# $\mathbb Q(2^{1/3}+3^{1/3})=\mathbb Q(2^{1/3},3^{1/3})$? [closed]

Is it true that $$\mathbb Q(2^{1/3}+3^{1/3})=\mathbb Q(2^{1/3},3^{1/3})$$? I’m looking for any hints. Thank you!

• You should try to prove that both the left-hand and right-hand sides are of degree $9$ over $\mathbb{Q}$; since the left-hand side is clearly a subextension of the right-hand side, that will suffice for equality. I suspect that it shouldn't be too difficult to show that the left-hand side is of degree $3$ over $\mathbb{Q}[\sqrt[3]{6}]$ and that the right-hand side is of degree $3$ over say $\mathbb{Q}[\sqrt[3]{2}]$.
– ΑΘΩ
Commented Dec 14, 2019 at 12:21

I think that Kummer theory provides the most "natural" approach. It allows to replace computational formulas by functorial arguments and, when needed, to make a clear distinction between the operations + and $$\times$$.

(1) Let $$\mu_3$$ be the group of 3-rd roots of unity, $$k=\mathbf Q(\mu_3), K=k(\sqrt [3] 2, \sqrt [3] 3)$$. Each of the two cubic extensions is galois cyclic of degree $$3$$, so $$K/k$$ is galois abelian. By Kummer theory, $$G=Gal(K/k)$$ is $$\cong Hom (R,\mu_3)$$, where $$R$$ is the subgroup of $$k^*/{k^*}^3$$ generated by the classes $$\bar 2, \bar 3$$ of $$2,3$$ mod $${k^*}^3$$. It will be convenient to view $$G, R$$ as vector spaces over $$\mathbf F_3$$ (the addition of vectors being written multipicatively). If they were a relation of linear dependence of the form $${\bar 2}^i.{\bar 3}^j=\bar 1$$, with $$i,j \in \mathbf F_3$$, or equivalently $$2^i.3^j\in {k^*}^3$$, with $$i,j=0,1,2$$, norming down from $$k$$ to $$\mathbf Q$$ would yied $$2^{2i}.3^{2j}\in {\mathbf Q^*}^3$$, in contradiction with the unique factorisation in $$\mathbf Z$$ unless $$i=j=0$$. This shows that $$R$$ has a basis {$$\bar 2, \bar 3$$}, $$G\cong C_2\times C_2, K=k(\sqrt [3] 2,\sqrt [3] 3)$$ has degree $$9$$, and all the cubic extensions of $$K/k$$ are $$k(\sqrt [3] 2), k(\sqrt [3] 3), k(\sqrt [3] 6)$$.

(2) The trick then is to intoduce the subextension $$k(\sqrt [3] S)$$, with $$S=\sqrt [3] 2+\sqrt [3] 3$$, and show that $$K=k(S)$$ (*). Since $$k(\sqrt [3] 2)\neq k(\sqrt [3] 3)$$, it is clear that $$S\notin k(\sqrt [3] 2)$$ nor $$k(\sqrt [3] 3)$$. To show that $$S\notin k(\sqrt [3] 6)$$ requires an extra non kummeriean argument because of the mix of + and $$\times$$. If $$S\in k(\sqrt [3] 6)$$, this subfield would contain both the sum $$S$$ and the product $$P=\sqrt [3] 6$$, hence $$K$$ would contain the roots $$\sqrt [3] 2, \sqrt [3] 3$$ of the quadratic polynomial $$X^2-SX+P$$, so that $$K/k(\sqrt [3] 6)$$ would have degree $$2$$ : contradiction. So $$K=k(\sqrt [3] 2+\sqrt [3] 3)$$.

(3) To determine $$G=Gal(K/\mathbf Q)$$, let us introduce the cyclic group $$\Delta=Gal(K/k)\cong C_2$$. By construction, any lift of an element of $$\Delta$$ to an embedding of $$K$$ into $$\bar{\mathbf Q}$$ stabilizes $$K$$, which means that $$K/\mathbf Q$$ is normal, visibly of Galois group $$\mathcal G\cong D_9$$. Hence $$\mathcal G$$ is a semi-direct product of the quotient $$G$$ and a non normal subgroup $$H$$ of order $$2$$ which fixes $$\mathbf Q(\sqrt [3] 2+\sqrt [3] 3)=\mathbf Q(\sqrt [3] 2,\sqrt [3] 3)$$.

