Is it possible that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$ when $\mathbb{Q}(\alpha)$ is a $p$th degree Galois extension of $\mathbb{Q}$?

($p$ is prime)

I got stuck with this problem while trying to construct polynomials whose Galois group is cyclic group of order $p$.

Edit: Okay, I got two nice answers for this question but to fulfill my original purpose(constructing polynomials with cyclic Galois group) I realized that I should ask for all primes $p$ if there is any such $\alpha$(depending on $p$) such that the above condition is satisfied. If the answer is no (i.e. there are primes $p$ for which there is no $\alpha$ such that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$) then I succeed up to certain stage.

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    $\begingroup$ Again, careful with the order of your quantifiers! You don't want an $\alpha$ that works for all primes, you want, for each prime $p$, and $\alpha$ that works for that $p$. That is, $\forall p\exists \alpha$, not $\exists \alpha\forall p$. $\endgroup$ – Arturo Magidin Nov 1 '11 at 19:42
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    $\begingroup$ Your edit doesn't make any sense, because $\alpha$ determines $p$. $\endgroup$ – Zev Chonoles Nov 1 '11 at 19:42
  • $\begingroup$ +1 for both. What both of you pointing is the same thing, right? $\endgroup$ – Dinesh Nov 1 '11 at 19:47
  • $\begingroup$ @Dinesh: Yes, we are. $\endgroup$ – Arturo Magidin Nov 1 '11 at 19:48
  • $\begingroup$ I hope now it makes sense, not taking for granted though :D $\endgroup$ – Dinesh Nov 1 '11 at 19:50

If I understood Dinesh's intent correctly he wants an example for each prime $p$. Henning and Zev have covered the case $p=2$.

Let first $p=3$.

Let $\zeta=e^{2\pi i/9}$. Then $L=\mathbf{Q}(\zeta)$ is the ninth cyclotomic field. It is well known that $Gal(L/\mathbf{Q})$ is cyclic of degree six. A generator $\sigma$ is determined by $\sigma(\zeta)=\zeta^2$. The subgroup generated by $\sigma^3$ has thus order two, and its fixed field $K=\mathbf{Q}(\zeta+\zeta^{-1})$ is a cyclic cubic Galois extension of $\mathbf{Q}$. I claim that $\alpha=\zeta+\zeta^{-1}=2\cos\left(2\pi/9\right)$ works. We can calculate that $$ N^K_{\mathbf{Q}}(\alpha)=(\zeta+\zeta^{-1})(\zeta^2+\zeta^{-2})(\zeta^4+\zeta^{-4})=-1, $$ so $\alpha$ is actually a unit of the ring of integers of $K$. But also note that $\alpha$ is a real number $>1$, so it cannot be a root of unity. Together these facts imply that no power of $\alpha$ will be rational. Namely, let $n>0$ be any integer.Then $\alpha^n$ is an algebraic integer of $K$, so if it were rational, it would have to be a rational integer $m$. On the other hand $N^K_{\mathbf{Q}}(\alpha^n)=(-1)^n$, so $m$ would have to be $\pm1$, and $\alpha$ would have to be a root of unity of order at most $2n$. A contradiction. Because $[K:\mathbf{Q}]=3$ is a prime, and $\alpha^n\in K\setminus\mathbf{Q}$, we have $K=\mathbf{Q}(\alpha^n)$ as desired.

Let then $p$ be any fixed prime $>3$. A way to generalize the above argument is to look at a bigger cyclotomic field. Let $\zeta=e^{2\pi i/p^2}$, and $L=\mathbf{Q}(\zeta)$. It is known that $L/\mathbf{Q}$ is a cyclic Galois extension of degree $\phi(p^2)=p(p-1)$. Let $H$ be the unique subgroup of this Galois group that is of order $p-1$. Then the fixed field $K$ of $H$ is a degree $p$ cyclic extension of $\mathbf{Q}$. Furthermore, the restriction of the usual complex conjugation to $L$ belongs to $H$, because it is of order two, and $2\mid p-1$. Therefore we can deduce that the numbers in the field $K$ are all real (the field $K$ is, in fact totally real). From Dirichlet's theorem of units we know that the group of units of the ring of integers of $K$ has a free abelian part of rank $p-1$. Let us pick a non-torsion element $u$ from that group. As $u$ is real, and not a root of unity, we can repeat the previous argument, and conclude that $K=\mathbf{Q}(u^n)$ for all positive integers $n$.

