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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...2012-04-12
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    There are plenty more. For example, if $u$ generates the extension, and $q$ is in $\mathbb{Q}$, then $u+q$ generates the extension.2012-04-12
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    @ChrisEagle: As also would be any nonzero rational multiple of one of those elements.2012-04-12
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    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 $(u+q)-q\in\mathbb{Q}(u+q)$.2012-04-12
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    @Bruno looking at this proof: http://planetmath.org/encyclopedia/ProofOfPrimitiveElementTheorem2.html, 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 $u$.2012-04-12
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    @DanielMontealegre I would not advise trying to bash algebra in terms of writing $\sqrt{2}$ and $\sqrt[3]{5}$ in terms of the basis. The algebra is long and painful and I don't think it leads you anywhere.2012-04-12
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    @BenjaminLim I totally agree, but I am a little scared that my grader will not think the same :P2012-04-12
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    @DanielMontealegre Try and copy my method below. You can cite it if you want to. I think that is pretty good in that you have shown that any rational linear combination of $\sqrt{2}$ and $\sqrt[3]{5}$ *will* work!!2012-04-12
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    @DanielMontealegre You will be row reducing a 6 by 6 matrix which is very ugly. I do not advise doing so (I did it once in an assignment and it was painfully long).2012-04-12
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    I'm trying to understand what the question is getting at. The intention seems to go beyond identifying a primitive element $v$ to finding all $w$ such that $\mathbb{Q}(u)=\mathbb{Q}(v)$.2012-04-12
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    @MarkBennet That seems to be the case, but I don't think it is so easy to do that. In my answer below I showed that any rational linear combination of $\sqrt{2}$ and $\sqrt[3]{5}$ generates that extension above.2012-04-12
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    @DanielMontealegre By the way I don't think it is clear that the degree of $\Bbb{Q}(u)$ for the $u$ that you found above has degree 6 over $6$. You say that we can check that the set $\{1, \ldots u^5\}$ is linearly independent over $\Bbb{Q}$ but have you checked it?2012-04-12
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    Since $a,b$ are just any integers that satisfy your congruences above I don't think one can even apply Eisenstein's Criterion to the polynomial $x^6 - \sqrt[6]{2^a5^b}$...2012-04-12
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    @ChrisEagle Perhaps the OP wants to find all such $u$ up to non-triviality, by that we mean excluding cases like adjoining extra elements that are already in the extension in question.2012-04-12
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    Basically what happens is the following. Galois theory tells you that (in cases like this) there are only finitely many subfields. The conclusion is then that any element that is **not** in any of those subfields will generate the entire field extension. See the answer by Gerry Myerson on how this approach plays out in this case. Galois theory also allows you to exhibit all the subfields, but that is a but more work. In general you will need to go via the normal closure. That may be covered in your class shortly.2012-04-13

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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) http://www.math.umn.edu/~garrett/m/v/linear_indep_roots.pdf2012-04-13
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    @paulgarrett I have not learned Galois Theory yet, but when I do I'll try to read up your article more closely.2012-04-13
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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 $b=d=f=0$.