Explicit family of number rings $\mathcal{O}_{K_n}$ requiring $n$ generators?

Could someone provide or point me to a family of number rings $$\mathcal{O}_{K_n}$$ that require $$n$$ generators (as $$\mathbb{Z}$$-algebra)? Second best would be a family requiring $$f(n)$$ generators for a strictly increasing and positive function $$f:\mathbb{N}\to\mathbb{N}$$.

I would also be interested in seeing several explicit examples of number rings requiring 3 or 4 generators

An earlier question on this type of topic was asked by Zev Chonoles in 2010 at Which number fields are monogenic? and related questions and I want to draw your attention to the comment there by BCnrd for a nice geometric analogy. When answering that question I did not address the part Zev asked about finding rings of integers needing lots of generators as a $$\mathbf Z$$-algebra, so I'll do that here.

Pick an integer $$r \geq 1$$. Here is a sufficient condition on a number field $$K$$ that forces $$\mathcal O_K$$ to need more than $$r$$ generators as a $$\mathbf Z$$-algebra: there is a prime $$p$$ such that (i) $$[K:\mathbf Q] > p^r$$ and (ii) $$p$$ splits completely in $$\mathcal O_K$$.

Example 1. Taking $$r = 1$$ and $$p=2$$, to find a number field $$K$$ such that $$\mathcal O_K$$ needs at least two generators as a $$\mathbf Z$$-algebra ($$K$$ is not "monogenic") it is sufficient to find a $$K$$ such that $$[K:\mathbf Q] > 2$$ and $$2$$ splits completely in $$\mathcal O_K$$, such as a cubic field in which $$2$$ splits completely. Dedekind found the first example of such a field: $$\mathbf Q(\alpha)$$ where $$\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$$. A method of constructing infinitely many such $$K$$ is to use the cubic subfield of the cyclotomic field $$\mathbf Q(\zeta_p)$$ for primes $$p$$ such that $$p \equiv 1 \bmod 3$$ and $$2^{(p-1)/3}\equiv 1 \bmod p$$; that means $$p$$ splits completely in $$\mathbf Q(\sqrt[3]{2},\zeta_3)$$, and there are infinitely many such $$p$$ (and hence infinitely many such cubic fields) since the density of such $$p$$ is $$1/6$$ by the Chebotarev density theorem. The first few such $$p$$ are $$31$$, $$43$$, $$109$$, and $$127$$. For example, using PARI and Galois theory, the cubic subfield of $$\mathbf Q(\zeta_{31})$$ is $$\mathbf Q(\alpha)$$ where $$\alpha$$ is a root of $$x^3 + x^2 - 10x - 8$$ and $$2$$ splits completely in this cubic field (e.g., PARI says this cubic polynomial splits completely over $$\mathbf Q_2$$) so the ring of integers of this cubic field needs at least $$2$$ generators as a $$\mathbf Z$$-algebra. (This cubic field is different from Dedekind's, e.g., Dedekind's cubic has discriminant -2012 -- too bad I didn't write about this 6 years ago! -- while this cubic has discriminant $$3844=62^2$$.)

Example 2. Taking $$r = 2$$, to find a number field $$K$$ such that $$\mathcal O_K$$ needs at least three generators as a $$\mathbf Z$$-algebra it is sufficient to find a $$K$$ such that $$[K:\mathbf Q] > 4$$ and $$2$$ splits completely in $$\mathcal O_K$$, such as a quintic field in which $$2$$ splits completely. A way to construct such fields is to use the quintic subfield of $$\mathbf Q(\zeta_p)$$ for primes $$p$$ such that $$p \equiv 1 \bmod 5$$ and $$2^{(p-1)/5}\equiv 1 \bmod p$$; that means $$p$$ splits completely in $$\mathbf Q(\sqrt[5]{2},\zeta_5)$$, and there are infinitely many such $$p$$ since their density is $$1/20$$ by Chebotarev. The first few such $$p$$ are $$151$$, $$241$$, $$251$$, and $$431$$. Using PARI and Galois theory as in the previous example, the quintic subfield of $$\mathbf Q(\zeta_{151})$$ is $$\mathbf Q(\alpha)$$ where $$\alpha$$ is a root of $$x^5 + x^4 - 60x^3 - 12x^2 + 784x + 128$$ and $$2$$ splits completely in this quintic field, so the integers of the quintic field need at least $$3$$ generators as a $$\mathbf Z$$-algebra.