I would love to give you an explicit $u$ as in the case $p=3$, but I can't find a suitable cyclotomic unit now that would belong to the field $K$.

  • $\begingroup$ Actually the construction given by Henning and Zev also fits into this family in a sense. After all $\alpha=1+\sqrt2$ is a unit in the real subfield of the eighth cyclotomic field. The difference is that in this case it is not enough to have conductor $p^2$. It often happens that the even prime needs a separate treatment. $\endgroup$ – Jyrki Lahtonen Nov 1 '11 at 21:22
  • $\begingroup$ My goodness! So we can find such an $\alpha$ for for every prime $p$?!! Its an excellent answer but I am very surprised of the result! $\endgroup$ – Dinesh Nov 2 '11 at 6:36
  • $\begingroup$ @Dinesh: I only used cyclotomic fields, because that is the quickest way to generate cyclic extensions of the rationals of a prime degree. That was the hard part. I think that examples meeting your criterion are plentiful in that almost any $\alpha$ works. $\endgroup$ – Jyrki Lahtonen Nov 2 '11 at 9:04
  • $\begingroup$ A small request. Can you immediately see from these kind of arguments or otherwise that there is an $\alpha$ for each $p$ such that $\alpha^{n}$ belongs to $\mathbb{Q}$ for at least one $n>1$? That would be very helpful too.Thanks. $\endgroup$ – Dinesh Nov 2 '11 at 9:28
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    $\begingroup$ ... and in the case $k=2$ we also get a contradiction. In that case the only conjugate of $z$ can be $-z$, so $z^2$ has to be rational, and we are back to the case $p=2$. $\endgroup$ – Jyrki Lahtonen Nov 2 '11 at 9:54

$\alpha=1+\sqrt 2$ (for $p=2$) should do the trick. By the binomial theorem, $(1+\sqrt 2)^n$ is always $a_n+b_n\sqrt2$ for some positive integer $b_n$.

  • $\begingroup$ I am sorry that my question is in such a way that it meant to ask for specific $\alpha$ and $p$, but what I meant to ask was for arbitrary $p$ and $\alpha$. $\endgroup$ – Dinesh Nov 1 '11 at 19:28
  • $\begingroup$ I may need to edit the question to make this clear $\endgroup$ – Dinesh Nov 1 '11 at 19:29

If you mean for some $\alpha$ and $p$, then yes: if $\alpha=1+\sqrt{2}$, then $\mathbb{Q}(\alpha)$ is of degree 2, which is prime, and $\alpha^n$ is never a rational number, so $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^n)$ for all $n>1$.

If you mean for all $\alpha$ such that the degree of $\mathbb{Q}(\alpha)$ is a prime number $p$, then no: if $\alpha=\sqrt{2}$, then $\mathbb{Q}(\alpha)$ is of degree 2, which is prime, but $\mathbb{Q}(\alpha^n)=\mathbb{Q}$ when $n$ is even.

  • $\begingroup$ here both $p$ and $\alpha$ are arbitrary, infact I got this $\alpha$ from some where else: let $q$ be a prime which is $1$mod$p$(such $q$ exists due to Dirichilet) then $Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$ is cyclic group of order $kp$ for some $k$ and there exists a subfield $K$ of $\mathbb{Q}(\zeta)$ such that $K$ is Galois and a $p$ extension, since Galois extensions are simple $K=\mathbb{Q}(\alpha)$ for some $\alpha$. So that is where the $\alpha$ comes from! $\endgroup$ – Dinesh Nov 1 '11 at 19:19
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    $\begingroup$ @Dinesh: Yes, but $\alpha$ and $p$ are fixed in the problem. That is, we have a fixed $\alpha$ and fixed $p$ with $\mathbb{Q}(\alpha)$ Galois of degree $p$, and you want to know if $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^n)$ for all $n$ is possible. Zev's second paragraph is really: "if $K$ is Galois of degree $p$, can it be true that for all $\alpha\in K$ such that $K=\mathbb{Q}(\alpha)$, and all $n$, we have $K=\mathbb{Q}(\alpha^n)$?" which is a stronger question (a stronger condition on $K$). $\endgroup$ – Arturo Magidin Nov 1 '11 at 19:28
  • $\begingroup$ @ArturoMagidin I got you. $\endgroup$ – Dinesh Nov 1 '11 at 19:32

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