# On Selmer's polynomial

Does this hold?

Let $p$ be an odd prime. If $\alpha$ is a root of $x^{p}-x-1$, prove that the ring of integers of $\mathbb{Q}[\alpha]$ is precisely $\mathbb{Z}[\alpha]$ and that this is a PID.

Is this true in general (without assuming $p$ to be prime, that is)?

We know that $x^{n}-x-1$ is irreducible over $\mathbb{Q}$ for all $n$.

I checked this for $p=3$ and $p=5$, but in general I got pretty stuck... we have a formula for the discriminant of $x^{n}+ax+b$ but I wasn't able to finish. Any help?

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Where did you come up with this? Is it based only on two data points (p = 3 and p = 5)? PARI claims the integers of ${\mathbf Q}(\alpha)$ are a PID for primes up to 37. I am very suspicious that the last part about being a PID could be proved in a simple way for all primes if it were true, because it would provide an explicit infinite set of number fields with class number 1, which is still an open problem. –  KCd Nov 15 '11 at 6:08
Seems like my computer just finished doing this up to p=53. I would guess finding a counterexample is hard (if it exists) and as KCd mentions, the latter claim implies an open problem. –  pki Nov 15 '11 at 6:19
Yes, I was just wondering about this since apparently the proof that $x^{p}-x-1$ is irreducible over $\mathbb{Q}$ is the same as in those particular cases, and I thought maybe we can do more... Didn't know we would cross an open problem though (heh!), sorry. –  Anna Nov 15 '11 at 6:36
A discriminant not being squarefree has no direct bearing on your question! If the discriminant is squarefree then you'd have a cheap way of knowing the ring of integers, but if the discriminant is not squarefree that does not mean the ring of integers can't be what you may think it is. For example, if $\zeta_p$ is a nontrivial $p$-th root of unity then ${\mathbf Z}[\zeta_p]$ is the ring of integers of ${\mathbf Q}(\zeta_p)$, but the discriminant is not squarefree (when $p > 2$). –  KCd Nov 15 '11 at 7:14
Dave Boyd, Mark Thom, and I have a manuscript in preparation that, among other things, gives a heuristic that predicts that the density of integers $n$ such that the discriminant of $x^n-x-1$ is squarefree is about $99.4$% (I don't have the exact number at hand). I wish we had the paper finished already, now that the issue has come up here! –  Greg Martin Nov 15 '11 at 7:34
For $p=257$, the ring $\mathbb{Z}[\alpha]/(\alpha^p-\alpha-1)$ is not integrally closed. Let $K$ be the field of fractions of this ring.
Make the change of variable $\alpha = \beta+55$. Then the minimal polynomial of $\beta$ looks like $\beta^{257} + \cdots + a \beta^2+b \beta + c$ where $c$ is divisible by $59$ twice, $b$ is divisible by $59$ once and $a$ is not divisible by $59$. So the $59$-adic Newton polygon for this polynomial ends in a segment of length $2$ and slope $1$. More precisely, $a = 27 \mod 59$, $b \equiv 13 \cdot 59 \mod 59^2$ and $c \equiv -1 \cdot 59^2 \mod 59^3$ and $27 x^2 +13 x -1$ factors as $27(x-36)(x-40) \mod 59$. So there is a place $\mathfrak{p}$ of $K$ lying above $59$ where $\alpha \equiv 55+36 \cdot 59 \mod \mathfrak{p}^2$ and another place $\mathfrak{q}$ where $\alpha \equiv 55+40 \cdot 59 \mod \mathfrak{q}^2$.
At this point, we know that the ring is not algebraically closed. $59^2$ divided the discriminant, but there were no ramified places of $K$ above $59$. (The other factors of $x^{257}-x-1$ modulo $59$ are all square free, since the discriminant was only divisible by $59$ twice.) Actually constructing one of the missing integral elements would still be a bit of a pain.