Is $\sqrt[3]{5}$ in $\mathbb Q(\sqrt[3]{2})$?

Question

I am trying to solve question 3.7 (b) from Chapter 15 in Artin's book "Algebra".

The problem is:

Is it true that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$?

It is clear by Eisenstein's criterion that both $\sqrt[3]{2}$ and $\sqrt[3]{5}$ have their minimal polynomails of egree $3$ over $\mathbb Q$.

Thus if we assume that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$, then we must have $\mathbb Q(\sqrt[3]{5})=\mathbb Q(\sqrt[3]{2})$.

This didn't get me anywhere.

Then I tried the most simple minded approach.

Say $\sqrt[3]{5}=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, for some $a,b,c\in \mathbb Q$.

By cubing and rearranging, it is likely that we will arrive at some absurdity.

But clearly this shouldn't be what the author must have intended.

Can somebody point me towards a slick solution?

Answer 1

It depends how much Galois Theory you want to use. I will show that $\sqrt[3]{5} \not \in \mathbb{Q}[\sqrt[3]{2}, \omega],$ where $\omega$ is a primitive complex cube of unity, and the cube roots are assumed to be the real ones. Suppose otherwise. Note that the last field (over $\mathbb{Q}$) has order $6$ and is generated by two automorphisms: one, say $\sigma,$ interchanges $\omega$ and $\omega^{2}$ and fixes every element of $\mathbb{Q}[\sqrt[3]{2}].$ The other, say $\tau$ sends $\sqrt[3]{2}$ to $\omega \sqrt[3]{2}$ and sends $\omega \sqrt[3]{2}$ to $\omega^{2} \sqrt[3]{2}$ (and therefore fixes $\omega$ and $\omega^{2})$ and sends $\omega^{2} \sqrt[3]{2}$ to $\sqrt[3]{2}.$

Let's see where $\tau$ can send $\sqrt[3]{5}$. It can't fix $\sqrt[3]{5}$ or it would fix the whole real subfield of $\mathbb{Q}[\sqrt[3]{2}, \omega],$ and it already fixes $\omega$ and $\omega^{2}$, so it would be the trivial automorphism. It must send $\sqrt[3]{5}$ to another cube root of $5$ though.

Suppose that $\tau(\sqrt[3]{5}) = \omega \sqrt[3]{5}.$ Then $\tau$ and $\sigma$ both fix $\frac{\sqrt[3]{5}}{\sqrt[3]{2}}$, so $\frac{\sqrt[3]{5}}{\sqrt[3]{2}} \in \mathbb{Q}$, which is quickly seen to contradict unique factorization in $\mathbb{Z}$ ( if $b\sqrt[3]{5} = a\sqrt[3]{2}$ for relatively prime integers $a$ and $b,$ then $5b^{3} = 2a^{3},$ so $5$ divides $a$ and then $5$ divides $b,$ a contradiction).

Suppose then that $\tau(\sqrt[3]{5}) = \omega^{2} \sqrt[3]{5}.$ Then $\tau(\sqrt[3]{25}) = \omega \sqrt[3]{25}.$ Now, however, $\tau$ and $\sigma$ both fix $\frac{\sqrt[3]{25}}{\sqrt[3]{2}}$, so $\frac{\sqrt[3]{25}}{\sqrt[3]{2}} \in \mathbb{Q}$, which is again quickly seen to contradict unique factorization in $\mathbb{Z}$ ( if $b\sqrt[3]{25} = a\sqrt[3]{2}$ for relatively prime integers $a$ and $b,$ then $25b^{3} = 2a^{3},$ so $5$ divides $a$ and then $5$ divides $b,$ a contradiction).

Hence the assumption that $\sqrt[3]{5} \in \mathbb{Q}[\sqrt[3]{2}, \omega]$ leads to a contradiction in any case.

Answer 2

Suppose $\Bbb Q(\sqrt[3]2) = \Bbb Q(\sqrt[3]5)$.

Let $R = \Bbb Z[\sqrt[3]2]$ and $O$ be the ring of integers of $\Bbb Q(\sqrt[3]2)$. We obviously have that $O$ contains $R$, and since $\sqrt[3]5$ is an algebraic integer , it is in $O$.

The fundamental volume for $R$ is the square root of the absolute value of the discriminant of $X^3-2$, so it is $6\sqrt 3$.

Since the fundamental volume for $O$ is the square root of an integer, the index of $O$ in $R$ is a divisor of $6$, and so $R \subset O \subset \frac 16 R$. Hence $\sqrt[3]5 \in \frac 16 R$, so $5 = \left(\frac{a+b\sqrt[3]2+c\sqrt[3]4}6\right)^3$ for some integers $a,b,c$.

However $31$ is a prime not dividing $6$ and modulo which $2$ is a cube and $5$ isn't, so we obtain a contradiction by looking at this relation modulo $31$.

Answer 3

Assuming that $\sqrt[3]{5}$ belongs to $\mathbb{Q}(\sqrt[3]{2})$, then for every prime $p\equiv 1\pmod{3}$ such that $2$ is a cubic residue, also $5$ is a cubic residue. However, there are an infinite number of primes for which that does not hold, the first of them being $31$. $x^3-2$ splits as $(x-4)(x-7)(x+11)$ over $\mathbb{F}_{31}$, while $x^3-5$ is an irreducible polynomial. It follows that $\sqrt[3]{5}\not\in\mathbb{Q}(\sqrt[3]{2})$, as expected.

Answer 4

Not an elementary answer, but $\Bbb Q(\root3\of2\,)$ is ramified over $\Bbb Q$ only at $2$ and $3$; $\root3\of5$ requires ramification at $5$.

EDIT: Perhaps a more convincing argument (I'll use the notation $\lambda=\root3\of2$):

It’s “well known” that $\Bbb Z[\lambda]$ is the integer ring of $\Bbb Q(\lambda)$, and that this ring is PID (the class number is $1$). But we have a clear factorization of $5$ there, namely $$ 5 = (1+\lambda^2)(1 + 2\lambda - \lambda^2)\,. $$ You can check that the first factor has norm $5$, the second one has norm $25$. In any event, neither is a unit, so by Eisenstein, $X^3-5$ is irreducible over $\Bbb Q(\root3\of2\,)$.

Answer 5

Here is an approach that is less elegant but more elementary than the answers so far: we examine the minimal polynomial of some algebraic combination of $\sqrt[3]{2}$ and $\sqrt[3]{5}$.

For example, $\sqrt[3]{2} + \sqrt[3]{5}$ satisfies the polynomial $x^9 - 21x^6 -123x^3 -343$, so we are done if we can show that this polynomial is irreducible. This is not so easy to do by hand, of course.