Is there an easy way to see that $$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}?$$ I know that $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})$ is a subfield of $\mathbb{Q}(\sqrt[3]{5})$ which is a field of degree 3. Since only 1 and 3 divides 3, $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})$ must be either $\mathbb{Q}$ or $\mathbb{Q}(\sqrt[3]{5})$. Hence, if $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})\neq\mathbb{Q}$ then $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}(\sqrt[3]{5})$, so $\mathbb{Q}(\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt[3]{2})$. But $\mathbb{Q}(\sqrt[3]{2})$ is spanned by $\{1,2^{1/3},2^{2/3}\}$, so $$5^{1/3}=a+b2^{1/3}+c2^{2/3},$$ for some $a,b,c\in\mathbb{Q}$. Cubing both sides would - I think - lead to a contradiction, but this is a very tedious computation. Is there a simpler approach?
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$\begingroup$ I think in a situation like this, one would like to show that $x^{3}-5$ is irreducible over the field $\mathbb{Q}(\sqrt[3]{2})$. But I don't know if that is any easier (Perhaps using some irreducibility criterion… For example, Eisenstein criterion, if only you can show $\sqrt[3]{5}$ is prime in $\mathbb{Q}[\sqrt[3]{2}]$). $\endgroup$– PrismMar 27, 2014 at 23:35
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2$\begingroup$ I don't know another method offhand but the computation might not be quite as bad as you think using linear independence of $1, 2^{1/3}, 2^{2/3}$ $\endgroup$– SethMar 27, 2014 at 23:42
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$\begingroup$ Glad you ask that question Spenser, I see we are both taking MATH371! I am also stuck at the same place. Cubing both sides didn't really help me out; it just modifies the problem to: Is there a rational solution $a,b,c \in \mathbb{Q}$ to the nonlinear system $a^3 +12abc+2b^3 +4c^3 = 5$ and $a^2b - a^2c-ab^2 + 2(ac^2 + b^2c - bc^2) = 0$... I know nothing about existence of solutions in this case. $\endgroup$– HubbleMar 28, 2014 at 2:19
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$\begingroup$ Well, I cubed the expression out, and looking at the constant term, I saw that $a$ had to be odd; at the next term that $b$ had to be even; and at the third that $c$ also had to be even. Then it seemed to me that you could prove inductively that $b$ and $c$ were divisible by every power of $2$. So they had to be zero, showing that $a^3=5$. Of course, I was assuming that $a$, $b$, and $c$ were integers! But a bit of thought ought to get you through that. Anyhow, I hafta go to bed. $\endgroup$– LubinMar 28, 2014 at 5:08
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$\begingroup$ @Lubin, would you mind explaining why they are integers? Clearing the denominators of the expression for $5^{1/3}$ does not help because the $5$ in the last expression becomes $5$ times a cube. $\endgroup$– angryavianMar 31, 2014 at 14:35
2 Answers
It’s probably a more advanced method than you want, but $p$-adic theory rides to the rescue here. If the two fields $\mathbb Q(2^{1/3})$ and $\mathbb Q(5^{1/3})$ were the same, then we’d have $\mathbb Q_2(2^{1/3})=\mathbb Q_2(5^{1/3})$. But $\mathbb Q_2$ already has a cube root of $5$ in it, as one easily sees from Hensel (factoring $X^3-5\equiv X^3-1\pmod2$ ) or from the Binomial expansion of $(1+X)^{1/3}$, which has only $3$’s in its coefficients’ denominators. Thus $\mathbb Q_2(5^{1/3})=\mathbb Q_2$. But of course $2^{1/3}\notin\mathbb Q_2$, in fact $2$ is irreducible in the integers $\mathbb Z_2$ of $\mathbb Q_2$.
Here's a different method, it's not exactly nice, but I thought I'd mention it because it generalises easily to any prime root, whereas the direct argument doesn't.
Suppose not. Then, as you mention, it must be that $K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{5})$. Then we have the three embeddings of $K$ into $\mathbb{C}$, $\sigma_i$ for $i = 1,2,3$ where $\sigma_i(\sqrt[3]{2})=\sqrt[3]{2}\zeta^i$ for $\zeta=e^{2\pi i/3}$. Then $\sigma_1(\sqrt[3]{5})=\sqrt[3]{5}\zeta^j$ for some $j=1$ or $2$.
But also $K=\mathbb{Q}(\sqrt[3]{10})=\mathbb{Q}(\sqrt[3]{20})$. But depending on the value of $j$, we either have $\sigma_1(\sqrt[3]{10})=\sqrt[3]{10}$ or $\sigma_1(\sqrt[3]{20})=\sqrt[3]{20}$, contradicting the fact that $\sigma_1$ isn't the identity.
