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

Question

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!

Answer 1

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) ./.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $\{u_{\iota}\}_{\iota \in I}=\{1\}$.