Let $n\geq 2$ and let $a,b\in{\mathbb Q}$. Suppose that both the polynomials $A=X^n-a$ and $B=X^n-b$ are irreducible. We want to know whether ( * ) there is a root $\alpha$ of $A$ and a root $\beta$ of $B$ such that ${\mathbb Q}(\alpha)={\mathbb Q}(\beta)$, or if you prefer whether the quotient rings $\frac{{\mathbb Q}[X]}{A(X)}$ and $\frac{{\mathbb Q}[X]}{B(X)}$ are isomorphic. An obvious sufficient condition for ( * ) is that $\frac{b^j}{a^i}$ be a perfect $n$-th power in $\mathbb Q$ for two positive integers $i,j$ coprime to $n$. Is it a necessary condition also ?

I also posted this question on MSE : https://math.stackexchange.com/questions/1036372/isomorphism-problem-for-two-radical-extensions-of-the-same-degree

  • 6
    $\begingroup$ No. Take $n = 3$, $a = 2$ and $b = 4$. $\endgroup$ Nov 26, 2014 at 21:57
  • $\begingroup$ But it is true when $n=2$. $\endgroup$ Nov 27, 2014 at 2:30
  • 2
    $\begingroup$ If the question is good enough to attract a response from User74230, it's good enough for this site. Voting to reopen. $\endgroup$
    – Lucia
    Nov 27, 2014 at 15:20
  • 6
    $\begingroup$ This question is more interesting than some of the exotic things I see on MO which are "research level" and it is too hard to be a homework problem (avoiding sign hypotheses). Why couldn't something like this come up as a toy version of a serious issue in studying a question about number fields? That is, the "on hold" is a bit surprising to see. Have any of those putting it "on hold" actually tried to figure it out? Perhaps Grunwald-Wang in its original form (without the Galois-character variant) is also not "research level"? :) $\endgroup$
    – user74230
    Nov 27, 2014 at 15:25
  • 1
    $\begingroup$ @Lucia and user74230. The answer to the original question was trivially "no", as pointed out by Jeremy in the first comment. The first two or three close votes were given before the question was edited. I am also puzzled by how user74230 managed to post an answer after the question was closed. $\endgroup$ Nov 27, 2014 at 15:36

2 Answers 2


It seems probably "yes", but the case $4|n$ is a pain. The following method covers all cases except when all of the following conditions hold: $4|n$, $a = -h^2$, and $b = -g^2$ for nonzero rational $h$ and $g$.

First, to save notation later, we note that it is harmless to initially replace $a$ with $a^i$ for any $i$ coprime to $n$, and likewise to replace $b$ with $b^j$ for any $j$ coprime to $n$. Such appropriate $i$ and $j$ can be found via Kummer theory over $\mathbf{Q}(\zeta_n)$; let's give that predictable argument now. By hypothesis the polynomials $X^n - a$ and $X^n - b$ over $\mathbf{Q}$ admit respective roots $\alpha$ and $\beta$ in an algebraic closure (or equivalently, in a sufficiently large finite Galois extension) $F$ of $\mathbf{Q}$ such that $\mathbf{Q}(\alpha) = \mathbf{Q}(\beta)$ as subfields of $F$. Consider the subfield $K = \mathbf{Q}(\zeta_n)$ of $F$ which contains a primitive $n$th root of unity. Then $K(\alpha) = K(\beta)$ inside $F$, yet $K(\alpha)$ is visibly a splitting field over $K$ of the (possibly reducible) polynomial $X^n-a \in K[X]$ and likewise $K(\beta)$ is visibly a splitting field over $K$ of $X^n - b \in K[X]$. Hence, by usual Kummer theory there exist $i, j$ coprime to $n$ such that $a^i/b^j$ is an $n$th power in $K$. Now taking advantage of our initial remark about adjustment of $a$ and $b$, we may and do assume that $a/b$ becomes an $n$th power in $\mathbf{Q}(\zeta_n)$.

It therefore is sufficient (though not necessary!) to prove for $n \ge 1$ that if a nonzero $q \in \mathbf{Q}$ becomes an $n$th power in $\mathbf{Q}(\zeta_n)$ then $q$ is an $n$th power in $\mathbf{Q}$.

This sufficient criterion fails for $n=2m$ with odd $m$, as we see via $m$th powers of non-squares whose square roots generate quadratic subfields of $\mathbf{Q}(\zeta_{n})$ for such $n$. We will prove this sufficient criterion for odd $n$, and then use that conclusion (and even method of proof) to address even $n$.