I hope from these examples you see the pattern by which, for each $$r$$, you can use a subfield of degree $$d$$ in $$\mathbf Q(\zeta_p)$$ for infinitely many primes $$p \equiv 1 \bmod d$$ to get infinitely many number fields whose ring of integers requires more than $$r$$ generators as a $$\mathbf Z$$-algebra.

It is time to prove (i) and (ii) above are sufficient conditions for $$\mathcal O_K$$ to require more than $$r$$ generators as a $$\mathbf Z$$-algebra. Let $$K$$ be a number field such that $$\mathcal O_K$$ has at most $$r$$ generators as a $$\mathbf Z$$-algebra. We will show either condition (i) or (ii) has to break down, or more simply if (ii) holds then (i) does not: if $$\mathcal O_K$$ has at most $$r$$ generators as a $$\mathbf Z$$-algebra and a prime $$p$$ splits completely in $$\mathcal O_K$$ then we will show $$[K:\mathbf Q] \leq p^r$$.

That $$\mathcal O_K$$ has at most $$r$$ generators as a $$\mathbf Z$$-algebra is the same as saying there is a surjective ring homomorphism $$\mathbf Z[x_1,\ldots,x_r] \twoheadrightarrow \mathcal O_K$$. Reducing both sides modulo $$p$$ for an arbitrary prime $$p$$, there is a surjective ring homomorphism $$\mathbf {\mathbf F}_p[x_1,\ldots,x_r] \twoheadrightarrow \mathcal O_K/(p)$$. Let $$I$$ be the kernel, so $$\mathbf {\mathbf F}_p[x_1,\ldots,x_r]/I \cong \mathcal O_K/(p)$$ as rings.

Assume $$p$$ splits completely in $$K$$ (infinitely many primes split completely in each number field, so this assumption is not crazy). Then $$\mathcal O_K/(p) \cong {\mathbf F}_p^n$$ as rings, where $$n = [K:\mathbf Q]$$, so there is a ring isomorphism $$\mathbf {\mathbf F}_p[x_1,\ldots,x_r]/I \cong {\mathbf F}_p^n$$. The product ring $$\mathbf F_p^n$$ has $$n$$ maximal ideals, each with residue field $$\mathbf F_p$$, so there are $$n$$ maximal ideals in $${\mathbf F}_p[x_1,\ldots,x_r]$$ that contain $$I$$ and have residue field $$\mathbf F_p$$. A maximal ideal in $${\mathbf F}_p[x_1,\ldots,x_r]$$ with residue field $$\mathbf F_p$$ has the form $$(x_1 - a_1,\ldots,x_r-a_r)$$ where $$a_i \in \mathbf F_p$$. There are $$p^r$$ such ideals in total, so necessarily $$n \leq p^r$$. That is exactly the negation of condition (i) at the start.

Update: This proof extends to the relative case. If $$E/F$$ is an extension of number fields then a sufficient condition for $$\mathcal O_E$$ to need more than $$r$$ generators as an $$\mathcal O_F$$-algebra is that there is a (nonzero) prime ideal $$\mathfrak p$$ in $$\mathcal O_F$$ such that (i) $$[E:F] > ({\rm N}\mathfrak p)^r$$ and (ii) $$\mathfrak p$$ splits completely in $$\mathcal O_E$$. In the above proof, "ring homomorphism" has to be replaced by "$$\mathcal O_F$$-algebra homomorphism".

• Thank you, Professor Conrad. I was familiar with that previous post, and your answer here will make a good complement. – Somatic Custard Sep 23 '18 at 19:47