Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?

Solution 1:

Here’s another proof, making light use of $p$-adic theory:

First, note that Eisenstein tells us that both $\Bbb Q(\sqrt[3]3\,)$ and $\Bbb Q(\sqrt[3]2\,)$ are cubic extensions of the rationals, and thus either they are the same field or their intersection is $\Bbb Q$. Therefore, a question equivalent to the stated one is, “Why is $\sqrt[3]2\notin\Bbb Q(\sqrt[3]3\,)$?” I’ll answer the new question.

Note that $\Bbb Q(\sqrt[3]3\,)$ may be embedded into $\Bbb Q_2$, the field of $2$-adic numbers, because $X^3-3\equiv(X+1)(X^2+X+1)\pmod2$, product of two relatively prime polynomials over $\Bbb F_2$, the residue field of $\Bbb Q_2$ (more properly the residue field of $\Bbb Z_2$). So Strong Hensel’s Lemma says that $X^3-3$ has a linear factor in $\Bbb Q_2[X]$, in other words, there’s a cube root of $3$ in $\Bbb Q_2$. But if there were a cube root of $2$ in $\Bbb Q(\sqrt[3]3\,)$, there’d be such a root in $\Bbb Q_2$, and there isn’t since it’d have to have (additive) valuation $v_2(\sqrt[3]2\,)=1/3$. No such thing, the value group of $\Bbb Q_2$ is $\Bbb Z$.

Solution 2:

I really like all of the 3 answers, and I think I found one more solution which uses Trace, not Norm. (I don't know why I didn't try to use trace map before.) So I'm going to write it down here.

Suppose $K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{3})$. Consider a trace map $$ \mathrm{Tr}_{K/\mathbb{Q}}:K\to \mathbb{Q}, \quad \alpha \mapsto \sum_{i=1}^{n}\sigma_{i}(\alpha) $$ where $\sigma_{1}(\alpha)=\alpha, \dots, \sigma_{n}(\alpha)$ is roots of minimal polynomial of $\alpha$. (We can consider trace map as a trace of a linear map $m_{\alpha}:K\to K, x\mapsto \alpha x$.) This is a $\mathbb{Q}$-linear map and we can check that $\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{2})=\sqrt[3]{2}+\sqrt[3]{2}w+\sqrt[3]{2}w^{2}=\sqrt[3]{2}(1+w+w^{2})=0$ where $w=e^{2\pi i /3}$ and $\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{3})=0$, too. If we assume $$ \sqrt[3]{3}=a+b\sqrt[3]{2}+c\sqrt[3]{4}\,\,\,\,(a, b, c\in \mathbb{Q}) $$ as above, by taking trace on both sides we get $0=3a$ and $a=0$. By multiplying $\sqrt[3]{2}$ and $\sqrt[3]{4}$ on both sides, we get $b=c=0$ and contradiction. I think this methods can be used to prove for higher power cases, such as $\sqrt[n]{3}\not\in \mathbb{Q}(\sqrt[n]{2})$ for $n\geq 2$.

Solution 3:

Let $\omega := e^{2 \pi i/3}$, and consider the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of $x^3 - 2$, and the unique automorphism $\sigma$ such that $\sigma(\sqrt[3]{2}) = \omega \sqrt[3]{2}$, $\sigma(\omega \sqrt[3]{2}) = \omega^2 \sqrt[3]{2}$, $\sigma(\omega^2 \sqrt[3]{2}) = \sqrt[3]{2}$. Suppose that $\sqrt[3]{3} \in \mathbb{Q}[\sqrt[3]{2}]$ is equal to $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ for $a, b, c \in \mathbb{Q}$. Then $\sigma(\sqrt[3]{3})$ is also a root of $x^3 - 3 = 0$, which implies it's equal to either $\sqrt[3]{3}$, $\omega \sqrt[3]{3}$, or $\omega^2 \sqrt[3]{3}$.

However, in the first case, $$\sigma(\sqrt[3]{3}) = \sqrt[3]{3} \Rightarrow a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4} = a + b \sqrt[3]{2} + c \sqrt[3]{4}.$$ Rewriting $\omega^2 = - \omega - 1$ and using the linear independence of the basis $\{ 1, \omega, \sqrt[3]{2}, \omega \sqrt[3]{2}, \sqrt[3]{4}, \omega \sqrt[3]{4} \}$ of the splitting field, we get $b = c = 0$, which gives a contradiction since $\sqrt[3]{3}$ is irrational.

Similarly, in the second case, $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{2}$, which is impossible since $\sqrt[3]{3/2}$ is irrational. And in the third case, $\sigma(\sqrt[3]{3}) = \omega^2 \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{4}$, which is impossible since $\sqrt[3]{4/3}$ is irrational.

Solution 4:

Somehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\mathbf Q (\omega)$, where $\omega$ is a primitive cubic root of $1$. Since $\mathbf Q(\sqrt[3] 2)$ and $ \mathbf Q(\sqrt[3]3)$ have degree $3$ over $\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\sqrt[3]2)/k$ and $k(\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\in (k^{*})^{3}$. Norming down to $\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$.