Case $n$ odd

Here are two methods. Firstly, this case can be handled by the method in Pace Nielsen's answer without sign constraints since (i) for odd primes $p$ the group $\mathbf{R}^{\times}$ is $p$-divisible, (ii) for $n$ odd (or twice an odd) and any nonzero rational $t$ the polynomial $X^n - t$ is irreducible over $\mathbf{Q}$ if and only if $t$ is not a rational $p$th-power for every prime $p|n$ (due to Theorem 9.1 in Chapter VI of the 3rd edition of Lang's "Algebra", which entails an extra constraint when $4|n$).

The following second method has the merit that it never uses the irreducibility hypothesis on $X^n-a$ and $X^n-b$ over $\mathbf{Q}$ (but we will need such irreducibility all over the place when handling even $n$, as we must since a root of the reducible $X^4+4 = (X^2 +2x+2)(X^2-2X+2)$ generates the field $\mathbf{Q}(i)$ that is generated by a root of the reducible $(X^2+1)(X^2-1) = X^4-1$).

I claim for any $n$ (allowing even $n$ too -- useful for later!) and odd positive $d|n$ that the natural map $\mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^d \rightarrow \mathbf{Q}(\zeta_n)^{\times}/{\mathbf{Q}(\zeta_n)^{\times}}^d$ is injective. Passing to $p$-primary parts for primes $p|d$, we may assume $d=p^j$ for $1 \le j \le e = {\rm{ord}}_p(n)$ with $p$ an odd prime. Since $\mathbf{Q}^{\times}$ has no nontrivial $p$th root of unity (as $p$ is odd), the right-exact sequence $$\mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^p \stackrel{x^{p^{j-1}}}{\rightarrow} \mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^{p^j} \rightarrow \mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^{p^{j-1}} \rightarrow 1$$ is also injective on the left and hence is short exact. (The informed reader will recognize this as expressing a fact about Galois cohomology formalism, but that is not logically necessary, though it certainly informs the motivation for what I am about to do.) Likewise, since the group of $p^{j-1}$th-roots of unity in $\mathbf{Q}(\zeta_n)$ is the image under $p$-power on the group of $p^j$th-roots of unity in $\mathbf{Q}(\zeta_n)$ (here we use that $p^j|n$ rather than that $p$ is odd), it follows by reasoning with the snake lemma or elementary Galois cohomology formalism that the analogous right-exact sequence for $\mathbf{Q}(\zeta_n)$ in place of $\mathbf{Q}$ is also injective on the left and hence is short exact. The two resulting short exact sequences fit into an evident commutative diagram, so by induction on $j$ the injectivity assertion is reduced to the case $j=1$.

Now it remains to show that if $q$ is a nonzero rational and $q = \theta^p$ for some $\theta \in \mathbf{Q}(\zeta_n)$ then $q$ is the $p$th-power of a rational number. If $q$ is not such a $p$th-power then $X^p - q$ is irreducible over $\mathbf{Q}$ and so the root $\theta$ generates a degree-$p$ extension of $\mathbf{Q}$ which must be abelian (and in particular Galois). Hence, this primitive extension has to split $X^p-q$, so it must contain a primitive $p$th root of unity. Being a degree-$p$ extension, it follows that the degree $p-1$ of $\mathbf{Q}(\zeta_p)$ over $\mathbf{Q}$ must divide $p$. That forces $p=2$, a contradiction.

This settles the case of odd $n$, and also shows for general $n$ that $b/a = c^{n'}$ for some rational $c$ with $n'$ the "odd part" of $n$ (and again we have not yet used irreducibility of $X^n-a$ and $X^n-b$ over $\mathbf{Q}$).

Case $n$ even

Now we may assume $n = 2^en'$ for odd $n'$ and $e \ge 1$. As we saw above, $b/a = c^{n'}$ for a nonzero rational $c$. To go further, we need a property of the field $\mathbf{Q}(\alpha)$ (which we will prove when $e=1$, and when $e > 1$ provided that $a$ and $b$ are not negatives of rational squares):

P: there is a unique quadratic subfield (so $\mathbf{Q}(\sqrt{a})$ is the unique one).

Here is an important consequence of P when $4|n$: for $m|n$ that is a power of 2, if $\mathbf{Q}(\zeta_m) \subset \mathbf{Q}(\alpha)$ then $m|4$. Indeed, otherwise $8|m$ yet $\mathbf{Q}(\zeta_8)$ does not have a unique quadratic subfield.

