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Suppose the polynomial $t^k - a$ has a root (hence splits) in $\mathbb{Q}(\zeta_k)$. For which $k$ does it follow that one of the roots of $t^k - a$ is rational? In particular, are there infinitely many such $k$?

A counting argument shows this is true whenever $k$ has the property that $\varphi(k)$ is a power of a prime relatively prime to $k$. Unfortunately, I think it's an open problem whether there are infinitely many such $k$.

Motivation: If enough $k$ have this property then I think I can complete my solution to "Is an integer uniquely determined by its multiplicative order mod every prime?"

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Can you give a sketch of the argument when $\varphi (k)$ is a power of a prime relatively prime to $k$? –  mixedmath May 29 '11 at 23:14
    
In that case, the Galois group is a $p$-group, and the set of roots of $t^k - a$ is a disjoint union of orbits of it, all of size a power of $p$. If none of them have size $1$ then this is a contradiction. –  Qiaochu Yuan May 29 '11 at 23:18
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If $k\gt2$ then $\phi(k)$ is even, so if $\phi(k)$ is a power of a prime then that prime is 2, and $k$ is a product of Fermat primes. –  Gerry Myerson May 30 '11 at 1:04
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up vote 10 down vote accepted

This is an amplification of Gerry Myerson's answer, which may be helpful.

You are asking about the kernel of the map $\mathbb Q^{\times}/(\mathbb Q^{\times})^k \to L^{\times}/(L^{\times})^k,$ where $L =\mathbb Q(\zeta_k)$.

In general, for any field $K$ of char. prime to $k$, there is a natural isomorphism $K^{\times}/(K^{\times})^n \cong H^1(G_K,\mu_k)$. (This is the content of Kummer theory, and follows from Hilbert's Thm. 90.)

Thus if $L$ is a Galois extension of $K$, the kernel of the map $K^{\times}/(K^{\times})^k \to L^{\times}/(L^{\times})^k$ is naturally identified with the kernel of the restriction map $H^1(G_K,\mu_k) \to H^1(G_L,\mu_k)$, which by the inflation-restriction exact sequence, is equal to $H^1(Gal(L/K),\mu_k(L))$ (where $\mu_k(L)$ denotes the subgroup of $\mu_k$ consisting of element which lie in $L$).

If we apply this with $K = \mathbb Q$ and $L = \mathbb Q(\zeta_k)$, we find that the kernel you are interested in is identified with $H^1((\mathbb Z/k)^{\times},\mu_k)$, which is not too hard to compute.

One approach to the computation is as follows: If we factor $k$ into a product of powers of distinct primes, say $k = \prod p^n,$ then $\mu_k = \oplus \mu_{p^n},$ and so we are reduced to computing $$H^1((\mathbb Z/m)^{\times} \times (\mathbb Z/p^n)^{\times}, \mu_{p^n})$$ for each $p$ (where, after having chosen a particular $p$, I have written $k = m p^n$, with $m$ coprime to $p$). One can compute this lots of ways, e.g. via the Kunneth formula.

The key facts are that if $p$ is odd then (since the mod $p$ cyclotomic character is distinct from the trivial character) $H^i((\mathbb Z/p^n)^{\times},\mu_{p^n})$ vanishes for all $i$, while if $p = 2$ and $n \geq 1$ (resp. $n \geq 2$), then $H^0((\mathbb Z/2^n)^{\times},\mu_{2^n})$ (resp. $H^1((\mathbb Z/2^n)^{\times},\mu_{2^n})$) has order two.

From these, one deduces that $H^1((\mathbb Z/k)^{\times},\mu_k)$ is trivial if $k$ is odd; is a product of $l$ cyclic groups of order $2$ if $k$ is exactly divisible by $2$, and is divisible by $l$ distinct odd primes; and is a product of $l+1$ cyclic groups of order $2$ if $k$ is divisible by $4$, and by $l$ distinct odd primes.

Here are the concrete interpretations:

  1. If $k$ is odd, then any element of $\mathbb Q^{\times}$ which becomes a $k$th power in $\mathbb Q(\zeta_k)$ was already a $k$th power in $\mathbb Q$.

  2. If $k = 2m,$ where $m$ is odd, divisible by primes $p_1,\ldots,p_l$, then any element of $\mathbb Q^{\times}$ which becomes a $k$th power in $\mathbb Q(\zeta_k) = \mathbb Q(\zeta_m)$ is a product of powers of $p_1^m,\ldots,p_l^m$.

  3. If $k = 2^n m,$ where $m$ is odd, divisible by primes $p_1,\ldots,p_l$, and $n \geq 4,$ then any element of $\mathbb Q^{\times}$ which becomes a $k$th power in $\mathbb Q(\zeta_k)$ is a product of powers of $p_1^{2^{n-1}m},\ldots,p_l^{2^{n-1}m}, (-4)^{2^{n-2}m}.$

Of course, one doesn't need group cohomology to work this out. The advantage of the group cohomology approach, though, is that it's completely systematic. (Except perhaps for the concrete interpretation part, which involves making Kummer theory effective; although in the particular case you are interested in, it's pretty easy to see directly that the specified elements become $k$th powers in $L$, and the problem is just to show that there are no other elements that do, for which the abstract cohomology computations suffice.)


Answer to your question: it follows for odd $k$ that $a$ was already a rational $k$th power.

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Great! This is enough to finish the other problem, and it's also more good motivation to learn group cohomology. –  Qiaochu Yuan May 30 '11 at 9:18
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I take it $a$ is to be a (rational) integer, otherwise you could take any old $\beta$ in ${\bf Q}(\zeta_k)$ and let $a=\beta^k$ and then $t^k-a$ would have a root in ${\bf Q}(\zeta_k)$ but, in general, not in $\bf Q$.

Now $t^k-a$ is irreducible over the rationals unless it is of the form $t^{pr}-b^p$ for some prime $p$ or else of the form $t^{4r}+4b^4$. And if it's irreducible, then its roots have degree $k$ over the rationals and thus can't be in ${\bf Q}(\zeta_k)$ which has degree at most $k-1$. So the hypothesis that $t^k-a$ has a root in ${\bf Q}(\zeta_k)$ already only holds true in a very few special cases.

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Thanks, Gerry. I remember knowing an elementary proof of the result you cite, but can't remember what it was. Could you sketch it and / or provide a reference? –  Qiaochu Yuan May 30 '11 at 9:17
    
I tried to find a reference before posting, but all I came up with was a review that started, "It is well-known that..." and then gave the result. I'll try again, tomorrow. –  Gerry Myerson May 30 '11 at 10:14
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