# Degree of $\sqrt{2}+\sqrt[3]{5}$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$

I'm self-studying field extensions. I ran over an exercise which I can't completely solve. (I haven't yet started studying Galois theory, and I think this exercise isn't meant to be solved using it, just in case):

The problem is:

a) Prove $\sqrt{2}+\sqrt[3]{5}$ is algebraic over $\mathbb{Q}$ of degree 6.

Done: I know it has degree $\leq 6$ because $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ which has degree 6; then I explicitly found the polynomial by solving a 6-equation linear system, and Wolfram Alpha proved it irreducible (btw: how can I prove it by hand?). The polynomial is $t^6-6t^4-10t^3+12t^2-60t+17$.

b) What's its degree over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{5})$?

It is this part b) which I can't solve. Of course its degree is $\leq 6$ in both cases, but I don't know what else to do.

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You can probably use Eisensteins criterion, that comes up a lot in Galois theory. – anon Oct 30 '10 at 22:47
It's not useful in this case. It would have to be 17, but also divide 12... – lentic catachresis Oct 30 '10 at 23:00
A systematic way of showing that it has degree 6 is to compute powers of $\alpha=\sqrt{2}+\sqrt[3]{5}$ as linear combinations of $\sqrt{2}^r\sqrt[3]{5}^s$ and solve for $\sqrt{2}$ and $\sqrt{3}$ in terms of $\alpha$. – George Lowther Oct 31 '10 at 1:21
(typo above. Meant $\sqrt[3]{5}$, not $\sqrt{3}$). – George Lowther Oct 31 '10 at 1:24
@Bruno: You solved to find a linear dependency. If instead you solved to show that $\sqrt{2},\sqrt[3]{5}$ is in the (rational) linear span of $\{\alpha^k:k=0,...,5\}$ then this would give $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$. So, the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ would be a multiple of 2 and 3, giving degree 6. – George Lowther Oct 31 '10 at 15:30

Bruno: The degree over ${\mathbb Q}(\sqrt2)$ is 3: Consider $(x-\sqrt 2)^3-5$. Since $\root 3\of 5$ is in ${\mathbb Q}(\sqrt 2)(\sqrt 2+\root 3\of 5)$, the only other option is that it already belongs to ${\mathbb Q}(\sqrt 2)$. But this is impossible, since its minimal polynomial is $x^3-5$ over ${\mathbb Q}$ and ${\mathbb Q}(\sqrt 2)$ is an extension of degree 2.

[ By the way, one can use this and the tower law, to show that ${\mathbb Q}(\sqrt 2+\root 3\of 5)$ has degree 6 over ${\mathbb Q}$. The irreducibility of your polynomial then follows for free. ]

Similarly, the degree over ${\mathbb Q}(\root 3\of 5)$ is 2.

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"...the only other option is that it already belongs to Q(√2)". I don't follow this. I get that the degree must be $\leq$ 3, but I don't understand why it can't be <. – lentic catachresis Oct 30 '10 at 23:56
@Bruno: because 3 is prime. The degree divides 3, so it must equal either 3 or 1. In the latter case, $\sqrt[3]{5}$ is in $\mathbb{Q}(\sqrt{2})$. – George Lowther Oct 31 '10 at 1:10
@Bruno: There is a way of arguing directly about the question in your comment: Only one of the roots of $t^3-5$ is real. Since ${\mathbb Q}(\sqrt 2)\subset{\mathbb R}$, if $(x-\sqrt 2)^3-5$ is not irreducible over ${\mathbb Q}(\sqrt 2)$ then, having degree 3, it must factor into a linear and a quadratic term, so one of its roots is in ${\mathbb Q}(\sqrt 2)$, and it must therefore be the real one. – Andrés Caicedo Oct 31 '10 at 15:33
@Andres Caicedo: Got it. Now, I don't understand your line between brackets. What tower do you consider? – lentic catachresis Oct 31 '10 at 18:26
First, I observe that $\mathbb{Q}(\sqrt{2})(\sqrt{2}+\sqrt[3]{5}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$. Now, arguing over the tower $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ and using what has just been proved, we get $|\mathbb{Q}(\sqrt{2},\sqrt[3]{5}):\mathbb{Q}|=6$. Then, using this in $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$, it remains to prove that $|\mathbb{Q}(\sqrt{2},\sqrt[3]{5}):\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})|=1$. How to do it? Is this the right way or did you have another idea in mind? – lentic catachresis Oct 31 '10 at 21:07

For part b, we can find the minimal polynomial in both cases.

Consider $a =\sqrt{2} + \sqrt[3]{5}$ over $\mathbb{Q}(\sqrt[3]{5})$. Notice that

$(x - \sqrt[3]{5})^2 -2 = 0$

is a degree two polynomial in $\mathbb{Q}(\sqrt[3]{5})[x]$ that $a$ is a root of. We know that $a \notin \mathbb{Q}(\sqrt[3]{5})$, so the degree of its minimal polynomial must be greater than 1, and now we have a polynomial of degree 2 that it is a root of, so its minimal polynomial must be that one above. Therefore $a$ has degree two over $\mathbb{Q}(\sqrt[3]{5})$.

Similarly, consider $a$ over $\mathbb{Q}(\sqrt{2})$. Notice that

$(x - \sqrt{2})^3 - 5 = 0$

is a degree 3 polynomial in $\mathbb{Q}(\sqrt{2})[x]$ that $a$ is a root of, and hence the minimal polynomial, which must divide this polynomial, can have degree either 2, or 3. But, if it has degree two, then it is attained from the above one by dividing by $x - a$. We get by polynomial division that

$(x - \sqrt{2})^3 - 5 = (x - \sqrt{2} - \sqrt[3]{5})(-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2)$

and $-2 - \sqrt{2} \sqrt[3]{5} + \sqrt[3]{25} + (-2\sqrt{2} + \sqrt[3]{5})x + x^2 \notin \mathbb{Q}(\sqrt{2})[x]$ because the $x$ coefficient is not in $\mathbb{Q}(\sqrt{2})$. If it were, then as $2\sqrt{2} \in \mathbb{Q}(\sqrt{2})$, we would get that

$2 \sqrt{2} + (-2\sqrt{2} + \sqrt[3]{5}) = \sqrt[3]{5} \in \mathbb{Q}(\sqrt{2})$

which is impossible. Thus that quadratic polynomial cannot be the minimal polynomial for $a$. Hence, the minimal polynomial of $a$ has degree 3.

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Eric, to argue like this in the second case you need to justify that either the constant term or the term for $x$ in your quadratic polynomial is not in ${\mathbb Q}(\sqrt 2)$ --granted, this is not too difficult, but it is needed here. – Andrés Caicedo Oct 31 '10 at 1:34
Yes you are right. I thought it was kind of self-evident, but since the person asking this may not have much experience with algebra I will edit my answer. Thanks. – Eric Haengel Oct 31 '10 at 2:44