Via Galois theory, P reduces to a group theory problem for subgroups $G$ of $\mathbf{Z}/n\mathbf{Z} \rtimes (\mathbf{Z}/n\mathbf{Z})^{\times}$ such that $p_2:G \rightarrow (\mathbf{Z}/n\mathbf{Z})^{\times}$ is surjective and $H := G \cap (\mathbf{Z}/n\mathbf{Z})^{\times}$ has index $n$ in $G$. Namely, is the only index-2 subgroup of $G$ containing $H$ precisely $G \cap (((n/2)\mathbf{Z}/n\mathbf{Z}) \rtimes (\mathbf{Z}/n\mathbf{Z})^{\times})$? If $e=1$ then the answer is affirmative. Indeed, in such cases $n=2n'$ with $n'$ odd, so $(\mathbf{Z}/n\mathbf{Z})^{\times} = (\mathbf{Z}/n'\mathbf{Z})^{\times}$ and $\mathbf{Z}/n\mathbf{Z} = \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/n'\mathbf{Z}$ with $\mathbf{Z}/2\mathbf{Z}$ the 2-part. Thus, $G$ is a subgroup of $$\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/n'\mathbf{Z} \rtimes (\mathbf{Z}/n'\mathbf{Z})^{\times}),$$ so the image $G'$ of $G$ in $\mathbf{Z}/n'\mathbf{Z} \rtimes (\mathbf{Z}/n'\mathbf{Z})^{\times}$ (corresponding to the irreducible $X^{n'}-a$) must be full (by applying Theorem 9.4 in Chapter VI of 3rd edition of Lang's "Algebra" to the odd $n'$). So either $G$ is full or of index 2. Hence, $G$ contains the odd-order subgroup $\mathbf{Z}/n'\mathbf{Z}$. Passing to the quotient by this brings us to an analogous question for the commutative group $\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/n'\mathbf{Z})^{\times}$ that is easy.

Suppose $c$ is not a rational square, so $\mathbf{Q}(\sqrt{c})$ is a quadratic subfield of $\mathbf{Q}(\alpha)$, but the only such field is $\mathbf{Q}(\sqrt{a})$ by P (which is proved when $e=1$), so $ac$ is a rational square. The same then holds for $ac^{n'}$ since $n'$ is odd, but $ac^{n'} = b$, contradicting that $X^n - b$ is irreducible over $\mathbf{Q}$ with $2|n$. Hence, $c$ must be a rational square, so $b/a$ is a $2n'$th-power of a rational number. This settles the case $e = 1$ unconditionally, so now assume $e \ge 2$.

This brings us to:

Additional tedium to handle $4|n$

We shall give a general argument when $4|n$ assuming P holds for the case under consideration. (This cannot happen if $a = -h^2$ for rational positive $h$, as then $\mathbf{Q}(\alpha^{n/4})$ is the biquadratic $\mathbf{Q}(i, \sqrt{h})$ when $h$ is a nonsquare and is $\mathbf{Q}(\zeta_8)$ when $h$ is a square.) Then we will prove that P does hold when $4|n$ away from the case when $a$ is the negative of a rational square (and so likewise away from the case when $b$ is the negative of a rational square); that will then leave unsettled only the exceptional cases mentioned at the start, for which P definitely is false. I will leave it to someone else to find an argument to handle such "biquadratic" situations.

Fix the choice of $\alpha$ in $\mathbf{C}$. The possibilities for $\beta$ in $\mathbf{C}$ take the form $\zeta \alpha c^{1/2^e}$ for an $n'$-th root of unity $\zeta$ and a $2^e$-th root of $c^{1/2^e}$ of $c$ in $\mathbf{C}$. Some such $\beta$ lies in $\mathbf{Q}(\alpha)$ by hypothesis, so this latter field must then contain $\beta/\alpha = \zeta c^{1/2^e}$. Hence, its $2^e$th-power $\zeta^{2^e} c$ lies in $\mathbf{Q}(\alpha)$, so $\zeta^{2^e}$ does since $c \in \mathbf{Q}^{\times}$. But $\zeta$ is a power of $\zeta^{2^e}$ since $\zeta$ is an odd-order root of unity, so $\zeta, c^{1/2^e} \in \mathbf{Q}(\alpha)$ too. We can then replace $\beta$ with $\beta/\zeta$, so now $\beta = c^{1/2^e}\alpha$ for a $2^e$-th root $c^{1/2^e}$ of $c$ that lies in $\mathbf{Q}(\alpha)$.

