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For distinct primes $p,q$, $\Bbb Q(\sqrt[3]{p})\cap\Bbb Q(\sqrt[3]{q}) =\Bbb Q$

This is equivalent to showing that $[\Bbb Q(\sqrt[3]{p})\cap\Bbb Q(\sqrt[3]{q}) :\Bbb Q] =1$. Since $p,q$ are primes, $[\Bbb Q(\sqrt[3]{p}):\Bbb Q] =[\Bbb Q(\sqrt[3]{q}):\Bbb Q] = 3$. So I need to show $\sqrt[3]{p}\notin \Bbb Q(\sqrt[3]{q})$ or $\sqrt[3]{q}\notin\Bbb Q(\sqrt[3]{p})$. How can I show this? Could you give any hints?

Edit : (1) There is a hint of this problem : Use field trace (field trace is defined in the problem). (2) I only know undergraduate level field theory.

one potato two potato
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  • I am unfamiliar with the syntax, so it is unclear whether this comment is relevant. If $\sqrt[3]{p} = \sqrt[3]{q} \times \frac{a}{b} ~: ~a,b \in \Bbb{Z^+}$, then $p = q\frac{a^3}{b^3} \implies pb^3 = qa^3.$ I suggest showing that it is impossible for there to exist $a,b \in \Bbb{Z^+},$ such that $pb^3 = qa^3.$ Here, you can assume, without loss of generality, that $a,b$ are relatively prime. – user2661923 Sep 25 '21 at 07:00
  • @user2661923 To prove that directly, I need to show $\sqrt[3]{p} \neq r_1+r_2\sqrt[3]{q}+r_3(\sqrt[3]{q})^2$ for any $r_1,r_2,r_3\in\Bbb Q$. But I guess this makes the problem harder. I want to use field theory maybe field trace? – one potato two potato Sep 25 '21 at 07:13
  • Then, as a first step, I would cube both sides, so that the problem reduces to showing that it is impossible for $r_1, r_2, r_3$ to exist such that $$p = \left(~ r_1 + r_2\sqrt[3]{q} + r_3\sqrt[3]{q^2} ~\right)^3.$$ – user2661923 Sep 25 '21 at 07:17
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    Because the intersection is again a field of degree $1$ or $3$ over $\mathbb{Q}$, it suffices to show that $\mathbb{Q}(\sqrt[3]p)\neq \mathbb{Q}(\sqrt[3]q)$. Easiest is probably to show that the two fields have different discriminants. – Mastrem Sep 25 '21 at 08:23
  • See https://math.stackexchange.com/q/3990185/72031 – Paramanand Singh Sep 26 '21 at 03:25

3 Answers3

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Let $K:=\mathbb{Q}(\sqrt[3]p)\cap \mathbb{Q}(\sqrt[3]q)$, then $$3=[\mathbb{Q}(\sqrt[3]p):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]p):K]\cdot [K:\mathbb{Q}],$$ Because $3$ is prime, one of the degrees one the right is $1$ and the other is $3$. Therefore, it suffices to show that $\mathbb{Q}(\sqrt[3]p)\neq\mathbb{Q}(\sqrt[3]q)$.

WLOG, $p\neq 3$, whence $p\nmid -27q^2$. Now, $p$ ramifies in the ring of integers of $\mathbb{Q}(\sqrt[3]p)$, but doesn't divide the discriminant of $\mathbb{Q}(\sqrt[3]q)$. Done.

Mastrem
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  • What is the discriminant of a field? I've never heard that before. I only know the discriminant of a polynomial – one potato two potato Sep 25 '21 at 10:14
  • @love_sodam It comes from algebraic number theory. Here's the wikipedia article: https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field, and here's a stackexchange post on what it intuitively means: https://math.stackexchange.com/questions/401733/what-does-the-discriminant-of-an-algebraic-number-field-mean-intuitively – Mastrem Sep 25 '21 at 10:16
  • I see. I should've mentioned that I only know undergraduate level field theory. So I'm looking for more elementary proof. The given hint of this problem is 'use field trace' but I don't know how to apply this. – one potato two potato Sep 25 '21 at 11:32
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Let $a, b$ denote the real cube roots of $p, q$ and $F$ the field of rationals.

Let us assume on the contrary that $b\in F(a) $ and let $n$ be the degree of $K=F(a, b) $ over $F$. Then we have $$b=r+sa+ta^2$$ for some rational $r, s, t$. Applying the trace $\text {tr} _F^K$ on the above equation we get $$0=nr+s\cdot 0+t\cdot 0$$ ie $r=0$ and we have $$b/a=s+ta$$ Applying trace again we get $$0=ns+t\cdot 0$$ ie $s=0$. Then we get $b=ta^2$ which is absurd as $p^2/q$ is not the cube of a rational number.

Paramanand Singh
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Let $j$ a primitive 3rd root of $1$. Since the degree of $\mathbb{Q}(\sqrt[3]{p})$ and $\mathbb{Q}(j)$ are coprime, they are linearly disjoint over $\mathbb{Q}$ (other argument: $\mathbb{Q}(j)/\mathbb{Q}$, and the intersection is clearly $\mathbb{Q}$).

Hence $[\mathbb{Q}(j)(\sqrt[3]{p}):\mathbb{Q}(j)]=3$, and similarly when we replace $p$ by $q$. We claim that $\mathbb{Q}(j)(\sqrt[3]{p})\cap\mathbb{Q}(j)(\sqrt[3]{q})=\mathbb{Q}(j)$. Otherwise, $\mathbb{Q}(j)(\sqrt[3]{p})=\mathbb{Q}(j)(\sqrt[3]{q})$. By Kummer theory, $p=q^m(a+bj)^3$, where $m=1$ or $2$ and $a,b\in\mathbb{Q}$.

So $p=q^m (a^3-3ab^2+b^3)$ and $3a^2b-3ab^2=0$ (expand and take into account that $j^2=-1-j$). In particular, $ab(a-b)=0$, so $a=0$, $b=0$ or $a=b$. In any case, you get that $p/q^m$ is the cube of rational, which is not possible since $p\neq q$.

Therefore, $\mathbb{Q}(j)(\sqrt[3]{p})\cap\mathbb{Q}(j)(\sqrt[3]{q})=\mathbb{Q}(j)$.

In particular, we get $\mathbb{Q}(\sqrt[3]{p})\cap\mathbb{Q}(\sqrt[3]{q})\subset \mathbb{Q}(j)$. But the left hand side is contained in $\mathbb{R}$, so the intersection is $\mathbb{Q}$.

GreginGre
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