To show $Q(\sqrt{p}, \sqrt{q})= Q(\sqrt{p} + \sqrt{q})$, where p and q are prime with gcd 1.

Asked Oct 13 '20 at 12:05

Active Oct 15 '20 at 01:45

Viewed 79 times

One-sided containment is clear i.e. $Q(\sqrt{p}, \sqrt{q}) \supseteq Q(\sqrt{p} + \sqrt{q})$. To show other side I tried in this manner, suppose $p > q$ then $\sqrt{p} + \sqrt{q} \in Q(\sqrt{p} + \sqrt{q})$ implies $({\sqrt{p} + \sqrt{q}})^{-1} = \frac{\sqrt{p} - \sqrt{q}}{p-q} \in Q(\sqrt{p} + \sqrt{q})$ implies $\sqrt{p} - \sqrt{q} \in Q(\sqrt{p} + \sqrt{q})$. Hence $\sqrt{p} = \frac{1}{2}((\sqrt{p} + \sqrt{q})+(\sqrt{p} -\sqrt{q})) \in Q(\sqrt{p} + \sqrt{q})$. Similiarly $\sqrt{q} \in Q(\sqrt{p} + \sqrt{q})$. From here we get $Q(\sqrt{p}, \sqrt{q}) \subseteq Q(\sqrt{p} + \sqrt{q})$.

I am not sure that it is correct or not, please tell me if any problem with my proof.

edited Oct 15 '20 at 01:45

asked Oct 13 '20 at 12:05

Kalas678

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That's quite a clever argument.... – Angina Seng Oct 13 '20 at 12:22
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See also this duplicate, or this one. Where did you use that $\gcd(p,q)=1$? – Dietrich Burde Oct 13 '20 at 14:24
Since gcd(p,q)=1, it means they are not same prime so the quotient $ \frac{p-q}{\sqrt{p} - \sqrt{q}}$ is well defined. – Kalas678 Oct 13 '20 at 15:15
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I do not think that $\sqrt{p} = \frac{1}{2}(\sqrt{p} + \sqrt{q})(\sqrt{p} -\sqrt{q}) $ is correct. And it is highly unususal to state that $gcd(p,q)=1$, in addition, the statement $\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p}+\sqrt{q})$ is true for any pair of primes, irrespective if they are different or not. – Nicky Hekster Oct 13 '20 at 16:46
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The proof (after correcting typos) seems to work for any distinct positive integers $p$ and $q$, and the result is obvious when $p=q$. – Derek Holt Oct 13 '20 at 18:54
Thanx, there was + instead of the product. - edited – Kalas678 Oct 13 '20 at 19:35
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You have the expression for $(\sqrt{p}+\sqrt{q})^{-1}$ upside down. – Derek Holt Oct 13 '20 at 21:56
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Yes, it's well known, e.g. see here for background – Bill Dubuque Oct 14 '20 at 00:04
@DerekHolt thanx, noticed it.....-corrected – Kalas678 Oct 15 '20 at 01:44

To show $Q(\sqrt{p}, \sqrt{q})= Q(\sqrt{p} + \sqrt{q})$, where p and q are prime with gcd 1.

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