Write $c = {c'}^{2^{e'}}$ for maximal $e' \le e$ and rational $c'$, so $1 \le e' \le e$. If $e' = e$ then we are done (as then $c^{n'} = {c'}^n$ is an $n$th power), so we assume $e' < e$. By maximality of $e'$, neither $c'$ nor $-c'$ is a rational square. Fix a $2^{e-e'}$th-root ${c'}^{1/2^{e-e'}}$ of $c'$, so the above distinguished $2^e$th root $c^{1/2^e} = \beta/\alpha$ has the form $z {c'}^{1/2^{e-e'}}$ for some $2^e$th-root of unity $z$. Raising the identity $\beta = z {c'}^{1/2^{e-e'}}\alpha$ to the $2^{e-e'}$th-power, we get $\beta^{2^{e-e'}} = \varepsilon c' \alpha^{2^{e-e'}}$ for $\varepsilon := z^{2^{e-e'}}$ a $2^{e'}$th-root of unity. But $\beta^{2^{e-e'}} \in \mathbf{Q}(\alpha)$, so the rational multiple $\varepsilon c'$ of $\varepsilon$ lies in $\mathbf{Q}(\alpha)$. Hence, $\varepsilon \in \mathbf{Q}(\alpha)$.

By a consequence of P noted earlier, the $2^{e'}$th-root of unity $\varepsilon$ is equal to $\pm 1$ or $\pm i$. In the $\pm i$ case, the unique quadratic subfield $\mathbf{Q}(\sqrt{a})$ (see P) would have to equal $\mathbf{Q}(i)$, forcing $a$ to be the negative of a rational square. Let us rule out the possibility $a = -h^2$ for rational $h$. In such cases the 4th roots of $a$ lies in the field $\mathbf{Q}(\zeta_8, \sqrt{h})$ whose Galois group over $\mathbf{Q}$ is a direct product of 2 or 3 copies of $\mathbf{Z}/2\mathbf{Z}$. But that is inconsistent with the fact that the 4th root $\alpha^{n/4}$ of $a$ generates a quartic field with a unique quadratic subfield (by P).

We conclude that $\varepsilon = \pm 1$. In particular, $\varepsilon c'$ is not a rational square. Since $e-e' > 0$ we see that the non-square $\varepsilon c'$ becomes a square in $\mathbf{Q}(\alpha)$, so $\mathbf{Q}(\sqrt{\varepsilon c'}) = \mathbf{Q}(\sqrt{a})$. In such cases $\varepsilon c' = a u^2$ for rational $u$. Write $\varepsilon c' = a^s w^{2^f}$ with odd $s$, rational $w$ and maximal $1 \le f \le e-e'$ (so $\pm w$ are not rational squares); equivalently, $\beta^{2^{e-e'}} = a^s w^{2^f} \alpha^{2^{e-e'}}$ with odd $s$, rational $w$, and maximal $1 \le f \le e-e'$. (Here and in some places below the exponent of $2^f$ looks like $2f$; not sure why.)

If $f = e-e'$ then raising both sides to the $2^{e'}n'$th-power gives $b = a^{2^{e'}n's}w^n a$, so $b/a^{1+2^{e'}n's}$ is an $n$th power with $1+2^{e'}n's$ coprime to $n$ (since $e' > 0$).

Assume instead $f < e-e'$. We will deduce a contradiction in such cases. Since $a = \alpha^n$ is a $2^{e-e'}$th-power in $\mathbf{Q}(\alpha)$, we conclude that $w^{2^f}$ is too. Writing $w^{2^f} = x^{2^{e-e'}}$ for some $x \in \mathbf{Q}(\alpha)$, the ratio $x^{2^{e-e'-f}}/w$ in $\mathbf{Q}(\alpha)^{\times}$ is a $2^f$th root of unity. We saw that such a root of unity has to be $\pm 1$ or $\pm i$, so raising to the 4th power yields $x^{2^{e-e'-f+2}} = w^4$ with $e-e'-f+2 \ge 3$. Thus, $x^{2^{e-e'-f+1}} = \pm w^2$ with $e-e'-f+1 \ge 2$. In the case of $-w^2$ it follows that $-1$ is a square in $\mathbf{Q}(\alpha)$, so the unique quadratic subfield $\mathbf{Q}(\sqrt{a})$ (see P) coincides with $\mathbf{Q}(i)$, which is to say $-a$ is a rational square, a case that we have ruled out. Hence, $x^{2^{e-e'-f+1}} = w^2$, so $x^{2^{e-e'-f}} = \pm w$ with $e-e'+f \ge 1$. Thus, one of $\pm w$ becomes a square in $\mathbf{Q}(\alpha)$, but we saw that neither is a rational square, so one of $\pm w/a$ is a rational square: $w = \pm a {w'}^2$ for some sign and some rational $w'$. It follows that $w^{2^f} = a^{2^f}{w'}^{2^{f+1}}$, so $\beta^{2^{e-e'}} = a^{s+2^f} {w'}^{2^{f+1}} \alpha^{2^{e-e'}}$. Since $s+2^f$ is odd and $w'$ is rational, this contradicts the maximality of $f$.

