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A number field $K$ is said to have a power basis if there is an $\alpha \in K$ such that the full ring of integers $O_K$ is the $\mathbb{Z}$-linear span of $1,\alpha,\alpha^2,\ldots,\alpha^{\deg{K}-1}$. In other words, $O_K = \mathbb{Z}[\alpha]$; another term for this is monogenic. This happens for example for the quadratic and the cyclotomic fields. The monogenic number fields are presumably very rare; results of Bhargava imply that they are a negligible fraction (zero density) among the number fields of degree $3, 4$, or $5$, and this is conjectured to hold in every fixed degree $d > 2$.

It is easily seen that totally $p$-adic number fields of degree $d$ are never monogenic for $d \gg p$. (To illustrate this, consider that the split primes in the cyclotomic field of level $N$ are the ones $\equiv 1 \mod{N}$; so they are in particular $> N > \phi(N)$.)

Question. Are there monogenic totally real number fields of arbitrarily high degree?

To put it differently: for totally real algebraic integers of arbitrarily high degree, may the ring $\mathbb{Z}[\alpha]$ be integrally closed?

Dummit and Kisilevsky [Indices in cyclic cubic fields, in "Number Theory and Algebra," 1977] have shown that infinitely many totally real cubic fields have a power basis. There exist totally real monogenic sextic fields; an example, taken at will from page 116 of [I. Gaal, Diophantine Equations and Power Integral Bases: New Computational Methods] is the field generated by a root of $$ X^6 - 5X^5 +2X^4 +18X^3 - 11X^2 -19X+1. $$ Are there examples of higher degree?

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up vote 6 down vote accepted

Kedlaya proves that there are infinitely many irreducible integer polynomials of every degree with real roots and square free discriminant; if $P$ has square free discriminant then $\mathbb{Z}[x]/P(x)$ is integrally closed. More precisely, he constructs $c N^{1/(n-1)}$ different fields of this sort with discriminant in $[-N,N]$.

According to Ash-Brakenhoff-Zarrabi, Hendrik Lenstra made the following conjecture in private communication: Let $n \geq 2$. If a degree $n$ monic integer polynomial $P$ is chosen at random, then $\mathbb{Z}[x]/P(x)$ is integrally closed with probability $6/\pi^2$. So, in some sense, monogenic fields are not that rare. We talked about these questions before here and here.

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Random monic integer polynomials and random rings of integers of number fields aren't really the same thing, since multiple polynomials could have a root generating the same monogenic ring of integers, so it's not clear if a heuristic for polynomials really translates into one for number fields. Although I read in Narkiwicz's massive tome on algebraic number theory that a monogenic ring of integers has the form $\mathbf Z[\alpha]$ for only finitely many $\alpha$ up to addition by integers. But I don't know how effective this finiteness is known. – KConrad Aug 8 '14 at 15:59
@KConrad I agree that they are not the same thing, but I am confused as to why monogenic polynomials should be MORE common than monogenic fields. Given a monogenic field, almost all elements of it are not monogenic generators. So I would naively expect that monogenic polynomials are rarer. (I am not claiming a logical contradiction, just confusion to my intuition.) In any case, Kedlaya's result answers the original question and Lenstra's conjecture (although it doesn't mention signature) suggests that there should be many more such. – David Speyer Aug 8 '14 at 16:03
Theorem 4.1 in the paper ( ) of Bhargava and Shankar shows that there are fewer than $X^{5/6+\epsilon}$ monogenic cubic fields with discriminant of abs. value $< X$, whereas the number of all cubic fields is $\asymp X$. So at least in the cubic case, this conjecture of Lenstra would mean that monogenic fields are much rarer than monogenic polynomials. – Vesselin Dimitrov Aug 8 '14 at 16:43
Trying to fix my intuition: Sample $x^3+rx^2+sx+t$ with $r \in \{ -1, 0, 1 \}$, $s \in [-X^{1/3}, X^{1/3}]$ and $t \in [-X^{1/2}, X^{1/2}]$. That should be pretty close to sampling according to $I \in [-X^{1/3}, X^{1/3}]$ and $J \in [-X^{1/2}, X^{1/2}]$ in Bhargava-Shankar's notation. If Lenstra still believes his conjecture when the size of the box varies for different coefficients of the cubic, he thinks that a positive proportion of these $O(X^{5/6})$ polynomials give monogenic orders. Davenport tells us that there are $\approx X$ cubic orders in this range. (continued) – David Speyer Aug 8 '14 at 17:32
So we have a Venn diagram: cubic orders, integrally closed cubic orders, monogenic cubic orders, integrally closed monogenic cubic orders. Each cubic order can be monogenized in at most 12 ways, so the difference between monogenic and monogenized is minor. I think our current belief is that the first two have size $\approx X$ and the other two are $\approx X^{5/6}$. Lenstra's conjecture is about (integrally closed monogenic)/(monogenic), there is no reason that should be similar to (integrally closed monogenic)/(integrally closed). – David Speyer Aug 8 '14 at 17:35

What about the maximal real subfield of cyclotomic fields?

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which are monogenic by Proposition 2.16 of Washington's Cyclotomic Fields. – Jeremy Rouse Aug 8 '14 at 13:14
Thanks... Indeed, the Chebyshev polynomials express $\zeta^k + \zeta^{-k} \in \mathbb{Z}[\zeta + \zeta^{-1}]$. I wonder if, excluding the cyclotomic fields and their maximal real subfields, there are any other abelian or totally real monogenic examples of high degree. – Vesselin Dimitrov Aug 8 '14 at 13:31
And this shows that the answer I gave earlier (claiming, in relation to a question of Serre, that a totally real field of degree $d$ and log discriminant $o(d^2)$ may never be monogenic), is wrong. Indeed, I have convinced myself now that unlike for its $p$-adic counterpart, the proof of Bilu's theorem uses the full strength of the hypothesis $h(\alpha) \to 0$, and not just $\log{|\mathrm{disc}(P)|} = o(d^2)$. So I am going to delete those two earlier answers. – Vesselin Dimitrov Aug 8 '14 at 13:55
Is there a reason to think there really should be not examples besides those in sufficiently high degree? That would be a pretty strong statement compared to expecting they have density 0. – KConrad Aug 8 '14 at 14:09
Not a particular reason, except that monogenic fields are rare, and abelian or totally real fields are special, so unless there is a really good reason, why should they have infinite intersection? (The kind of situation that Zannier in his book calls "a problem of unlikely intersections"). The cyclotomic fields have a "really good reason" for being monogenic (are there other CM fields with such property?). The $p$-adic analog could be some motiation for asking such a question. For instance, may the compositum of sufficiently many real quadratic fields ever be monogenic? – Vesselin Dimitrov Aug 8 '14 at 14:23

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