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So we want to find an $u$ such that $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$. I obtained that if $u$ is of the following form: $$u=\sqrt[6]{2^a5^b}$$Where $a\equiv 1\pmod{2}$, and $a\equiv 0\pmod{3}$, and $b\equiv 0\pmod{2}$ and $ b\equiv 1\pmod{3}$. This works since $$u^3=\sqrt{2^a5^b}=2^{\frac{a-1}{2}}5^{\frac{b}{2}}\sqrt{2}$$and also, $$u^2=\sqrt[3]{2^a5^b}=2^{\frac{a}{3}}5^{\frac{b-1}{3}}\sqrt[3]{5}$$Thus we have that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq \mathbb{Q}(u)$. Note that $\sqrt{2}$ has degree of $2$ (i.e., $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$) and alsothat $\sqrt[3]{5}$ has degree $3$. As $\gcd(2,3)=1$, we have that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{5}),\mathbb{Q}]=6$. Note that this is also the degree of the extension of $u$, since one could check that the set $\{1,u,...,u^5\}$ is $\mathbb{Q}$-independent. Ergo, we must have equality. That is, $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$.

My question is: How can I find all such $w$ such that $\mathbb{Q}(w)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$? This is homework so I would rather hints rather hints than a spoiler answer. I believe that They are all of the form described above, but apriori I do not know how to prove this is true.

My idea was the following, since $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ has degree $6$, then if $w$ is such that the desired equality is satisfied, then $w$ is a root of an irreducible polynomial of degree $6$, moreover, we ought to be able to find rational numbers so that $$\sqrt{2}=\sum_{i=0}^5q_iw^i$$ and $$\sqrt[3]{5}=\sum_{i=0}^5p_iw^i$$But from here I do not know how to show that the $u$'s described above are the only ones with this property (It might be false, apriori I dont really know).

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Hint: look at the proof of the Primitive Element Theorem... – Bruno Joyal Apr 12 '12 at 21:15
There are plenty more. For example, if $u$ generates the extension, and $q$ is in $\mathbb{Q}$, then $u+q$ generates the extension. – Chris Eagle Apr 12 '12 at 21:40
@ChrisEagle: As also would be any nonzero rational multiple of one of those elements. – hardmath Apr 12 '12 at 21:56
But I meant different $u$ up to linear $\mathbb{Q}$ combinations of them. I know that $u+q$ also generates it, because we must have that $(u+q)-q\in\mathbb{Q}(u+q)$. – Daniel Montealegre Apr 12 '12 at 21:58
@Bruno looking at this proof:…, and in particular I am looking at the second half of the proof. They are basically doing theorically what I want to do computationally. They have $F(\alpha,\beta)$, and they find a $u$ such that that field equals $F(u)$, which is precisely what I want to do. The way they do it is that they identify $a$ and $b$ such that $\alpha+a\beta$ and $\alpha+b\beta$ extend the same field, but I do not know how to find such an $a$ and $b$, and moreover, how do we know we get all posible values for $u$. – Daniel Montealegre Apr 12 '12 at 22:05
up vote 2 down vote accepted

If we take $u = \sqrt{2} + \sqrt[3]{5}$, such a $u$ almost always turns out to work. In fact let's try if a rational linear combination of $\sqrt{2}$ and $\sqrt[3]{5}$ will work. Let us now write $u$ as $u = a\sqrt{2} + b\sqrt[3]{5}$ for rationals $a$ and $b$.

Clearly we have that $\Bbb{Q}(u)\subseteq \Bbb{Q}(\sqrt{2},\sqrt[3]{5})$. To show the other inclusion, we just need to show that say $\sqrt{2} \in \Bbb{Q}(u)$ for then $\sqrt[3]{5} = \frac{a\sqrt{2} + b\sqrt[3]{5} - a\sqrt{2}}{b}$ will be in $\Bbb{Q}(u)$. Here is a quick and easy way of doing this:

Write $u = a\sqrt{2} + b\sqrt[3]{5}$ so that $(\frac{u - a\sqrt{2}}{b})^3 = 5$. We can assume that $a$ and $b$ are simultaneously not zero for then the proof becomes redundant. Then expanding the left hand side by the binomial theorem we get that

$$ u^3 - 3\sqrt{2}u^2a + 6ua^2 - 2a^3\sqrt{2} = 5.$$

Rearranging, we get that

$$\sqrt{2} = \frac{u^3 + 6ua^2 -5}{ 3u^2a + 2a^3 }.$$

Since $\Bbb{Q}(u)$ is a field the right hand side is in $\Bbb{Q}(u)$ so that $\sqrt{2}$ is in here. Done!

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Indeed, one can prove something quite generally. Suggested by various other MSE and MO discussions, I wrote up (with references to the prior discussions and some historical detail) – paul garrett Apr 13 '12 at 0:09
@paulgarrett I have not learned Galois Theory yet, but when I do I'll try to read up your article more closely. – user38268 Apr 13 '12 at 0:15

The field has degree 6 over the rationals. Any element $w$ of degree 6 will generate the field.

Now, every element of the field has degree 1, 2, 3, or 6. The only elements of degree 1 are the rationals. The only elements of degree 2 are those of the form $a+b\sqrt2$ (although it takes some work to check this). The only elements of degree 3 are those of the form $a+b\root3\of5+c\root3\of{25}$ (again, this takes some checking). It follows that the generators are all the elements $a+b\sqrt2+c\root3\of5+d\sqrt2\root3\of5+e\root3\of{25}+f\sqrt2\root3\of{25}$ except those with $b=c=d=e=f=0$, those with $c=d=e=f=0$, and those with $b=d=f=0$.

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