Verifying P when $4|n$ and $a$ is not negative of a rational square:

Since $4|n$ and $X^n-a$ is irreducible, so is $X^4-a$. It is classical that the Galois group for the irreducible $X^4-a$ is either $D_4 = (\mathbf{Z}/4\mathbf{Z}) \rtimes (\mathbf{Z}/4\mathbf{Z})^{\times}$ or $(\mathbf{Z}/2\mathbf{Z})^2$, with the latter happening precisely when $a = -h^2$ for rational $h$ (in which case a single root generates the biquadratic splitting field that is $\mathbf{Q}(i, \sqrt{h})$ when $h$ is a non-square and $\mathbf{Q}(\zeta_8)$ when $h$ is a square). Looking at the subgroup structure of $D_4$ and what corresponds to $\mathbf{Q}(\alpha)$ in cases with $n=4$, another intrinsic characterization is that for $n=4$ the $D_4$-case is precisely the one for which $\mathbf{Q}(\alpha)$ admits a unique quadratic subfield.

Consider a general rational $c$ such that $X^n - c$ is irreducible (with $4|n$), and let $\mathbf{Q}(\gamma)$ be the field generated by a root of $X^n - c$. The Galois group $G$ for the splitting field of $\mathbf{Q}(\gamma)$ over $\mathbf{Q}$ is a subgroup of $$G(n) := (\mathbf{Z}/n\mathbf{Z}) \rtimes (\mathbf{Z}/n\mathbf{Z})^{\times}$$ (see the discussion just above Theorem 9.4 in Ch. VI of the 3rd edition of Lang's "Algebra"), where this injection $G \hookrightarrow G(n)$ is well-defined up to conjugation (given $c$). The composite map $G \hookrightarrow G(n) \rightarrow G(8) = D_4$ has image that corresponds exactly to the Galois group of the splitting field of $X^4-c$, so this map is surjective if and only if $X^4-c$ has Galois group $D_4$. We claim that in such $D_4$-cases, $\mathbf{Q}(\gamma)$ has a unique quadratic subfield.

Once this is proved, it follows that if $X^4 - a$ is in the $D_4$-case (i.e., $a$ is not the negative of a rational square) then $\mathbf{Q}(\alpha)$ has a unique quadratic subfield, so likewise for $\mathbf{Q}(\beta) \simeq \mathbf{Q}(\alpha)$, and hence the subfield $\mathbf{Q}(\beta^{n/4})$ has only one quadratic subfield. But then $X^4 - b$ must also be in the "$D_4$-case" too (as otherwise $\mathbf{Q}(\beta^{n/4})$ would be a biquadratic field). In other words, once the above general claim is proved we would have unconditionally settled all cases except those for which $4|n$ and $a = -h^2$ and $b = -g^2$ for nonzero rational $h$ and $g$.

The subfield $\mathbf{Q}(\gamma)$ corresponds to $G \cap (\{0\} \times (\mathbf{Z}/n\mathbf{Z})^{\times})$, so our aim is to prove that this lies inside a unique index-2 subgroup of $G$ (as is clear when $G=G(n)$). The hypothesis of being in the $D_4$-case says exactly that the reduction map $G \rightarrow G(4)$ is surjective, so $n=4$ is settled and we now assume $n > 4$. Either $n = 2^e$ with $e \ge 3$ or $n = mp$ for an odd prime $p$ with $4|m$.
Consider the latter cases, so $X^m - c$ fall into the $D_4$-case (i.e., $X^4 - c$ has $D_4$ splitting field). Let $\gamma' = \gamma^p$, a root of $X^m - c$, so by induction $\mathbf{Q}(\gamma')$ has a unique quadratic subfield. Since $[\mathbf{Q}(\gamma):\mathbf{Q}] = n = mp = p[\mathbf{Q}(\gamma'):\mathbf{Q}]$, $\mathbf{Q}(\gamma)$ is a degree-$p$ extension of $\mathbf{Q}(\gamma')$. But $p$ is an odd prime, so if $F$ is a quadratic subfield of $\mathbf{Q}(\gamma)$ then it must lie inside $\mathbf{Q}(\gamma')$ and so is unique by induction.

