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If $p$, $q$ are two distinct prime numbers in $\Bbb N$ then

$$\Bbb Q(\sqrt p, \sqrt q) = \Bbb Q(\sqrt p + \sqrt q)$$

How to prove that the first is included in the second, and why is it ligical to assume the equality

Robert Lewis
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user568442
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  • Since $\sqrt{p} + \sqrt{q} \in \mathbb{Q}(\sqrt{q},\sqrt{p})$, it follows that $\mathbb{Q}(\sqrt{p} + \sqrt{q}) \subseteq \mathbb{Q}(\sqrt{q},\sqrt{p})$. Showing the reverse inclusion is how you get equality. – nilradical1 Dec 02 '20 at 01:32
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    Please use Mathjax rather than extended characters. – Arturo Magidin Dec 02 '20 at 01:44
  • https://math.stackexchange.com/questions/3921711/let-alpha-be-a-root-of-x2-a-and-beta-be-a-root-of-x2-b-provide?noredirect=1#comment8087617_3921711

    Look to this answer, and tell me if can help you... Is more general then your question but is regarding the same point

     –  Dec 02 '20 at 02:41
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    Your question has been asked (and answered) many times. See e.g. https://math.stackexchange.com/a/3325514/300700 for what I think is the minimal amount of calculations. – nguyen quang do Dec 02 '20 at 16:57
  • You have been asking a lot of questions in introductory field theory. Please try to give your own thoughts/attempts and context (like which book you are studying, what are the definitions you know). Otherwise most of these questions will get closed for lack of context. – Paramanand Singh Dec 03 '20 at 06:51
  • Here is a hint. $a=\sqrt{p} +\sqrt{q}, b=(p-q) /a$ are members of $\mathbb{Q} (a) $. Express $\sqrt{p}, \sqrt{q} $ in terms of $a, b$ and you are done. You may as well update your question with an attempt based on this. – Paramanand Singh Dec 03 '20 at 07:02

1 Answers1

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Let $a=\sqrt p$ and $b=\sqrt q$.

Let $c=(a^2+3b^2)a+(3a^2+b^2)b.$ Then $c=a^3+3a^2b+3ab^2+b^3=(a+b)^3\in \Bbb Q(a+b).$

Let $d=(a^2+3b^2)(a+b).$ Then $d=(p+3q)(a+b)\in \Bbb Q(a+b).$

So $2(p-q)b=2(a^2-b^2)b=c-d\in \Bbb Q(a+b).$

So $\sqrt q=b\in \Bbb Q(a+b)=\Bbb Q(\sqrt p +\sqrt q).$

The rest is up to you.

DanielWainfleet
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