Non-trivial gap in a field extension problem proof

Question

I am not asking the following problem, but merely stating it for context.

Prove that $\sqrt[3]{2} \not\in \mathbb{Q}[\sqrt[3]{5}]$.

The answer I saw then goes on to imply that it is sufficient to instead prove that $\sqrt[3]{2} \not\in \mathbb{Z}[\sqrt[3]{5}]$. (†)

From this, it is trivial by considering the projection of the underlying ring $\mathbb{Z} \to \mathbb{Z}/13\mathbb{Z}$. ($5$ is a cube, but $2$ is not a cube in this ring)

I do not immediately see why (†) would be true, at least from the perspective of a beginner's course in field and ring theory.

What am I missing here?

Answer 1

This relies on the non trivial fact that the set of elements of $\mathbb{Q}[\sqrt[3]{5}]$ which are integral over $\mathbb{Z}$ (that is, which are roots of a monic polynomial of $\mathbb{Z}[X]$) is precisely $\mathbb{Z}[\sqrt[3]{5}]$. If you know a bit of theory of algebraic integers, you should know that it is equivalent to show that elements of $\mathbb{Q}[\sqrt[3]{5}]$ whose minimal polynomial lie in $\mathbb{Z}[X]$ are exactly the elements of $\mathbb{Z}[\sqrt[3]{5}]$. If you admit this, it is easy to conclude since $\sqrt[3]{2}$ is a root of $X^3-2$.

Another way to conclude would be to show that $\mathbb{Z}[\sqrt[3]{5}]$ is integrally closed in $\mathbb{Q}[\sqrt[3]{5}]$, meaning that any element of $\mathbb{Q}[\sqrt[3]{5}]$ which is the root of a monic polynomial of $\mathbb{Z}[\sqrt[3]{5}][X]$ lies in $\mathbb{Z}[\sqrt[3]{5}]$. I think this is not that easy to prove by hand... However, verifying this last property is trivial if you know that $\mathbb{Z}[\sqrt[3]{5}]$ is a UFD (a UFD is always integrally closed in is field of fractions). Once again, this is not trivial to establish.

Answer 2

$5\equiv (-2)^3\bmod 13$ so $(13,5^{1/3}+2)=(5^{1/3}+2)$ is a maximal ideal of $\Bbb{Z}[5^{1/3}]$.

Then you should look at the ring $$R=\{ \frac{a}{b}, a,b\in \Bbb{Z}[5^{1/3}],b\not \in (5^{1/3}+2)\}$$ Note that every non-zero element of $\Bbb{Q}[5^{1/3}]$ is of the form $\frac{a}{(5^{1/3}+2)^n}$ for some $n\in \Bbb{Z}$ and $a\in R^\times$.

It is not hard to see that for $a\in R^\times,n>0$ then $A=\frac{a}{(5^{1/3}+2)^n}$ can't be a root of a monic integer polynomial:

$A^k+\sum_{m=0}^{k-1} c_m A^m=\frac{a^k+d(5^{1/3}+2)}{(5^{1/3}+2)^{nk}}$ for some $d\in R$, and $a^k+d(5^{1/3}+2)$ can't be zero as it won't be zero in $R/(5^{1/3}+2)$.

Whence, if $2^{1/3}\in \Bbb{Q}[5^{1/3}]$ then $2^{1/3}\in R$.

This is a contradiction because it would give a root of $x^3-2$ in $R/(5^{1/3}+2)\cong \Bbb{F}_{13}$.