It remains to treat the case $n = 2^e$ with $e \ge 3$. First consider $e=3$. Then $G$ is a subgroup of $G(8)$ that maps onto $G(4)$ (so its index divides $[G(8):G(4)] = 4$) and onto $(\mathbf{Z}/8\mathbf{Z})^{\times}$. By inspection, the element $(2,1) \in G(4)$ of order 2 does not admit an order-2 lift in $G(8)$, so $G \twoheadrightarrow G(4)$ has nontrivial kernel. Thus, $G$ has index 1 or 2 in $G(8)$. If index 1 then we are done for $e=3$, so assume the index is 2. Thus, $G$ contains $2\mathbf{Z}/8\mathbf{Z}$. Since $G$ maps onto $G(4)=D_4$ and onto $(\mathbf{Z}/8\mathbf{Z})^{\times}$, it is not hard to deduce that $G$ is the preimage of the graph inside $\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/8\mathbf{Z})^{\times}$ of one of the two order-2 quotient $q: (\mathbf{Z}/8\mathbf{Z})^{\times} \twoheadrightarrow \mathbf{Z}/2\mathbf{Z}$ which is not the reduction onto $(\mathbf{Z}/4\mathbf{Z})^{\times}$. In both cases, by inspection $G$ has a unique index-2 subgroup containing $G \cap (\mathbf{Z}/8\mathbf{Z})^{\times}$. This settles $e = 3$.

Suppose $e \ge 4$. We proceed separately depending on whether the image of $G \rightarrow G(8)$ is full or one of the two index-2 subgroups obtained in the study of the case $e=3$ (applied to $X^8-c$). First assume $G \rightarrow G(8)$ is surjective. In such cases we claim that $G = G(2^e)$ (so in such cases there would be a unique quadratic subfield). Proceeding by induction on $e\ge 4$, we can assume $G \rightarrow G(2^{e-1})$ is surjective (as holds when $e=4$ by hypothesis). But $\#G(2^{e-1}) = (1/4)\#G(2^e)$, so $G$ has index 1, 2, or 4 in $G(2^e)$. If the index is 4 then $G$ maps isomorphically onto $G(2^{e-1})$, so all elements of $G(2^{e-1})$ of order 2 would admit an order-2 lift in $G(2^e)$. But inspection shows that none of the elements of $G(2^e)$ of order 2 reduce to the element $(2^{e-2},1) \in G(2^{e-1})$ of order 2. Hence, the index of $G$ in $G(2^e)$ is 1 or 2. Thus, $G \cap ((\mathbf{Z}/2^e\mathbf{Z}) \times \{1\})$ has index 1 or 2 in $\mathbf{Z}/2^e \mathbf{Z}$. If the latter index is 1 then $G$ contains the entire {\em normal} subgroup $\mathbf{Z}/2^e\mathbf{Z}$ of $G(2^e)$ yet maps onto the quotient by it, so we would get $G=G(2^e)$ as desired.

Suppose instead that $G$ has index 2 in $G(2^e)$; we will show that $G$ cannot then map onto $G(8)$ (and we will identify two possibilities for its image in $G(8)$). Certainly $G$ corresponds to an index-2 subgroup $H$ of the quotient $$G(2^e)/((2\mathbf{Z}/2^e \mathbf{Z}) \times \{1\}) = (\mathbf{Z}/2\mathbf{Z}) \times (\mathbf{Z}/2^e\mathbf{Z})^{\times}$$ that maps {\em onto} $\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/4\mathbf{Z})^{\times}$ since we are in the $D_4$-case. Likewise, since the projection $H \rightarrow (\mathbf{Z}/2^e\mathbf{Z})^{\times}$ is surjective it must be an isomorphism for size reasons. In other words, $H$ has to be the graph of a quotient map $q:(\mathbf{Z}/2^e\mathbf{Z})^{\times} \twoheadrightarrow \mathbf{Z}/2\mathbf{Z}$, so $q$ factors through the quotient $(\mathbf{Z}/8\mathbf{Z})^{\times}$ modulo squares, and $q$ is one of the two such quotients which is not the reduction map onto $(\mathbf{Z}/4\mathbf{Z})^{\times}$ (since $H$ maps onto $(\mathbf{Z}/2\mathbf{Z}) \times (\mathbf{Z}/4\mathbf{Z})^{\times}$). Hence, $$G = H_{q,e} := \{(x, u) \in (\mathbf{Z}/2^e\mathbf{Z}) \rtimes (\mathbf{Z}/2^e\mathbf{Z})^{\times}\,|\,q(u) = x \bmod 2\},$$ so the image of $G$ in $G(2^{e-1})$ has to be index 2 (since $e-1 \ge 3$ and $q(u)$ only depends on $u \bmod 8$), showing that this situation cannot arise when $G \rightarrow G(8)$ is surjective. This settles the case $e \ge 4$ when $G \rightarrow G(8)$ is surjective.