(*) NB : For biquadratic extensions, see e.g. https://math.stackexchange.com/a/3475943/300700o . I have not checked thoroughly, but the above method should work when replacing $$3$$ by any odd prime $$p$$ . In particular, (2) should apply to show that the equality $$\mathbf Q(\sqrt [p] x+\sqrt [p] y)=\mathbf Q(\sqrt [p] x.\sqrt [p] y)$$ has no rational solutions s.t. $$xy^{-1} \notin {\mathbf Q^*}^p$$ (a kind of "inverted Fermat property) ./.

There is a little trick for showing $$\sqrt[3]{3}\not\in\mathbb{Q}(\sqrt[3]{2}),$$ i.e. finding some prime $$p\equiv 1\pmod{3}$$ such that $$2$$ is a cubic residue but $$3$$ is not a cubic residue. $$31$$ is one of these primes, and you are done: if $$\sqrt[3]{3}$$ were in $$\mathbb{Q}(\sqrt[3]{2})$$, it would be in $$\mathbb{F}_{31}(\sqrt[3]{2})=\mathbb{F}_{31}$$ too, but this is not the case.

• The idea is very elegant but the notations used seem to suggest the notion of extending $\mathbb{Z}_{31}$ by the real number $\sqrt[3]{2}$ as though it were contained in some extension of $\mathbb{Z}_{31}$....notion which is not very pleasant.
– ΑΘΩ
Commented Dec 14, 2019 at 15:05
• @ΑΘΩ: in more comfortable terms, $x^3-2$ completely factors over $\mathbb{F}_{31}$, while $x^3-3$ does not. Commented Dec 14, 2019 at 15:07
• Despite my bickering on matters of notation, the method you propose is indeed profoundly related to higher notions of algebraic number theory, so may I say that a more elaborate presentation of the proof you hinted at would be more than welcome. If you happen to have the time to spare.
– ΑΘΩ
Commented Dec 15, 2019 at 9:20
• For more general results, see math.stackexchange.com/questions/1657374 Commented Dec 22, 2019 at 14:51

Here's a fairly straightforward proof that the two fields are equal, relying mainly on the binomial expansions for cubes and quartics.

It's clear that $$\mathbb{Q}(2^{1/3}+3^{1/3})\subseteq \mathbb{Q}(2^{1/3},3^{1/3})$$, so we need only show that $$2^{1/3}$$ and $$3^{1/3}$$ are in $$\mathbb{Q}(2^{1/3}+3^{1/3})$$. It's enough, of course, to show it for just of them; we'll wind up with $$3^{1/3}\in\mathbb{Q}(2^{1/3}+3^{1/3})$$.

From

$$(2^{1/3}+3^{1/3})^3=2+3\cdot2^{2/3}\cdot3^{1/3}+3\cdot2^{1/3}\cdot3^{2/3}+3=5+3(12^{1/3}+18^{1/3})$$

we see that

$$12^{1/3}+18^{1/3}\in\mathbb{Q}(2^{1/3}+3^{1/3})\quad(*)$$

Putting this together with

\begin{align}(2^{1/3}+3^{1/3})^4&=2\cdot2^{1/3}+4\cdot12^{1/3}+6\cdot36^{1/3}+4\cdot18^{1/3}+3\cdot3^{1/3}\\&=2(2^{1/3}+3^{1/3})+4(12^{1/3}+18^{1/3})+6\cdot36^{1/3}+3^{1/3}\end{align}

we see that

$$3^{1/3}+6\cdot36^{1/3}\in\mathbb{Q}(2^{1/3}+3^{1/3})\quad(**)$$

Now from

\begin{align} (3^{1/3}+6\cdot36^{1/3})^3 &=3+3(6\cdot(9\cdot36)^{1/3}+36\cdot(3\cdot36^2)^{1/3})+6^3\cdot36\\ &=3+6^3\cdot36+54(12^{1/3}+12\cdot18^{1/3}) \end{align}

we find

$$12^{1/3}+12\cdot18^{1/3}\in\mathbb{Q}(2^{1/3}+3^{1/3})\quad(***)$$

The inclusions $$(*)$$ and $$(***)$$ together imply $$12^{1/3},18^{1/3}\in\mathbb{Q}(2^{1/3}+3^{1/3})$$, and we now see from $$(**)$$ that

$$3^{1/3}={3^{1/3}+6\cdot36^{1/3}\over1+6\cdot12^{1/3}}\in\mathbb{Q}(2^{1/3}+3^{1/3})$$

and we're done.

Remark: It took me a couple of tries to do all the arithmetic correctly. At least I think it's all correct now. I'd appreciate any remaining errors being pointed out.

After all I see myself driven to give a full answer to the problem which is not so difficult conceptually but perhaps slightly annoying on the computational side.

