# Forcing the discriminant of an integral basis to be a Carmichael number.

I was thinking about the following lemma recently.

Lemma: Let $K=\mathbb{Q}(\theta)$ for some algebraic number $\theta$ and let $n=[K:\mathbb{Q}]$. If $\{\tau_1, \,\dots\,, \tau_n\}$ consists of algebraic integers and the discriminant $\Delta_{K} [\tau_1 ,\, \dots \,, \tau_n]$ is squarefree, then $\{\tau_1,\, \dots\, ,\,\tau_n\}$ is an integral basis for the ring of algebraic integers $\mathcal{O}_{K}$.

[NB: The converse is not true: consider, for example, $\mathbb{Q}(\sqrt{-5})$ with integral basis $\{1, \sqrt{-5}\}$ s.t. $\Delta_{\mathbb{Q}(\sqrt{5})} (1, \sqrt{-5})=-20$.]

It got me thinking about Carmichael numbers (simply because they're squarefree) and so I came up with this question:

How do we restrict $K$ and an integral basis $\mathcal{B}=\{\tau_1,\, \dots\, ,\, \tau_n\}$ for $K$ so that the discriminant $\Delta_{K} (\tau_1,\, \dots \,,\, \tau_n)=C$ is a Carmichael number?

I don't have any coherent ideas, I'm afraid; I hope you don't mind. I've looked around online but I couldn't find anything.

Thoughts: Looking at the definition of $\Delta_{K}$, we might have to restrict the conjugation maps somewhat to make the prime divisors of $C$ pop out.

A Frustrated Sub-question:

Do such discriminants even exist?

I'm looking at $\sqrt{\pm 561}$ and similar numbers to see if I can find one. See my comments below for a description of a similar attempt that might require a computer programme.

Why is this interesting? I don't know: I'm just curious $\ddot\smile$

• I've thrown in the soft-question tag for good measure. This question doesn't exactly admit a definite answer. – Shaun Feb 6 '14 at 19:41
• Maybe I could take the prime divisors of a Carmichael number, square-root integers close to each of them, then see what I can make of adjoining what I get to $\mathbb{Q}$. That seems like a reasonable approach to finding an example . . . – Shaun Feb 8 '14 at 10:31
• I think I'd need a computer programme to implement this approach more effectively :/ – Shaun Feb 8 '14 at 12:44
• Stickelberger's theorem forces $C\equiv 1\pmod{4}$ since Carmichael numbers are all odd. – Shaun Feb 11 '14 at 16:12
• Just to break the silence: I'd be curious too. But it's way outside my ken, I can't even guess how the answer might look like. – Daniel Fischer Feb 14 '14 at 15:15

As you say, by Stickelberger's Theorem, the discriminant of any number field is $0$ or $1$ modulo $4$.
Conversely, if $d \equiv 1 \pmod{4}$ is squarefree, then the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$. An integral basis for the ring of integers in this case is $\{1$, $\frac{1+\sqrt{d}}{2}\}$.
It follows that a Carmichael number is a discriminant of a number field if and only if it is $1$ modulo $4$. Certainly plenty of Carmichael numbers are $1$ modulo $4$ but not all of them: I believe the first counterexample is $8911$. It is known that there are infinitely many Carmichael numbers (this is one of the more famous results to have come out of my mathematics department), but I am not a Carmichael expert and I don't know whether there are infinitely many which are $1$ modulo $4$. Is that part of your question?
• Thank you, Pete! Yeah, I think that's enough, although I did have in mind arbitrary integral bases for arbitrary $\theta$, not necessarily $\theta =\sqrt{C}$ (in my notation). Perhaps that was asking too much; besides, that restriction on $\theta$ is, of course, a restriction on $K$ :) – Shaun Feb 18 '14 at 11:03