Assume instead that the image of $G$ in $G(8)$ is $H_{q,3}$ for one of the two quotients $q:(\mathbf{Z}/8\mathbf{Z})^{\times} \twoheadrightarrow \mathbf{Z}/2\mathbf{Z}$ that is not the quotient $(\mathbf{Z}/4\mathbf{Z})^{\times}$ (as we have seen is the case when $e=3$). We claim that $G = H_{q,e}$, as has been shown above when $e=3$.
Let's first show that this would do the job by proving $H := H_{q,e}$ has a unique index-2 subgroup containing $H \cap (\mathbf{Z}/2^e\mathbf{Z})^{\times} = 1 \times \ker q$. Since $H$ contains $(2\mathbf{Z}/2^e\mathbf{Z}) \times \{1\}$, any index-2 subgroup contains $4\mathbf{Z}/2^e\mathbf{Z}$, so we seek index-2 subgroups of $H = H_{q,e}$ containing the subgroup $$(4 \mathbf{Z}/2^e \mathbf{Z}) \rtimes \ker q \supseteq (4 \mathbf{Z}/2^e \mathbf{Z}) \rtimes (1 + 8\mathbf{Z})/(1 + 2^e\mathbf{Z}).$$ Hence, the problem is reduced to the same for the image $H_{q,3}$ in $G(8)$ which we already handled by inspection.

It remains to prove that necessarily $G = H_{q,e}$ when $e \ge 4$. Proceeding by induction, we may assume that the image of $G$ in $G(2^{e-1})$ is the index-2 subgroup $H_{q,e-1}$ (as we know to hold when $e-1=3$; i.e., when $e=4$) Hence, $G \subset H_{q,e}$ and it has index 1, 2, or 4 in $H_{q,e}$; we want to show that its index is 1. If the index is 4 then $G$ maps isomorphically onto $H_{q,e-1}$, so all order-2 elements of $H_{q,e-1}$ admit order-2 lifts in $G(2^e)$. Determination of all order-2 elements shows that the image under the reduction map $G(2^e) \rightarrow G(2^{e-1})$ of the order-2 elements consists of precisely the elements $(y,-1)$ with $y \in 2\mathbf{Z}/2^{e-1}\mathbf{Z}$. But at least one of the order-2 elements $(0, \pm (1 + 2^{e-2})) \in G(2^{e-1})$ lies in $H_{q,e-1}$ (treat $e=4$ separately), and neither has the form $(y,-1)$. Hence, $G$ has index 1 or 2 in $H_{q,e}$. We will rule out index 2, so assume $G$ has index 2 in $H_{q,e}$.

The quotient map $H_{q,e} \rightarrow H_{q,e-1}$ is 4-to-1 with subgroup $G \subset H_{q,e}$ of index 2 that must be 2-to-1 onto $H_{q,e-1}$. Let $C_e$ be the normal subgroup $2\mathbf{Z}/2^e\mathbf{Z} \subset H_{q,e}$, so $C_e$ is 2-to-1 onto $C_{e-1} \subset H_{q,e-1}$. The induced quotient map $H_{q,e}/C_e \rightarrow H_{q,e-1}/C_{e-1}$ is the 2-to-1 natural map $\Gamma_{q,e} \rightarrow \Gamma_{q,e-1}$ between graphs, so it is identified with the reduction map $\pi:(\mathbf{Z}/2^e\mathbf{Z})^{\times} \rightarrow (\mathbf{Z}/2^{e-1}\mathbf{Z})^{\times}$. In particular, if $G$ contains $C_e$ then $G/C_e$ provides a splitting to $\pi$, which does not exist. Thus, $G \cap C_e$ has index 2 in $C_e$, so passing to the quotient by the normal subgroup $4\mathbf{Z}/2^{e-1}\mathbf{Z}$ that is 2-to-1 onto its image in $C_{e-1}$ gives an index-2 subgroup $\overline{H}_{q,e} \subset (\mathbf{Z}/4\mathbf{Z}) \rtimes (\mathbf{Z}/2^e\mathbf{Z})^{\times}$ mapping 2-to-1 onto the analogous $\overline{H}_{q,e-1}$ and we have an index-2 subgroup $\overline{G} \subset \overline{H}_{q,e}$ that maps onto $\overline{H}_{q,e-1}$ and hence provides a splitting. Thus, it is enough to show that the natural map $\overline{H}_{q,e} \rightarrow \overline{H}_{q,e-1}$ has no splitting. Passing to the quotient of each side by the normal subgroup $2\mathbf{Z}/4\mathbf{Z}$ gives the reduction map $\Gamma_{q,e} \rightarrow \Gamma_{q,e-1}$ that we have already noted has no splitting.