Let us first treat the right-hand side $$\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3})$$. Clearly $$X^3-3$$ is an annihilating polynomial of $$\sqrt[3]{3}$$ over $$\mathbb{Q}(\sqrt[3]{2})$$ and the question is whether it is irreducible or not; if it is indeed so, then by the degree theorem you have right away that $$[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=9$$ (the irreducibility of $$X^3-3$$ over $$\mathbb{Q}$$ being quite a standard and well-known fact). Assume by contradiction that our cubic polynomial were not irreducible over $$\mathbb{Q}(\sqrt[3]{2})$$ and hence would have a root in this extension; as this extension is a real subfield and the cubic polynomial in question has only one real root, this would amount to claiming that $$\sqrt[3]{3} \in \mathbb{Q}(\sqrt[3]{2})$$. Explicitly, this would entail the existence of rational coefficients $$a, b, c$$ such that

$$\sqrt[3]{3}=a \sqrt[3]{4}+b \sqrt[3]{2}+c$$

One raises this relation to the third power and makes use of the linear independence of $$1, \sqrt[3]{2}, \sqrt[3]{4}$$ over $$\mathbb{Q}$$ to obtain, after a bit of patience with the calculations, the following relations:

$$2a^2b+ac^2+b^2c=0 \tag{1}$$ $$2ab^2+2a^2c+bc^2=0 \tag{2}$$ $$4a^3+2b^3+c^3+12abc-3=0 \tag{3}$$

By multiplying equation (1) with $$c$$, equation (2) with $$b$$ and comparing the results we infer that $$a(c^3-2b^3)=0$$ which has one of the two following consequences, to be treated disjunctively:

• $$a=0$$, which from either equation (1) or (2) leads to $$b=0$$ or $$c=0$$ and thus from equation (3) to either $$c^3=3$$ when $$b=0$$ or $$2b^3=3$$ when $$c=0$$, both of which cases are absurd (neither $$3$$ nor $$\frac{3}{2}$$ being rational cubes).
• $$c^3=2b^3$$ which by the same token (that $$3$$ is not a rational cube) forces $$b=c=0$$ and leads to $$4a^3=3$$ from equation (3). However, $$\frac{3}{4}$$ is also not a rational cube, so this case is also absurd.

This settles the degree of the right-hand side extension, and allows us to also draw the conclusion that the family $$(\sqrt[3]{2}^k \sqrt[3]{3}^l)_{0 \leqslant k \leqslant 2 \\ 0 \leqslant l \leqslant 2}$$ is linearly independent over $$\mathbb{Q}$$ (more elegantly put the subextensions $$\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\sqrt[3]{3})$$ are linearly disjoint over $$\mathbb{Q}$$).

We deal next with the left-hand extension and abbreviate $$a=\sqrt[3]{2}+\sqrt[3]{3}$$; a straightforward calculation reveals that $$a$$ is annihilated by the polynomial $$f=X^3-3\sqrt[3]{6}X-5 \in \mathbb{Q}(\sqrt[3]{6})[X]$$, and again we ask ourselves whether this polynomial is irreducible over the specified subextension $$\mathbb{Q}(\sqrt[3]{6})$$ or not; if it is, the irreducibility of $$X^3-6$$ over $$\mathbb{Q}$$ ($$6$$ also fails to be a rational cube) combined with the degree theorem tell us that $$[\mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{3}):\mathbb{Q}]=9$$, since it is easy to see that $$\mathbb{Q}(\sqrt[3]{6}) \subseteq \mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{3})$$. The observation that $$\mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{3}) \subseteq \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3})$$ is then the one more detail we need to show the two subextensions are equal.

As to the irreducibility of $$f$$, we proceed in a fashion similar to the previous time we argued irreducibility above; first of all we introduce $$\epsilon=\frac{-1+\mathrm{i} \sqrt{3}}{2}$$ and we remark that the other two roots of $$f$$ are $$\epsilon \sqrt[3]{2}+\epsilon^2 \sqrt[3]{3}, \epsilon \sqrt[3]{3}+\epsilon^2 \sqrt[3]{2}$$ neither of them real (assuming they were real would lead by considering conjugates to the absurd relation $$(\epsilon-\epsilon^2)(\sqrt[3]{2}-\sqrt[3]{3})=0$$). Therefore, assuming by contradiction the reducibility of the cubic $$f$$ over the specified subextension would entail the existence of a root of $$f$$ in that subextension, which as a real subfield can only contain the unique real root of $$f$$, namely $$a$$; this would entail a relation of the form:

$$\sqrt[3]{2}+\sqrt[3]{3}=p \sqrt[3]{6}^2+q \sqrt[3]{6}+r$$ with rational coefficients $$p, q, r$$. However, the existence of such a relation is precluded by the observation of linear independence made above.

