1/4}$ etc form an ascending chain (see here).

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