  • $\begingroup$ At the beginning of the even case you say "the preceding methods show that we may arrange $b=ac^{n′}$ for some nonzero rational $c$." Could you expand that a bit? $\endgroup$ Nov 27, 2014 at 16:31
  • $\begingroup$ Also, there seems to be an error in the odd case. Consider the field extension $\mathbb{Q}(\zeta_{p^2})/\mathbb{Q}(\zeta_{p})$. This extension is of degree $p$, and so the norm (over this extension) of $q\in \mathbb{Q}$ should be $q^{p}$ not $q^{p-1}$. (Or am I making a silly mistake myself?) $\endgroup$ Nov 27, 2014 at 16:38
  • $\begingroup$ @PaceNielsen: I have made an edit near the start of the even case to address your first question above. $\endgroup$
    – user74230
    Nov 27, 2014 at 18:34
  • $\begingroup$ @PaceNielsen: When addressing the absurd error I made in connection with your second question I came across a variety of other embarrassing glitches for the case of even n, but I think I have fixed up the case of even n now, and for odd n I have explained why your approach takes care of that without sign restrictions and I also give another approach inspired by (but not logically requiring) elementary Galois cohomology formalism. Please let me know if it now looks OK to you, or if mistakes still lurk within. $\endgroup$
    – user74230
    Nov 27, 2014 at 20:35
  • 2
    $\begingroup$ @GerryMyerson: Ha-ha. This website seems like a good home for it. $\endgroup$
    – user74230
    Dec 7, 2014 at 5:59

With the recent change, I believe the answer to your question is now yes, at least in the case of positive rationals.

Let $a,b\in\mathbb{Q}_{>0}$ (I'll leave the cases when one of $a$ or $b$ is negative for you to work out) and assume $X^n-a$ and $X^n-b$ are irreducible over $\mathbb{Q}[X]$, for some $n\geq 2$. Let $\alpha,\beta$ be positive real roots of the respective polynomials, and further assume $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$. This means we can write $$ r_0+r_1\alpha+r_2\alpha^2+\cdots + r_{m-1}\alpha^{n-1}=\beta $$ for some elements $r_0,r_1,\ldots, r_{n-1}\in \mathbb{Q}$. After dividing by enough copies of $\alpha$ we have $$ (\ast) \qquad s_0 + s_1\alpha+\cdots + s_{n-1}\alpha^{n-1} = \beta/\alpha^i $$ where $s_0\neq 0$, $0\leq i\leq n-1$, and $s_0,s_1,\ldots, s_{n-1}\in\mathbb{Q}$.

Now, I claim that ${\rm Tr}(\alpha^j)=0$ for all $1\leq j\leq n-1$, which will follow from the fact that $\alpha^j$ is a root of the irreducible polynomial $X^{n/\gcd(n,j)}-a^{j/\gcd(n.j)}$.

Important Note: Since $X^n-a$ is irreducible, $a$ is not a $p$th power for any prime $p|n$. Thus, $a^{j/\gcd(n,j)}$ is also not a $p$th power for any prime $p|(n/\gcd(n.j))$. By the result mentioned in the answer to this question, we know that the polynomial above is irreducible. (This is one place where you will have to be more careful if $a<0$.)

Since $s_0\neq 0$, the trace (computed from $\mathbb{Q}(\alpha)$ down to $\mathbb{Q}$) of the left-hand side of $(\ast)$ is nonzero. Hence ${\rm Tr}(\beta/\alpha^i)={\rm Tr}(\sqrt[n]{b/a^i})\neq 0$. Writing $b/a^i=c^m$ where $m|n$ and $c>0$ is not a $p$th power for any prime $p|(n/m)$, we see that the minimal polynomial for $\beta/\alpha^i$ is $X^{n/m}-c$. (This is the more crucial place where we use positivity.) The only way the trace can be nonzero is if $n=m$, hence $b/a^i$ is an $n$th power.

  • $\begingroup$ The case where one or both of $a$ and $b$ is negative is OK too, but seems to require rather more effort. If that alternative method can be simplified a lot, please let me know. $\endgroup$
    – user74230
    Nov 27, 2014 at 15:23
  • $\begingroup$ The cases with $4|n$ have now been settled unconditionally except when $a$ and $b$ are both negatives of rational squares (a situation that is easy when $n=4$ by inspection of biquadratic fields, but is going to require an entirely different method when $n > 4$ if it is true). $\endgroup$
    – user74230
    Dec 7, 2014 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.