In office hours yesterday my instructor said $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$.

I know $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ because $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ is a field so it is closed under addition which means since $\sqrt{2}$ and $\sqrt[3]{5}$ are members then their sum $\sqrt{2}+\sqrt[3]{5}$ is a member.

I am not sure how to show $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$. A friend suggested I could find a linear combination of $\sqrt{2}+\sqrt[3]{5}$ and its inverse that yields $\sqrt{2}$. However I have to admit that I am not sure how to rationalize $\frac{1}{\sqrt{2}+\sqrt[3]{5}}$ in order to even find the inverse of $\sqrt{2}+\sqrt[3]{5}$.

I have found problems similar to this (but simpler) where both members of the sum in question are square roots (instead of one square root and one cube root).


  • $\begingroup$ you could go through the proof of the primitive element theorem which states that there are $\alpha,\beta\in \mathbb Q$ such that $\mathbb Q(\alpha \sqrt2 + \beta \sqrt[3]5) = \mathbb Q(\sqrt 2,\sqrt[3]5)$ and check whether $\alpha=\beta=1$ does the job. $\endgroup$ – Claudius May 25 '16 at 2:38
  • $\begingroup$ We definitely have not talked about or mentioned that in class. This is only the first section introducing field extensions. $\endgroup$ – fullyhip May 25 '16 at 2:52
  • $\begingroup$ It's possible to check that if $\alpha=\sqrt2+\sqrt[3]5$, then $$\frac{\alpha^3+6\alpha-5}{3\alpha^2+2} = \sqrt2.$$ However, I know no way of finding this expression without using some more advanced knowledge of these number fields. (I found it by finding the minimal polynomial of $\alpha$ over $\Bbb Q$—itself a nontrivial task—and then factoring it over $\Bbb Q(\sqrt 2)$.) $\endgroup$ – Greg Martin May 25 '16 at 3:43
  • $\begingroup$ OK ... so this is not exactly one of those "clearly" statements. She simply stated it and moved on like it was obvious. $\endgroup$ – fullyhip May 25 '16 at 3:56
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    $\begingroup$ See my answer here: math.stackexchange.com/questions/1745628/… $\endgroup$ – Steve D May 25 '16 at 4:34

$K = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) $ is seen to have degree 6 over $ \mathbb{Q} $ by the observation that it has subfields of degree 2 and 3. Take the normal closure $ L $ of $ K $ and consider its Galois group. $ K $ has $ 6 $ distinct $\mathbb{Q}$-embeddings into its normal closure, and these embeddings extend to $\mathbb{Q}$-automorphisms of $ L $; which means that $ \sqrt{2} + \sqrt[3]{5} $ has 6 $ \mathbb{Q} $-conjugates. By the Tower Law, we then have $$ 6 = [\mathbb{Q}(\sqrt{2}, \sqrt[3]{5}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt[3]{5}):\mathbb{Q}(\sqrt{2} + \sqrt[3]{5})][\mathbb{Q}(\sqrt{2} + \sqrt[3]{5}):\mathbb{Q}] = 6[\mathbb{Q}(\sqrt{2}, \sqrt[3]{5}):\mathbb{Q}(\sqrt{2} + \sqrt[3]{5})]$$

so that $ \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) = \mathbb{Q}(\sqrt{2} + \sqrt[3]{5})$. The result then follows easily.

  • $\begingroup$ I don't get your argument. $[K:Q] = [K:L][L:Q]$ whenever $Q \subseteq L \subseteq K$. and $[K:Q] = 6$. but how do you know $[K:L]$ and $[L:Q]$ ? $\endgroup$ – reuns May 27 '16 at 21:25
  • $\begingroup$ I don't know $ [K:L] $, that is what I am trying to find. $ [L:Q] = 6 $ because $ L = Q(\alpha) $ where $ \alpha $ has 6 distinct $ Q$-conjugates, so that its minimal polynomial over $ Q $ has degree 6. $\endgroup$ – Starfall May 27 '16 at 21:26
  • $\begingroup$ yes I just saw this. so those conjugates are based on the idea that $L$ is fixed by $Gal(K/Q)$ ? $\endgroup$ – reuns May 27 '16 at 21:27
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    $\begingroup$ No, $ L $ would not be fixed by the Galois group. The idea is that I can extend an embedding $ \sigma : K \to L $ to an automorphism of $ L $ by the isomorphism extension theorem, and then note that the $ \mathbb{Q} $-automorphisms of $ L $ have to permute roots of irreducible polynomials in $ \mathbb{Q}[X] $. $\endgroup$ – Starfall May 27 '16 at 21:29
  • $\begingroup$ those $Q$-conjugates $\{\sigma(a) \ \mid \ \sigma \in Gal(K/Q) \ \}$ are supposed to be all in $Q(a)$, right ? hence it means that $\sigma(Q(a)) = Q(a)$ for every $\sigma \in Gal(K/Q)$ ? (I did not mean each element of $L$ is fixed by $\sigma$, only that the whole field is sent to itself) $\endgroup$ – reuns May 27 '16 at 21:32

You know that $$\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ and I suppose you know also that the degree of $$\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ is equal to $2\cdot 3=6$. Well, the minimal polynomial of $x=\sqrt{2}+\sqrt[3]{5}$ being of degree $6$ you can ensure that the degree of $x=\sqrt{2}+\sqrt[3]{5}$ is $6$ and consequently $$\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ This give you an easy way to verify that $$\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$$ because obviously $\sqrt 2\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$

  • $\begingroup$ I disagree ... I think the degree of $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ is $6$. After some computations you find it satisfies the polynomial $x^6-6x^4-10x^3+12x^2-60x+17$ but all that guarantees is the degree is at most 6. $\endgroup$ – fullyhip May 25 '16 at 3:28
  • $\begingroup$ Yes, absolutely. I am tired and i am wrong.I worked with the 5 root of 2.... $\endgroup$ – Piquito May 25 '16 at 3:31
  • $\begingroup$ @fullyhip: Not "at most 6" but exactly 6. All element in an extension has the degree of its minimal polynomial. Is for that I have chosen this way avoiding the obvious direct way using the primitive element for you. See at the comment of user218931 (in which all rational couple such $\alpha \beta \ne 0$ satisfies) $\endgroup$ – Piquito May 25 '16 at 14:39
  • $\begingroup$ Finding "a" polynomial does not guarantee it is "the" minimum polynomial. There needs to be some evidence that it is irreducible. $\endgroup$ – fullyhip May 26 '16 at 14:19
  • $\begingroup$ Then if $x=\sqrt 2+\sqrt 3+\sqrt 5$ and you find the first polynomial $x^8-40x^6+352x^4-960x^2+576$ of $\mathbb Q[x]$ involving your $x$ you are still not sure that it is the minimal polynomial of $x$? Do you need to verify the irreducibility before? Well. If I had known it, my answer to you would have been very different. I thought you really were a beginner. Sorry by bad English. Regards. $\endgroup$ – Piquito May 26 '16 at 23:00

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