• Could you please develop why $Q(\sqrt [3] 6)$ is contained in $Q(\sqrt [3] 2 + \sqrt [3] 3)$ ? I got lost in calculation. Commented Dec 15, 2019 at 14:52
• @nguyenquangdo: look once more at the paragraph where $a$ and $f$ are introduced; saying that $f(a)=0$ means that $a^3-3\sqrt[3]{6}a-5=0$ whence $\sqrt[3]{6}=\frac{a^3-5}{3a} \in \mathbb{Q}(a)$.
– ΑΘΩ
Commented Dec 15, 2019 at 14:56
• Got it, thank you. Commented Dec 15, 2019 at 15:51

Following the treatment given by Nguyen Quang Do in the previous answer, may I add a few details to clarify the situation and establish the relation of equality between subextensions that the original poster inquires about.

Set $$\epsilon=\frac{1+\mathrm{i}\sqrt{3}}{2}, K=\mathbb{Q}(\epsilon)$$ and $$L=K(\sqrt[3]{2}, \sqrt[3]{3}), E=\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}), E'=\mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{3})$$ together with the Galois groups $$\Gamma=\mathrm{Aut}_{\mathbb{Q}}(L), \Delta=\mathrm{Aut}_{K}(L)$$.

We note that $$L/\mathbb{Q}$$ is a splitting extension for the polynomials $$X^3-2, X^3-2$$ and its subextension $$K/\mathbb{Q}$$ is a cyclotomic extension (of level $$3$$), hence both these extensions are indeed Galois and we have $$\Delta \trianglelefteq \Gamma$$. By traditional Kummer theory, as indicated in the previous answer, one sees that $$\Delta=\langle \alpha, \beta \rangle$$, where the mentioned automorphisms are given by: $$\alpha(\sqrt[3]{2})=\epsilon \sqrt[3]{2}, \alpha(\sqrt[3]{3})=\sqrt[3]{3}\\ \beta(\sqrt[3]{2})=\sqrt[3]{2}, \beta(\sqrt[3]{3})=\epsilon \sqrt[3]{3}$$

We notice that $$X^2+X+1$$ is also the minimal polynomial of $$\epsilon$$ over $$E$$ ($$\epsilon$$ and its conjugate-neither of which are real-cannot belong to the real subfield $$E$$) and that $$L=E(\epsilon)=E(\epsilon^2)$$; hence $$L/E$$ is a quadratic extension and since the two cubic roots of unity have the same minimal polynomial over $$E$$, there must exist $$\gamma \in \mathrm{Aut}_E(L)$$ such that $$\gamma(\epsilon)=\epsilon^2$$; this automorphism necessarily is an involution. By introducing $$\kappa=_{K|}\gamma_{|K}$$ we have $$\mathrm{Aut}_{\mathbb{Q}}(K)=\langle \kappa \rangle$$ and we may also remark that the exact sequence

$$\{1\} \rightarrow \Delta \xrightarrow{\mathrm{i}_{\Gamma}^{\Delta}} \Gamma \xrightarrow{\mathrm{res_{KL}}} \mathrm{Aut}_{\mathbb{Q}}(K) \rightarrow \{1\}$$

splits, since the natural morphism of restriction to $$K$$ has a section, namely the map: $$\mathrm{Aut}_{\mathbb{Q}}(K) \to \Gamma\\ \mathbf{1}_K \mapsto \mathbf{1}_L \\ \kappa \mapsto \gamma$$

Thus, $$\Gamma \approx (\mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2$$ is indeed a semidirect product generated by $$\alpha, \beta, \gamma$$ with the commutation relations $${}^{\gamma}\alpha=\alpha^{-1}\\ {}^{\gamma}\beta=\beta^{-1}$$

From here on it is not difficult to ascertain that $$\mathrm{Aut}_{E}L=\mathrm{Aut}_{E'}L=\langle \gamma \rangle$$ which automatically entails $$E=E'$$ by the fundamental theorem of Galois theory.

As a side-note, instrumental in obtaining the above equality between Galois groups is the following (where $$\mathbb{U}=\{z \in \mathbb{C}|\ |z|=1\}$$ denotes the unit circle):

Lemma. Let $$I$$ be a nonempty finite set, $$a \in (0, \infty)^I$$ and $$u \in \mathbb{U}^I$$ such that $$\sum_{\iota \in I}(a_{\iota}u_{\iota})=\sum_{\iota \in I}a_{\iota}$$ Then $$\{u_{\iota}\}_{\iota \in I}=\{1\}$$.