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I have to solve the following problem:

Find $\gamma$ such that $\mathbb{Q}(\gamma)=\mathbb{Q}(\sqrt{-5},\sqrt{2})$

I am a bit confused about how to do it but as $\mathbb{Q}(\sqrt{-5},\sqrt{2})$ has degree $4$,then $\mathbb{Q}(\gamma)$ must also have degree $4$. I think this means that we must be able to write:

$$a+b\gamma + c \gamma^2 + d\gamma^3=\sqrt{-5}\\a'+b'\gamma + c' \gamma^2 + d'\gamma^3=\sqrt{2}$$

And find a suitable $\gamma$, I suspect $\gamma=\sqrt{-5}\sqrt{2}$ but I still can't solve it for $a,b,c,d$.

Red Banana
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    Your first thought does not work because $\sqrt{-5}\sqrt2=\sqrt{-10}$ satisfies a quadratic polynomial. Have you tried $\gamma=\sqrt2+\sqrt{-5}$? What would $1/\gamma$ be in that case? – Jyrki Lahtonen Mar 25 '21 at 04:14
  • @JyrkiLahtonen $1/ \gamma = 1/(\sqrt2+\sqrt{-5})$? – Red Banana Mar 25 '21 at 04:31
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    Rationalize the denominator, @BillyRubina. – kobe Mar 25 '21 at 04:36
  • @kobe $\gamma = \frac{\sqrt{2}-i \sqrt{5}}{7}?$ – Red Banana Mar 25 '21 at 04:42
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    No, $1/\gamma = (\sqrt{2} - \sqrt{-5})/7$. Then $\sqrt{2} - \sqrt{-5} \in \Bbb Q(\gamma)$. – kobe Mar 25 '21 at 04:44
  • @kobe I am confused: What does this means? – Red Banana Mar 25 '21 at 04:47
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    Since $1/\gamma \in \Bbb Q(\gamma)$, then $7/\gamma \in \Bbb Q(\gamma)$. That's why $\sqrt{2} - \sqrt{-5}\in \Bbb Q(\gamma)$. But you also have $\sqrt{2} + \sqrt{-5} \in \Bbb Q(\gamma)$ as well. Use these two facts to show that $\sqrt{2}$ and $\sqrt{-5}$ are in $\Bbb Q(\gamma)$. – kobe Mar 25 '21 at 04:50
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    @BillyRubina On why taking the sum (like $\sqrt{2} + \sqrt{-5}$ here) works usually: there is the theorem called the primitive element theorem, that gives the existence of a $\gamma$ satisfying the conditions above. The proof of that theorem involves taking elements $\gamma_k$ which are basically one element plus $k$ times the other element (like in our case, $\gamma_k = \sqrt{2} + k \sqrt{-5}$). Then the proof shows that one of the $\gamma_k$ can be taken as $\gamma$. So when we test, we start with small $k$ like $k=1,-1$ etc. and see if they work. In simple cases like here, $k=1$ works out. – Sarvesh Ravichandran Iyer Mar 25 '21 at 04:57
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    Here is another example where the sum of the elements ends up generating the extension. In general, you can remember this approach if and when you encounter these questions. – Sarvesh Ravichandran Iyer Mar 25 '21 at 05:00
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    After you have covered basic Galois theory this becomes rather easy. Galois theory tells us that ALL the intermediate fields of $\Bbb{Q}(\sqrt2,\sqrt{-5})$ are exactly $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt{-5})$, $\Bbb{Q}(\sqrt{-10})$. So if $\gamma$ is NOT of one of the forms $q_1+q_2\sqrt{2}$, $q_1+q_2\sqrt{-5}$, $q_1+q_2\sqrt{-10}$, then $\gamma$ will work. I'm afraid you are not "allowed" to use this result yet. Posting it for near future reference :-) – Jyrki Lahtonen Mar 25 '21 at 05:35
  • @JyrkiLahtonen Yeah, I kinda noticed that sometimes answers to problems like this use concepts from other parts of the book I still didn't read. – Red Banana Mar 25 '21 at 06:22
  • No matter. At this point a good (possibly the best) answer is the one outlined by kobe. The element $$\gamma+\frac7{\gamma}=2\sqrt2$$ is surely in $\Bbb{Q}(\gamma)$. Hence so is $\sqrt2$, and thereafter also $\sqrt{-5}=\gamma-\sqrt2$. Remember that $\Bbb{Q}(\gamma)$ is a field, so closed under division by non-zero elements also. – Jyrki Lahtonen Mar 25 '21 at 08:57

1 Answers1

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Consider $\gamma = \sqrt{2} + \sqrt{-5}$. By rationalizing the denominator, it follows that $$\frac{7}{\gamma} = \sqrt{2} - \sqrt{-5}$$ and hence the field $\Bbb Q(\gamma)$ contains $\sqrt{2} - \sqrt{-5}$. Now $$\sqrt{2} = \frac{1}{2}\left(\gamma + \frac{7}{\gamma}\right) \in \Bbb Q(\gamma)$$ and so $\Bbb Q(\gamma)$ contains $\gamma - \sqrt{2}$, or $\sqrt{-5}$. Therefore $\Bbb Q(\gamma) \supset \Bbb Q(\sqrt{2}, \sqrt{-5})$. As $\gamma \in \Bbb Q(\sqrt{2}, \sqrt{-5})$, the reverse containment $\Bbb Q(\gamma) \subset \Bbb Q(\sqrt{2}, \sqrt{-5})$ follows.

kobe
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